2. Motion Along a Straight Line =============================== 2-1 POSITION, DISPLACEMENT, AND AVERAGE VELOCITY Learning Objectives After reading this module, you should be able to … 2.01 Identify that if all parts of an object move in the same direction and at the same rate, we can treat the object as if it were a (point-like) particle. (This chapter is about the motion of such objects.) 2.02 Identify that the position of a particle is its location as read on a scaled axis, such as an x axis. 2.03 Apply the relationship between a particle’s displacement and its initial and final positions. 2.04 Apply the relationship between a particle’s average velocity, its displacement, and the time interval for that displacement. 2.05 Apply the relationship between a particle’s average speed, the total distance it moves, and the time interval for the motion. 2.06 Given a graph of a particle’s position versus time, determine the average velocity between any two particular times Key Ideas * The position x of a particle on an x axis locates the particle with respect to the origin, or zero point, of the axis. * The position is either positive or negative, according to which side of the origin the particle is on, or zero if the particle is at the origin. The positive direction on an axis is the direction of increasing positive numbers; the opposite direction is the negative direction on the axis. * The displacement :math:`\Delta x` of a particle is the change in its position: .. math:: \Delta x = x_2 - x_1. * Displacement is a vector quantity. It is positive if the particle has moved in the positive direction of the x axis and negative if the particle has moved in the negative direction. * When a particle has moved from position x1 to position x2 during a time interval :math:`\delta t_2 - t_1`, its average velocity during that interval is .. math:: v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x_2-x_1}{t_2-t_1} * The algebraic sign of vavg indicates the direction of motion (:math:`v_{avg}` is a vector quantity). Average velocity does not depend on the actual distance a particle moves, but instead depends on its original and final positions. * On a graph of x versus t, the average velocity for a time in- terval :math:`\Delta t` is the slope of the straight line connecting the points on the curve that represent the two ends of the interval. * The average speed :math:`s_{avg}` of a particle during a time interval :math:`\Delta t` depends on the total distance the particle moves in that time interval: .. math:: s_{avg} = \frac{total distance}{\Delta t} What Is Physics? ---------------- One purpose of physics is to study the motion of objects—how fast they move, for example, and how far they move in a given amount of time. NASCAR engineers are fanatical about this aspect of physics as they determine the performance of their cars before and during a race. Geologists use this physics to measure tectonic-plate motion as they attempt to predict earthquakes. Medical researchers need this physics to map the blood flow through a patient when diagnosing a partially closed artery, and motorists use it to determine how they might slow sufficiently when their radar detector sounds a warning. There are countless other examples. In this chapter, we study the basic physics of motion where the object (race car, tectonic plate, blood cell, or any other object) moves along a single axis. Such motion is called one-dimensional motion. Motion ------ The world, and everything in it, moves. Even seemingly stationary things, such as a roadway, move with Earth’s rotation, Earth’s orbit around the Sun, the Sun’s orbit around the center of the Milky Way galaxy, and that galaxy’s migration relative to other galaxies. The classification and comparison of motions (called kinematics) is often challenging.What exactly do you measure, and how do you compare? Before we attempt an answer, we shall examine some general properties of motion that is restricted in three ways. 1. The motion is along a straight line only. The line may be vertical, horizontal, or slanted, but it must be straight. 2. Forces (pushes and pulls) cause motion but will not be discussed until Chapter 5. In this chapter we discuss only the motion itself and changes in the motion. Does the moving object speed up, slow down, stop, or reverse direction? If the motion does change, how is time involved in the change? 3. The moving object is either a particle (by which we mean a point-like object such as an electron) or an object that moves like a particle (such that every portion moves in the same direction and at the same rate). A stiff pig slipping down a straight playground slide might be considered to be moving like a par- ticle; however, a tumbling tumbleweed would not. Position and Displacement To locate an object means to find its position relative to some reference point, of- ten the origin (or zero point) of an axis such as the x axis in Fig. 2-1. The positive direction of the axis is in the direction of increasing numbers (coordinates), which is to the right in Fig. 2-1. The opposite is the negative direction. For example, a particle might be located at x = 5 m, which means it is 5 m in the positive direction from the origin. If it were at x = -5 m, it would be just as far from the origin but in the opposite direction. On the axis, a coordinate of $5 m is less than a coordinate of -1 m, and both coordinates are less than a coordinate of +5 m. A plus sign for a coordinate need not be shown, but a minus sign must always be shown. A change from position x1 to position x2 is called a displacement -x, where .. math:: \Delta x = x_2 - x_1. (2-1) (The symbol :math:`\Delta`, the Greek uppercase delta, represents a change in a quantity, and it means the final value of that quantity minus the initial value.) When numbers are inserted for the position values x1 and x2 in Eq. 2-1, a displacement in the positive direction (to the right in Fig. 2-1) always comes out positive, and a displacement in the opposite direction (left in the figure) always comes out negative. For example, if the particle moves from x1 " 5 m to x2 " 12 m, then the displacement is :math:`\Delta x = (12 m) - (5 m) = +7 m`. The positive result indicates that the motion is in the positive direction. If, instead, the particle moves from x1 " 5 m to x2 " 1 m, then :math:`\Delta x = (1 m) - (5 m) = -4 m`. The negative result in- dicates that the motion is in the negative direction. The actual number of meters covered for a trip is irrelevant; displacement in- volves only the original and final positions. For example, if the particle moves from x = 5 m out to x = 200 m and then back to x = 5 m, the displacement from start to finish is :math:`\Delta x = (5 m) - (5 m) = 0`. **Signs.** A plus sign for a displacement need not be shown, but a minus sign must always be shown. If we ignore the sign (and thus the direction) of a displace- ment, we are left with the magnitude (or absolute value) of the displacement. For example, a displacement of :math:`\Delta x = -4 m` has a magnitude of 4 m. Displacement is an example of a vector quantity, which is a quantity that has both a direction and a magnitude. We explore vectors more fully in Chapter 3, but here all we need is the idea that displacement has two features: (1) Its magnitude is the distance (such as the number of meters) between the original and final po- sitions. (2) Its direction, from an original position to a final position, can be repre- sented by a plus sign or a minus sign if the motion is along a single axis. Here is the first of many checkpoints where you can check your understanding with a bit of reasoning. The answers are in the back of the book. Checkpoint 1 Here are three pairs of initial and final positions, respectively, along an x axis. Which pairs give a negative displacement: (a) -3 m, +5 m; (b) -3 m, -7 m; (c) 7 m, -3 m? Average Velocity and Average Speed ---------------------------------- A compact way to describe position is with a graph of position x plotted as a func- tion of time t—a graph of x(t). (The notation x(t) represents a function x of t, not the product x times t.) As a simple example, Fig. 2-2 shows the position function x(t) for a stationary armadillo (which we treat as a particle) over a 7 s time inter- val. The animal’s position stays at x = -2 m. Figure 2-3 is more interesting, because it involves motion. The armadillo is apparently first noticed at t = 0 when it is at the position x=-5 m. It moves toward x = 0, passes through that point at t = 3 s, and then moves on to increasingly larger positive values of x. Figure 2-3 also depicts the straight-line motion of the armadillo (at three times) and is something like what you would see. The graph in Fig. 2-3 is more abstract, but it reveals how fast the armadillo moves. Actually, several quantities are associated with the phrase “how fast.” One of them is the average velocity :math:`v_{avg}`, which is the ratio of the displacement :math:`\Delta x` that occurs during a particular time interval :math:`\Delta t` to that interval: .. math:: v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1} (2-2) The notation means that the position is x1 at time t1 and then x2 at time t2. A com- mon unit for vavg is the meter per second (m/s). You may see other units in the problems, but they are always in the form of length/time. **Graphs.** On a graph of x versus t, vavg is the slope of the straight line that connects two particular points on the x(t) curve: one is the point that corresponds to x2 and t2, and the other is the point that corresponds to x1 and t1. Like displace- ment, vavg has both magnitude and direction (it is another vector quantity). Its magnitude is the magnitude of the line’s slope. A positive vavg (and slope) tells us that the line slants upward to the right; a negative vavg (and slope) tells us that the line slants downward to the right. The average velocity vavg always has the same sign as the displacement -x because -t in Eq. 2-2 is always positive. Figure 2-4 shows how to find vavg in Fig. 2-3 for the time interval t " 1 s to t " 4 s. We draw the straight line that connects the point on the position curve at the begin- ning of the interval and the point on the curve at the end of the interval.Then we find the slope :math:`\Delta x/\Delta t` of the straight line. For the given time interval, the average velocity is .. math: v_{avg} = \frac{6m}{3s} = 2 m/s. **Average speed** savg is a different way of describing “how fast” a particle moves. Whereas the average velocity involves the particle’s displacement :math:`\Delta x`, the average speed involves the total distance covered (for example, the number of meters moved), independent of direction; that is, .. math:: s_{avg} = \frac{total distance}{\Delta x} Because average speed does not include direction, it lacks any algebraic sign. Sometimes savg is the same (except for the absence of a sign) as vavg. However, the two can be quite different. Sample Problem 2.01 Average velocity, beat-up pickup truck You drive a beat-up pickup truck along a straight road for 8.4 km at 70 km/h, at which point the truck runs out of gasoline and stops. Over the next 30 min, you walk another 2.0 km farther along the road to a gasoline station. (a) What is your overall displacement from the beginning of your drive to your arrival at the station? KEY IDEA Assume, for convenience, that you move in the positive di- rection of an x axis, from a first position of x1 " 0 to a second position of x2 at the station. That second position must be at x2 = 8.4 km + 2.0 km = 10.4 km. Then your displacement -x along the x axis is the second position minus the first position. Calculation: From Eq. 2-1, we have :math:`\Delta x = x_2 - x_1 = 10.4 km - 0 = 10.4 km`. (Answer) Thus, your overall displacement is 10.4 km in the positive direction of the x axis. (b) What is the time interval :math:`\Delta t` from the beginning of your drive to your arrival at the station? KEY IDEA We already know the walking time interval :math:`\Delta t_{wlk}` (= 0.50 h), but we lack the driving time interval :math:`\Delta t_{dr}`. However, we know that for the drive the displacement :math:`\Delta x_{dr}` is 8.4 km and the average velocity vavg,dr is 70 km/h. Thus, this average velocity is the ratio of the displacement for the drive to the time interval for the drive. Calculations: We first write .. math:: v_{avg,dr} = \frac{\Delta x_{dr}}{\Delta t_{dr}} Rearranging and substituting data then give us .. math:: \Delta t_{dr} = \frac{\Delta x_{dr}}{v_{avg,_dr}} = \frac{8.4 km}{70 km/h} = 0.12 h (Answer) So, :math:`\Delta t = \Delta t_{dr} + \Delta t_{wlk}` = 0.12 h + 0.50 h = 0.62 h. (c) What is your average velocity vavg from the beginning of your drive to your arrival at the station? Find it both numerically and graphically KEY IDEA From Eq. 2-2 we know that vavg for the entire trip is the ratio of the displacement of 10.4 km for the entire trip to the time interval of 0.62 h for the entire trip. Calculation: Here we find .. math:: v_{avg} = \frac{\Delta x}{\Delta t} = \frac{10.4 km}{0.62 h} To find vavg graphically, first we graph the function x(t) as shown in Fig. 2-5, where the beginning and arrival points on the graph are the origin and the point labeled as “Station.”Your average velocity is the slope of the straight line connecting those points; that is, vavg is the ratio of the rise (:math:`\Delta x = 10.4 km`) to the run (:math:`\Delta t = 0.62 h`), which gives us vavg " 16.8 km/h. (d) Suppose that to pump the gasoline, pay for it, and walk back to the truck takes you another 45 min. What is your average speed from the beginning of your drive to your return to the truck with the gasoline? KEY IDEA Your average speed is the ratio of the total distance you move to the total time interval you take to make that move. Calculation: The total distance is 8.4 km + 2.0 km + 2.0 km = 12.4 km. The total time interval is 0.12 h + 0.50 h + 0.75 h = 1.37 h. Thus, Eq. 2-3 gives us .. math:: s_{avg} = \frac{12.4 km}{1.37 h} = 9.1 km/h Figure 2-5 The lines marked “Driving” and “Walking” are the position – time plots for the driving and walking stages. (The plot for the walking stage assumes a constant rate of walking.) The slope of the straight line joining the origin and the point labeled “Station” is the average velocity for the trip, from the beginning to the station. 2-2 INSTANTANEOUS VELOCITY AND SPEED ------------------------------------ Learning Objectives After reading this module, you should be able to . . . 2.07 Given a particle’s position as a function of time, calculate the instantaneous velocity for any particular time. 2.08 Given a graph of a particle’s position versus time, deter- mine the instantaneous velocity for any particular time. 2.09 Identify speed as the magnitude of the instantaneous velocity. Key Ideas * The instantaneous velocity (or simply velocity) v of a moving particle is .. math:: v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt} where :math:`\Delta x = x_2 - x_1` and :math:`\Delta t = t_2 - t_1`. * The instantaneous velocity (at a particular time) may be found as the slope (at that particular time) of the graph of x versus t. * Speed is the magnitude of instantaneous velocity. Instantaneous Velocity and Speed -------------------------------- You have now seen two ways to describe how fast something moves: average velocity and average speed, both of which are measured over a time interval -t. However, the phrase “how fast” more commonly refers to how fast a particle is moving at a given instant—its instantaneous velocity (or simply velocity) v. The velocity at any instant is obtained from the average velocity by shrinking the time interval :math:`\Delta t` closer and closer to 0. As :math:`\Delta t` dwindles, the average velocity approaches a limiting value, which is the velocity at that instant: .. math:: v = \lim_{\Delta t \to 0} = \frac{\Delta x}{\Delta t} = \frac{dx}{dt} Note that v is the rate at which position x is changing with time at a given instant; that is, v is the derivative of x with respect to t. Also note that v at any instant is the slope of the position – time curve at the point representing that instant. Velocity is another vector quantity and thus has an associated direction. **Speed** is the magnitude of velocity; that is, speed is velocity that has been stripped of any indication of direction, either in words or via an algebraic sign. (Caution: Speed and average speed can be quite different.) A velocity of #5 m/s and one of $5 m/s both have an associated speed of 5 m/s. The speedometer in a car measures speed, not velocity (it cannot determine the direction). Checkpoint 2 The following equations give the position x(t) of a particle in four situations (in each equation, x is in meters, t is in seconds, and t > 0): (1) x = 3t - 2; (2) :math:`x = 4t^2 - 2`; (3) :math:`x = 2/t^2`; and (4) x = -2. (a) In which situation is the velocity v of the particle con- stant? (b) In which is v in the negative x direction? Sample Problem 2.02 Velocity and slope of x versus t, elevator cab Figure 2-6a is an x(t) plot for an elevator cab that is initially stationary, then moves upward (which we take to be the pos- itive direction of x), and then stops. Plot v(t). KEY IDEA We can find the velocity at any time from the slope of the x(t) curve at that time. Calculations: The slope of x(t), and so also the velocity, is zero in the intervals from 0 to 1 s and from 9 s on, so then the cab is stationary. During the interval bc, the slope is con- stant and nonzero, so then the cab moves with constant velocity. We calculate the slope of x(t) then as .. math:: \frac{\Delta x}{\Delta t} = v = \frac{24 m - 4-0 m}{8.0 s - 3.0 s} = + 4.0 m/s (2-5) The plus sign indicates that the cab is moving in the posi- tive x direction. These intervals (where v " 0 and v " 4 m/s) are plotted in Fig. 2-6b. In addition, as the cab ini- tially begins to move and then later slows to a stop, v varies as indicated in the intervals 1 s to 3 s and 8 s to 9 s. Thus, Fig. 2-6b is the required plot. (Figure 2-6c is consid- ered in Module 2-3.) Given a v(t) graph such as Fig. 2-6b, we could “work backward” to produce the shape of the associated x(t) graph (Fig. 2-6a). However, we would not know the actual values for x at various times, because the v(t) graph indicates only changes in x. To find such a change in x during any interval, we must, in the language of calculus, calculate the area “under the curve” on the v(t) graph for that interval. For example, during the interval 3 s to 8 s in which the cab has a velocity of 4.0 m/s, the change in x is .. math:: \Delta x = (4.0 m/s)(8.0 s $ 3.0 s) = +20 m. (2-6) (This area is positive because the v(t) curve is above the t axis.) Figure 2-6a shows that x does indeed increase by 20 m in that interval. However, Fig. 2-6b does not tell us the values of x at the beginning and end of the interval. For that, we need additional information, such as the value of x at some instant. 2-3 ACCELERATION ----------------- Learning Objectives After reading this module, you should be able to . . . 2.10 Apply the relationship between a particle’s average acceleration, its change in velocity, and the time interval for that change. 2.11 Given a particle’s velocity as a function of time, calcu- late the instantaneous acceleration for any particular time. 2.12 Given a graph of a particle’s velocity versus time, deter- mine the instantaneous acceleration for any particular time and the average acceleration between any two particular times Key Ideas * Average acceleration is the ratio of a change in velocity :math:`\Delta v` to the time interval :math:`\Delta t` in which the change occurs: .. math:: a_{avg} = \frac{\Delta v}{\Delta t} The algebraic sign indicates the direction of aavg. * Instantaneous acceleration (or simply acceleration) a is the first time derivative of velocity v(t) and the second time deriv- ative of position x(t): .. math:: a = \frac{dv}{dt} = \frac{d^2 x}{dt^2} * On a graph of v versus t, the acceleration a at any time t is the slope of the curve at the point that represents t. Acceleration ------------ When a particle’s velocity changes, the particle is said to undergo acceleration (or to accelerate). For motion along an axis, the average acceleration aavg over a time interval :math:`\Delta t` is .. math:: a_{avg} = \frac{v_2 - v_1}{t_2 - t_1} = \frac{\Delta v}{\Delta t} (2-7) where the particle has velocity v1 at time t1 and then velocity v2 at time t2. The instantaneous acceleration (or simply acceleration) is .. math:: a= \frac{dv}{dt} (2-8) In words, the acceleration of a particle at any instant is the rate at which its velocity is changing at that instant. Graphically, the acceleration at any point is the slope of the curve of v(t) at that point.We can combine Eq. 2-8 with Eq. 2-4 to write .. math:: a = \frac{dv}{dt} = \frac{d}{dt} (\frac{dx}{dt}) = \frac{d^2 x}{dt^2} In words, the acceleration of a particle at any instant is the second derivative of its position x(t) with respect to time. A common unit of acceleration is the meter per second per second: :math:`m/(s \cdot s)` or :math:`m/s^2`. Other units are in the form of length/(time ! time) or length/time2. Acceleration has both magnitude and direction (it is yet another vector quan- tity). Its algebraic sign represents its direction on an axis just as for displacement and velocity; that is, acceleration with a positive value is in the positive direction of an axis, and acceleration with a negative value is in the negative direction. Figure 2-6 gives plots of the position, velocity, and acceleration of an ele- vator moving up a shaft. Compare the a(t) curve with the v(t) curve — each point on the a(t) curve shows the derivative (slope) of the v(t) curve at the corresponding time. When v is constant (at either 0 or 4 m/s), the derivative is zero and so also is the acceleration. When the cab first begins to move, the v(t) curve has a positive derivative (the slope is positive), which means that a(t) is positive. When the cab slows to a stop, the derivative and slope of the v(t) curve are negative; that is, a(t) is negative. Next compare the slopes of the v(t) curve during the two acceleration peri- ods. The slope associated with the cab’s slowing down (commonly called “decel- eration”) is steeper because the cab stops in half the time it took to get up to speed. The steeper slope means that the magnitude of the deceleration is larger than that of the acceleration, as indicated in Fig. 2-6c. **Sensations.** The sensations you would feel while riding in the cab of Fig. 2-6 are indicated by the sketched figures at the bottom. When the cab first accelerates, you feel as though you are pressed downward; when later the cab is braked to a stop, you seem to be stretched upward. In between, you feel nothing special. In other words, your body reacts to accelerations (it is an accelerometer) but not to velocities (it is not a speedometer). When you are in a car traveling at 90 km/h or an airplane traveling at 900 km/h, you have no bodily awareness of the motion. However, if the car or plane quickly changes velocity, you may become keenly aware of the change, perhaps even frightened by it. Part of the thrill of an amusement park ride is due to the quick changes of velocity that you undergo (you pay for the accelerations, not for the speed). A more extreme example is shown in the photographs of Fig. 2-7, which were taken while a rocket sled was rapidly accelerated along a track and then rapidly braked to a stop. **g Units**. Large accelerations are sometimes expressed in terms of g units, with .. math:: 1g = 9.8 m/s^2 (g unit). (2-10) (As we shall discuss in Module 2-5, g is the magnitude of the acceleration of a falling object near Earth’s surface.) On a roller coaster, you may experience brief accelerations up to 3g, which is (3)(9.8 m/s2), or about 29 m/s2, more than enough to justify the cost of the ride. Signs. In common language, the sign of an acceleration has a nonscientific meaning: positive acceleration means that the speed of an object is increasing, and negative acceleration means that the speed is decreasing (the object is decelerat- ing). In this book, however, the sign of an acceleration indicates a direction, not whether an object’s speed is increasing or decreasing. For example, if a car with an initial velocity v " $25 m/s is braked to a stop in 5.0 s, then aavg " #5.0 m/s2. The acceleration is positive, but the car’s speed has decreased. The reason is the differ- ence in signs: the direction of the acceleration is opposite that of the velocity. Here then is the proper way to interpret the signs: If the signs of the velocity and acceleration of a particle are the same, the speed of the particle increases. If the signs are opposite, the speed decreases. Checkpoint 3 A wombat moves along an x axis. What is the sign of its acceleration if it is moving (a) in the positive direction with increasing speed, (b) in the positive direction with decreasing speed, (c) in the negative direction with increasing speed, and (d) in the negative direction with decreasing speed? Sample Problem 2.03 Acceleration and dv/dt A particle’s position on the x axis of Fig. 2-1 is given by .. math:: x = 4 - 27t + t^3, with x in meters and t in seconds. (a) Because position x depends on time t, the particle must be moving. Find the particle’s velocity function v(t) and ac- celeration function a(t). KEY IDEAS (1) To get the velocity function v(t), we differentiate the po- sition function x(t) with respect to time. (2) To get the accel- eration function a(t), we differentiate the velocity function v(t) with respect to time. Calculations: Differentiating the position function, we find .. math:: v = -27 + 3t^2, (Answer) with v in meters per second. Differentiating the velocity function then gives us .. math:: a = +6t, (Answer) with a in meters per second squared. b) Is there ever a time when v = 0? Calculation: Setting v(t) = 0 yields .. math:: 0 = -27 + 3t^2, which has the solution .. math:: t = \pm 3 s Thus, the velocity is zero both 3 s before and 3 s after the clock reads 0. (c) Describe the particle’s motion for :math:`t \geq 0`. Reasoning: We need to examine the expressions for x(t), v(t), and a(t). At t = 0, the particle is at x(0) = +4 m and is moving with a velocity of v(0) " $27 m/s — that is, in the negative direction of the x axis. Its acceleration is a(0) " 0 because just then the particle’s velocity is not changing (Fig. 2-8a). For 0 < t < 3 s, the particle still has a negative velocity, so it continues to move in the negative direction. However, its acceleration is no longer 0 but is increasing and positive. Because the signs of the velocity and the acceleration are opposite, the particle must be slowing (Fig. 2-8b). Indeed, we already know that it stops momentarily at t " 3 s. Just then the particle is as far to the left of the origin in Fig. 2-1 as it will ever get. Substituting t " 3 s into the expression for x(t), we find that the particle’s position just then is x " $50 m (Fig. 2-8c). Its acceleration is still positive. For t > 3 s, the particle moves to the right on the axis. Its acceleration remains positive and grows progressively larger in magnitude. The velocity is now positive, and it too grows progressively larger in magnitude (Fig. 2-8d). 2-4 CONSTANT ACCELERATION ------------------------- Learning Objectives After reading this module, you should be able to . . . 2.13 For constant acceleration, apply the relationships between position, displacement, velocity, acceleration, and elapsed time (Table 2-1). 2.14 Calculate a particle’s change in velocity by integrating its acceleration function with respect to time. 2.15 Calculate a particle’s change in position by integrating its velocity function with respect to time. Key Ideas * The following five equations describe the motion of a particle with constant acceleration: .. math:: v = v_0 + at, x - x_0 = v_0 t + \frac{1}{2}at^2, v^2 = v_0^2 + 2a (x-x_0), x - x_0 = \frac{1}{2} (v_0 + v)t, x-x_0 = vt - \frac{1}{2} at^2 These are not valid when the acceleration is not constant. Constant Acceleration: A Special Case -------------------------------------- In many types of motion, the acceleration is either constant or approximately so. For example, you might accelerate a car at an approximately constant rate when a traffic light turns from red to green. Then graphs of your position, velocity, and acceleration would resemble those in Fig. 2-9. (Note that a(t) in Fig. 2-9c is constant, which requires that v(t) in Fig. 2-9b have a constant slope.) Later when you brake the car to a stop, the acceleration (or deceleration in common language) might also be approximately constant. Such cases are so common that a special set of equations has been derived for dealing with them. One approach to the derivation of these equations is given in this section. A second approach is given in the next section. Throughout both sections and later when you work on the homework problems, keep in mind that these equations are valid only for constant acceleration (or situations in which you can approximate the acceleration as being constant). **First Basic Equation.** When the acceleration is constant, the average accel- eration and instantaneous acceleration are equal and we can write Eq. 2-7, with some changes in notation, as .. math:: a = a_{avg} = \frac{v - v_0}{t - 0} Here v0 is the velocity at time t " 0 and v is the velocity at any later time t. We can recast this equation as .. math:: v = v_0 + at As a check, note that this equation reduces to v " v0 for t " 0, as it must. As a fur- ther check, take the derivative of Eq. 2-11. Doing so yields dv/dt " a, which is the definition of a. Figure 2-9b shows a plot of Eq. 2-11, the v(t) function; the function is linear and thus the plot is a straight line. **Second Basic Equation.** In a similar manner, we can rewrite Eq. 2-2 (with a few changes in notation) as .. math:: v_{avg} = \frac{x-x_0}{t -0} and then as .. math:: x = x_0 + v_{avg}t, (2-12) in which x0 is the position of the particle at t " 0 and vavg is the average velocity between t " 0 and a later time t. For the linear velocity function in Eq. 2-11, the average velocity over any time interval (say, from t " 0 to a later time t) is the average of the velocity at the be- ginning of the interval (" v0) and the velocity at the end of the interval (" v). For the interval from t " 0 to the later time t then, the average velocity is .. math:: v_{avg} = \frac{1}{2} (v_0 + v). (2-13) Substituting the right side of Eq. 2-11 for v yields, after a little rearrangement, .. math:: v_{avg} = v_0 + \frac{1}{2} at. (2-14) Finally, substituting Eq. 2-14 into Eq. 2-12 yields .. math:: x - x_0 = v_0 t + \frac{1}{2} at^2. (2-15) As a check, note that putting t " 0 yields x " x0, as it must. As a further check, taking the derivative of Eq. 2-15 yields Eq. 2-11, again as it must. Figure 2-9a shows a plot of Eq. 2-15; the function is quadratic and thus the plot is curved. Three Other Equations. Equations 2-11 and 2-15 are the basic equations for constant acceleration; they can be used to solve any constant acceleration prob- lem in this book. However, we can derive other equations that might prove useful in certain specific situations. First, note that as many as five quantities can possi- bly be involved in any problem about constant acceleration — namely, x $ x0, v, t, a, and v0. Usually, one of these quantities is not involved in the problem, either as a given or as an unknown. We are then presented with three of the remaining quantities and asked to find the fourth. Equations 2-11 and 2-15 each contain four of these quantities, but not the same four. In Eq. 2-11, the “missing ingredient” is the displacement x $ x0. In Eq. 2-15, it is the velocity v. These two equations can also be combined in three ways to yield three additional equations, each of which involves a different “missing variable.” First, we can eliminate t to obtain .. math:: v^2 = v^2_0 + 2a(x - x_0). (2-16) This equation is useful if we do not know t and are not required to find it. Second, we can eliminate the acceleration a between Eqs. 2-11 and 2-15 to produce an equation in which a does not appear: .. math:: x - x_0 = \frac{1}{2}(v_0 + v)t Finally, we can eliminate v0, obtaining .. math:: x - x_0 = vt - \frac{1}{2}a t^2 Note the subtle difference between this equation and Eq. 2-15. One involves the initial velocity v0; the other involves the velocity v at time t. Table 2-1 lists the basic constant acceleration equations (Eqs. 2-11 and 2-15) as well as the specialized equations that we have derived.To solve a simple constant ac- celeration problem, you can usually use an equation from this list (if you have the list with you). Choose an equation for which the only unknown variable is the vari- able requested in the problem. A simpler plan is to remember only Eqs. 2-11 and 2-15, and then solve them as simultaneous equations whenever needed. Checkpoint 4 The following equations give the position x(t) of a particle in four situations: (1) :math:`x = 3t - 4; (2) x = -5t^3 + 4t^2 + 6; (3) x = 2/t^2 - 4/t; (4) x = 5t^2 - 3`. To which of these situations do the equations of Table 2-1 apply? Sample Problem 2.04 Drag race of car and motorcycle A popular web video shows a jet airplane, a car, and a mo- torcycle racing from rest along a runway (Fig. 2-10). Initially the motorcycle takes the lead, but then the jet takes the lead, and finally the car blows past the motorcycle. Here let’s focus on the car and motorcycle and assign some reasonable values to the motion. The motorcycle first takes the lead because its (constant) acceleration am " 8.40 m/s2 is greater than the car’s (constant) acceleration ac " 5.60 m/s2, but it soon loses to the car because it reaches its greatest speed vm " 58.8 m/s before the car reaches its greatest speed vc " 106 m/s. How long does the car take to reach the motorcycle? KEY IDEAS We can apply the equations of constant acceleration to both vehicles, but for the motorcycle we must consider the mo- tion in two stages: (1) First it travels through distance xm1 with zero initial velocity and acceleration am " 8.40 m/s2, reaching speed vm " 58.8 m/s. (2) Then it travels through dis- tance xm2 with constant velocity vm " 58.8 m/s and zero ac- celeration (that, too, is a constant acceleration). (Note that we symbolized the distances even though we do not know their values. Symbolizing unknown quantities is often help- ful in solving physics problems, but introducing such un- knowns sometimes takes physics courage.) Calculations: So that we can draw figures and do calcula- tions, let’s assume that the vehicles race along the positive di- rection of an x axis, starting from x " 0 at time t " 0. (We can choose any initial numbers because we are looking for the elapsed time, not a particular time in, say, the afternoon, but let’s stick with these easy numbers.) We want the car to pass the motorcycle, but what does that mean mathematically? It means that at some time t, the side-by-side vehicles are at the same coordinate: xc for the car and the sum xm1 # xm2 for the motorcycle. We can write this statement mathe- matically as .. math:: x_c = x_{m1} + x_{m2} (2-19) xc " xm1 # xm2. (Writing this first step is the hardest part of the problem. That is true of most physics problems. How do you go from the problem statement (in words) to a mathematical expres- sion? One purpose of this book is for you to build up that ability of writing the first step — it takes lots of practice just as in learning, say, tae-kwon-do.) Now let’s fill out both sides of Eq. 2-19, left side first. To reach the passing point at xc, the car accelerates from rest. From Eq. 2-15 (:math:`x - x_0 = v_0t + \frac{1}{2}at^2`), with x0 and v0 = 0, we have .. math:: x_c = \frac{1}{2} a_c t^2. (2-20) To write an expression for xm1 for the motorcycle, we first find the time tm it takes to reach its maximum speed vm, using Eq. 2-11 (v " v0 # at). Substituting v0 " 0, v " vm " 58.8 m/s, and a " am " 8.40 m/s2, that time is .. math:: t_m = \frac{v_m}{a_m} = \frac{58.8 m/s}{8.40 m/s^2} = 7.00 s To get the distance xm1 traveled by the motorcycle during the first stage, we again use Eq. 2-15 with x0 " 0 and v0 " 0, but we also substitute from Eq. 2-21 for the time. We find .. math:: x_{m1} = \frac{1}{2} a_m t_m^2 =\frac{1}{2}a_m (\frac{v_m}{a_m})^2 = \frac{1}{2} \frac{v_m^2}{a_m} For the remaining time of t $ tm , the motorcycle travels at its maximum speed with zero acceleration. To get the distance, we use Eq. 2-15 for this second stage of the motion, but now the initial velocity is v0 " vm (the speed at the end of the first stage) and the acceleration is a " 0. So, the dis- tance traveled during the second stage is .. math:: x_{m2} = v_m(t - t_m) = v_m(t - 7.00 s). (2-23) To finish the calculation, we substitute Eqs. 2-20, 2-22, and 2-23 into Eq. 2-19, obtaining .. math:: \frac{1}{2} a_c t^2 = \frac{1}{2} \frac{v_m^2}{a_m} + v_m (t - 7.00 s) This is a quadratic equation. Substituting in the given data, we solve the equation (by using the usual quadratic-equa- tion formula or a polynomial solver on a calculator), finding t " 4.44 s and t " 16.6 s. But what do we do with two answers? Does the car pass the motorcycle twice? No, of course not, as we can see in the video. So, one of the answers is mathematically correct but not physically meaningful. Because we know that the car passes the motorcycle after the motorcycle reaches its maxi- mum speed at t " 7.00 s, we discard the solution with t ( 7.00 s as being the unphysical answer and conclude that the passing occurs at .. math:: t = 16.6 s. (Answer) Figure 2-11 is a graph of the position versus time for the two vehicles, with the passing point marked. Notice that at t " 7.00 s the plot for the motorcycle switches from being curved (because the speed had been increasing) to being straight (because the speed is thereafter constant). Another Look at Constant Acceleration* -------------------------------------- The first two equations in Table 2-1 are the basic equations from which the others are derived. Those two can be obtained by integration of the acceleration with the condition that a is constant. To find Eq. 2-11, we rewrite the definition of ac- celeration (Eq. 2-8) as .. math:: dv = a dt. We next write the indefinite integral (or antiderivative) of both sides: .. math:: \int dv = \int a dt Since acceleration a is a constant, it can be taken outside the integration. We obtain .. math:: \int dv = a \int dt or v = at + C To evaluate the constant of integration C, we let t " 0, at which time v " v0. Substituting these values into Eq. 2-25 (which must hold for all values of t, including t " 0) yields .. math:: v_0 = (a)(0) + C = C Substituting this into Eq. 2-25 gives us Eq. 2-11. To derive Eq. 2-15, we rewrite the definition of velocity (Eq. 2-4) as .. math:: dx = v dt and then take the indefinite integral of both sides to obtain .. math:: \int dx = \int v dt Next, we substitute for v with Eq. 2-11: .. math:: \int dx = \int (v_0 + at) dt Since v0 is a constant, as is the acceleration a, this can be rewritten as .. math:: \int dx = v_0 \int dt + a \int t dt Integration now yields .. math:: x = v_0t + \frac{1}{2} at^2 + C', (2-26) where C' is another constant of integration. At time t = 0, we have x = x0. Substituting these values in Eq. 2-26 yields x0 = C'. Replacing C' with x0 in Eq. 2-26 gives us Eq. 2-15. 2-5 FREE-FALL ACCELERATION Learning Objectives After reading this module, you should be able to . . . 2.16 Identify that if a particle is in free flight (whether upward or downward) and if we can neglect the effects of air on its motion, the particle has a constant downward acceleration with a magnitude g that we take to be 9.8 m/s2. 2.17 Apply the constant-acceleration equations (Table 2-1) to free-fall motion. Key Ideas * An important example of straight-line motion with constant acceleration is that of an object rising or falling freely near Earth’s surface. The constant acceleration equations de- scribe this motion, but we make two changes in notation: (1) we refer the motion to the vertical y axis with #y vertically up; (2) we replace a with -g, where g is the magnitude of the free-fall acceleration. Near Earth’s surface, .. math:: g = 9.8 m/s^2 = 32 ft/s^2. Free-Fall Acceleration ---------------------- If you tossed an object either up or down and could somehow eliminate the effects of air on its flight, you would find that the object accelerates downward at a certain constant rate. That rate is called the free-fall acceleration, and its magni- tude is represented by g. The acceleration is independent of the object’s charac- teristics, such as mass, density, or shape; it is the same for all objects. Two examples of free-fall acceleration are shown in Fig. 2-12, which is a series of stroboscopic photos of a feather and an apple. As these objects fall, they accelerate downward — both at the same rate g. Thus, their speeds increase at the same rate, and they fall together. The value of g varies slightly with latitude and with elevation. At sea level in Earth’s midlatitudes the value is 9.8 m/s2 (or 32 ft/s2), which is what you should use as an exact number for the problems in this book unless otherwise noted. The equations of motion in Table 2-1 for constant acceleration also apply to free fall near Earth’s surface; that is, they apply to an object in vertical flight, either up or down, when the effects of the air can be neglected. However, note that for free fall: (1) The directions of motion are now along a vertical y axis instead of the x axis, with the positive direction of y upward. (This is important for later chapters when combined horizontal and vertical motions are examined.) (2) The free-fall acceleration is negative — that is, downward on the y axis, toward Earth’s center — and so it has the value $g in the equations. The free-fall acceleration near Earth’s surface is a = -g = -9.8 m/s2, and the magnitude of the acceleration is g " 9.8 m/s2. Do not substitute $9.8 m/s2 for g. Suppose you toss a tomato directly upward with an initial (positive) velocity v0 and then catch it when it returns to the release level. During its free-fall flight (from just after its release to just before it is caught), the equations of Table 2-1 apply to its motion. The acceleration is always a " $g " $9.8 m/s2, negative and thus down- ward. The velocity, however, changes, as indicated by Eqs. 2-11 and 2-16: during the ascent, the magnitude of the positive velocity decreases, until it momentarily be- comes zero. Because the tomato has then stopped, it is at its maximum height. During the descent, the magnitude of the (now negative) velocity increases. Checkpoint 5 (a) If you toss a ball straight up, what is the sign of the ball’s displacement for the ascent, from the release point to the highest point? (b) What is it for the descent, from the highest point back to the release point? (c) What is the ball’s acceleration at its highest point? Sample Problem 2.05 Time for full up-down flight, baseball toss In Fig. 2-13, a pitcher tosses a baseball up along a y axis, with an initial speed of 12 m/s. (a) How long does the ball take to reach its maximum height? KEY IDEAS (1) Once the ball leaves the pitcher and before it returns to his hand, its acceleration is the free-fall acceleration a=-g. Because this is constant, Table 2-1 applies to the motion. (2) The velocity v at the maximum height must be 0. Calculation: Knowing v, a, and the initial velocity v0 = 12 m/s, and seeking t, we solve Eq. 2-11, which contains those four variables. This yields .. math:: t = \frac{v - v_0}{a} = \frac{0 - 12 m/s}{-9.8 m/s^2} = 1.2 s (Answer) (b) What is the ball’s maximum height above its release point? Calculation: We can take the ball’s release point to be y0 = 0. We can then write Eq. 2-16 in y notation, set y - y0 = y and v = 0 (at the maximum height), and solve for y.We get .. math:: t = \frac{v^2 - v_0^2}{2a} = \frac{0 - 12 m/s}{2(-9.8 m/s^2)} = 7.3 m (Answer) (c) How long does the ball take to reach a point 5.0 m above its release point? Calculations: We know v0, a = -g, and displacement y -y0 = 5.0 m, and we want t, so we choose Eq. 2-15. Rewriting it for y and setting y0 = 0 give us .. math:: y = v0 t - \frac{1}{2} gt^2, or 5.0 m = (12 m/s)t - (12)(9.8 m/s2)t 2. If we temporarily omit the units (having noted that they are consistent), we can rewrite this as .. math:: 4.9t^2 - 12t + 5.0 = 0. Solving this quadratic equation for t yields .. math:: t = 0.53 s and t = 1.9 s. (Answer) There are two such times! This is not really surprising because the ball passes twice through y = 5.0 m, once on the way up and once on the way down. 2-6 GRAPHICAL INTEGRATION IN MOTION ANALYSIS --------------------------------------------- Learning Objectives After reading this module, you should be able to . . . 2.19 Determine a particle’s change in position by graphical integration on a graph of velocity versus time. 2.18 Determine a particle’s change in velocity by graphical integration on a graph of acceleration versus time. Key Ideas * On a graph of acceleration a versus time t, the change in the velocity is given by .. math:: v_1 - v_0 = \int_{t_0}^{t_1} The integral amounts to finding an area on the graph: .. mayh:: \int_{t_0}^{t_1} = (area between acceleration curve an time axis , from t0 to t1) * On a graph of velocity v versus time t, the change in the position is given by .. math:: x_1 - x_0 = \int_{t_0}^{t_1} v dt where the integral can be taken from the graph as \int_{t_0}^{t_1} v dt = (area between velocity curve an time axis , from t0 to t1) Graphical Integration in Motion Analysis ----------------------------------------- **Integrating Acceleration.** When we have a graph of an object’s acceleration a ver- sus time t, we can integrate on the graph to find the velocity at any given time. Because a is defined as a " dv/dt, the Fundamental Theorem of Calculus tells us that .. math:: v_1 - v_0 = \int_{t_0}^{t_1} a dt (2-27) The right side of the equation is a definite integral (it gives a numerical result rather than a function), v0 is the velocity at time t0, and v1 is the velocity at later time t1.The def- inite integral can be evaluated from an a(t) graph,such as in Fig.2-14a.In particular, .. math:: \int_{t_0}^{t_1} a dt = (area beteen acceleration curve and time axis, from t_0 to t_1) (2-28) If a unit of acceleration is 1 m/s2 and a unit of time is 1 s, then the correspon- ding unit of area on the graph is .. math (1 m/s^2)(1 s) = 1 m/s, which is (properly) a unit of velocity. When the acceleration curve is above the time axis, the area is positive; when the curve is below the time axis, the area is negative. **Integrating Velocity.** Similarly, because velocity v is defined in terms of the posi- tion x as v = dx/dt, then .. math:: x_1 - x_0 = \int_{t_0}^{t_1} v dt where x0 is the position at time t0 and x1 is the position at time t1. The definite integral on the right side of Eq. 2-29 can be evaluated from a v(t) graph, like that shown in Fig. 2-14b. In particular, .. math:: \int_{t_0}^{t_1} = (area between velocity curve and time axis, from t_0 to t_1) If the unit of velocity is 1 m/s and the unit of time is 1 s, then the corre- sponding unit of area on the graph is .. math:: (1 m/s)(1 s) = 1 m, which is (properly) a unit of position and displacement. Whether this area is posi- tive or negative is determined as described for the a(t) curve of Fig. 2-14a. Sample Problem 2.06 Graphical integration a versus t, whiplash injury “Whiplash injury” commonly occurs in a rear-end collision where a front car is hit from behind by a second car. In the 1970s, researchers concluded that the injury was due to the occupant’s head being whipped back over the top of the seat as the car was slammed forward. As a result of this finding, head restraints were built into cars, yet neck injuries in rear- end collisions continued to occur. In a recent test to study neck injury in rear-end collisions, a volunteer was strapped to a seat that was then moved abruptly to simulate a collision by a rear car moving at 10.5 km/h. Figure 2-15a gives the accelerations of the volun- teer’s torso and head during the collision, which began at time t " 0. The torso acceleration was delayed by 40 ms because during that time interval the seat back had to compress against the volunteer. The head acceleration was delayed by an additional 70 ms. What was the torso speed when the head began to accelerate? KEY IDEA We can calculate the torso speed at any time by finding an area on the torso a(t) graph. Calculations: We know that the initial torso speed is v0 = 0 at time t0 = 0, at the start of the “collision.” We want the torso speed v1 at time t1 " 110 ms, which is when the head begins to accelerate. Combining Eqs. 2-27 and 2-28, we can write .. math: v_1 - v_0 = (area between acceleration curve and time axis, from t0 to t1) For convenience, let us separate the area into three regions (Fig. 2-15b). From 0 to 40 ms, region A has no area: .. math:: area_A = 0 From 40 ms to 100 ms, region B has the shape of a triangle, with area .. math:: areaB " 12(0.060 s)(50 m/s2) " 1.5 m/s. From 100 ms to 110 ms, region C has the shape of a rectan- gle, with area .. math:: areaC " (0.010 s)(50 m/s2) " 0.50 m/s. Substituting these values and v0 = 0 into Eq. 2-31 gives us .. math:: v_1 - 0 = 0 + 1.5 m/s + 0.50 m/s or v1 = 2.0 m/s = 7.2 km/h. (Answer) Comments: When the head is just starting to move forward, the torso already has a speed of 7.2 km/h. Researchers argue that it is this difference in speeds during the early stage of a rear-end collision that injures the neck. The backward whip- ping of the head happens later and could, especially if there is no head restraint, increase the injury. Review & Summary ---------------- **Position** The position x of a particle on an x axis locates the par- ticle with respect to the origin, or zero point, of the axis.The position is either positive or negative, according to which side of the origin the particle is on, or zero if the particle is at the origin. The positive direction on an axis is the direction of increasing positive numbers; the opposite direction is the negative direction on the axis. **Displacement** The displacement :math:`\Delta x` of a particle is the change .. math:: \Delta = x_2 - x_1 Displacement is a vector quantity. It is positive if the particle has moved in the positive direction of the x axis and negative if the particle has moved in the negative direction. **Average Velocity** When a particle has moved from position x1 to position x2 during a time interval :math:`\Delta t = t_2 - t_1`, its average velocity during that interval is .. math:: v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x_2-x_1}{t_2-t_1} The algebraic sign of vavg indicates the direction of motion (vavg is a vector quantity). Average velocity does not depend on the actual distance a particle moves, but instead depends on its original and final positions. On a graph of x versus t, the average velocity for a time interval :math:`\Delta t` is the slope of the straight line connecting the points on the curve that represent the two ends of the interval. **Average Speed** The average speed savg of a particle during a time interval :math:`\Delta t` depends on the total distance the particle moves in that time interval: .. math:: s_{avg} = \frac{total distance}{\Delta t} (2-3) **Instantaneous Velocity** The instantaneous velocity (or sim- ply velocity) v of a moving particle is .. math:: v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt} where :math:`Delta x` and :math:`\Delta t` are defined by Eq. 2-2. The instantaneous velocity (at a particular time) may be found as the slope (at that particular time) of the graph of x versus t. **Speed** is the magnitude of instanta- neous velocity. **Average Acceleration** Average acceleration is the ratio of a change in velocity :math:`\Delta v` to the time interval :math:`Delta t` in which the change occurs: .. math:: a_{avg} = \frac{\Delta v}{\Delta t} The algebraic sign indicates the direction of aavg. **Instantaneous Acceleration** Instantaneous acceleration (or simply acceleration) a is the first time derivative of velocity v(t) and the second time derivative of position x(t): .. math:: a = \frac{dv}{dt} = \frac{d^2 x}{dt^2} On a graph of v versus t, the acceleration a at any time t is the slope of the curve at the point that represents t. **Constant Acceleration** The five equations in Table 2-1 describe the motion of a particle with constant acceleration: .. math:: v = v_0 + at x-x_0 = v_0t + \frac{1}{2} at^2 v^2 = v_0^2 + 2a(x - x_0) x - x_0 = \frac{1}{2} (v_0 + v)t x - x_0 = vt - \frac{1}{2}at^2 These are not valid when the acceleration is not constant. **Free-Fall Acceleration** An important example of straight- line motion with constant acceleration is that of an object rising or falling freely near Earth’s surface. The constant acceleration equa- tions describe this motion, but we make two changes in notation: (1) we refer the motion to the vertical y axis with #y vertically up; (2) we replace a with $g, where g is the magnitude of the free-fall acceleration. Near Earth’s surface, g = 9.8 m/s2 (= 32 ft/s2). Questions --------- 1. Figure 2-16 gives the velocity of a particle moving on an x axis. What are (a) the initial and (b) the final di- rections of travel? (c) Does the parti- cle stop momentarily? (d) Is the ac- celeration positive or negative? (e) Is it constant or varying? 2. Figure 2-17 gives the accelera- tion a(t) of a Chihuahua as it chases Figure 2-16 Question 1. a German shepherd along an axis. In which of the time periods indicated does the Chihuahua move at constant speed? 3. Figure 2-18 shows four paths along which objects move from a starting point to a final point, all in the same time interval. The paths pass over a grid of equally spaced straight lines. Rank the paths according to (a) the av- erage velocity of the objects and (b) the average speed of the objects, great- est first. 4. Figure 2-19 is a graph of a particle’s position along an x axis versus time. (a) At time t = 0, what is the sign of the particle’s position? Is the particle’s velocity positive, negative, or 0 at (b) t = 1 s, (c) t = 2 s, and (d) t = 3 s? (e) How many times does the particle go through the point x = 0? 5. Figure 2-20 gives the velocity of a particle moving along an axis. Point 1 is at the highest point on the curve; point 4 is at the lowest point; and points 2 and 6 are at the same height. What is the direction of travel at (a) time t " 0 and (b) point 4? (c) At which of the six numbered points does the particle reverse its direction of travel? (d) Rank the six points according to the magnitude of the acceleration, greatest first. 6 At t = 0, a particle moving along an x axis is at position x0 " $20 m. The signs of the particle’s initial velocity v0 (at time t0) and constant acceleration a are, respectively, for four situations: (1) #, #; (2) #, $; (3) $, #; (4) $, $. In 0 which situations will the particle (a) stop momentarily, (b) pass through the origin, and (c) never pass through the origin? 7 Hanging over the railing of a bridge, you drop an egg (no initial ve- locity) as you throw a second egg downward. Which curves in Fig. 2-21 give the velocity v(t) for (a) the dropped egg and (b) the thrown egg? (Curves A and B are parallel; so are C, D, and E; so are F and G.) 8 The following equations give the velocity v(t) of a particle in four situations: (a) v " 3; (b) v " 4t 2 # 2t $ 6; (c) v " 3t $ 4; (d) v " 5t 2 $ 3. To which of these situations do the equations of Table 2-1 apply? 9 In Fig. 2-22, a cream tangerine is thrown di- rectly upward past three evenly spaced windows of equal heights. Rank the windows according to (a) the average speed of the cream tangerine while passing them, (b) the time the cream tan- gerine takes to pass them, (c) the magnitude of the acceleration of the cream tangerine while passing them, and (d) the change -v in the speed of the cream tangerine during the pas- sage, greatest first. 10 Suppose that a passenger intent on lunch during his first ride in a hot-air balloon accidently drops an apple over the side during the balloon’s liftoff. At the moment of the apple’s release, the balloon is accelerating upward with a magni- tude of 4.0 m/s2 and has an upward velocity of magnitude 2 m/s. What are the (a) magnitude and (b) direction of the acceleration of the apple just after it is released? (c) Just then, is the apple moving upward or downward, or is it stationary? (d) What is the magni- tude of its velocity just then? (e) In the next few moments, does the speed of the apple increase, decrease, or remain constant? 11 Figure 2-23 shows that a particle moving along an x axis un- dergoes three periods of acceleration. Without written computa- tion, rank the acceleration periods according to the increases they produce in the particle’s velocity, greatest first. Problems -------- Module 2-1 Position, Displacement, and Average Velocity 1 While driving a car at 90 km/h, how far do you move while your eyes shut for 0.50 s during a hard sneeze? 2 Compute your average velocity in the following two cases: (a) You walk 73.2 m at a speed of 1.22 m/s and then run 73.2 m at a speed of 3.05 m/s along a straight track. (b) You walk for 1.00 min at a speed of 1.22 m/s and then run for 1.00 min at 3.05 m/s along a straight track. (c) Graph x versus t for both cases and indicate how the average velocity is found on the graph. 3 SSM WWW An automobile travels on a straight road for 40 km at 30 km/h. It then continues in the same direction for an- other 40 km at 60 km/h. (a) What is the average velocity of the car during the full 80 km trip? (Assume that it moves in the positive x direction.) (b) What is the average speed? (c) Graph x versus t and indicate how the average velocity is found on the graph. 4 A car moves uphill at 40 km/h and then back downhill at 60 km/h. What is the average speed for the round trip? 5 SSM The position of an object moving along an x axis is given by x " 3t $ 4t 2 # t 3, where x is in meters and t in seconds. Find the position of the object at the following values of t: (a) 1 s, (b) 2 s, (c) 3 s, and (d) 4 s. (e) What is the object’s displacement between t " 0 and t " 4 s? (f) What is its average velocity for the time interval from t " 2 s to t " 4 s? (g) Graph x versus t for 0 0 t 0 4 s and indi- cate how the answer for (f) can be found on the graph. 6 The 1992 world speed record for a bicycle (human-powered vehicle) was set by Chris Huber. His time through the measured 200 m stretch was a sizzling 6.509 s, at which he commented, “Cogito ergo zoom!” (I think, therefore I go fast!). In 2001, Sam Whittingham beat Huber’s record by 19.0 km/h. What was Whittingham’s time through the 200 m? •7 Two trains, each having a speed of 30 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h flies off the front of one train when they are 60 km apart and heads directly for the other train. On reaching the other train, the (crazy) bird flies directly back to the first train, and so forth. What is the to- tal distance the bird travels before the trains collide? 8 Panic escape. Figure 2-24 shows a general situation in which a stream of people attempt to escape through an exit door that turns out to be locked. The people move toward the door at speed vs " 3.50 m/s, are each d " 0.25 m in depth, and are sepa- rated by L " 1.75 m. The arrangement in Fig. 2-24 occurs at time t " 0. (a) At what average rate does the layer of people at the door increase? (b) At what time does the layer’s depth reach 5.0 m? (The answers reveal how quickly such a situation becomes dangerous.) 9 ILW In 1 km races, runner 1 on track 1 (with time 2 min, 27.95 s) appears to be faster than runner 2 on track 2 (2 min, 28.15 s). However, length L2 of track 2 might be slightly greater than length L1 of track 1. How large can L2 $ L1 be for us still to conclude that runner 1 is faster? 10 To set a speed record in a measured (straight-line) distance d, a race car must be driven first in one direction (in time t1) and then in the opposite direction (in time t2). (a) To eliminate the ef- fects of the wind and obtain the car’s speed vc in a windless situation, should we find the average of d/t1 and d/t2 (method 1) or should we di- vide d by the average of t1 and t2? (b) What is the fractional difference in the two methods when a steady wind blows along the car’s route and the ratio of the wind speed vw to the car’s speed vc is 0.0240? 11 You are to drive 300 km to an interview. The interview is at 11!15 A.M. You plan to drive at 100 km/h, so you leave at 8!00 A.M. to allow some extra time. You drive at that speed for the first 100 km, but then construction work forces you to slow to 40 km/h for 40 km. What would be the least speed needed for the rest of the trip to arrive in time for the interview? 12 Traffic shock wave. An abrupt slowdown in concen- trated traffic can travel as a pulse, termed a shock wave, along the line of cars, either downstream (in the traffic direction) or up- stream, or it can be stationary. Figure 2-25 shows a uniformly spaced line of cars moving at speed v " 25.0 m/s toward a uni- formly spaced line of slow cars moving at speed vs " 5.00 m/s. Assume that each faster car adds length L " 12.0 m (car length plus buffer zone) to the line of slow cars when it joins the line, and as- sume it slows abruptly at the last instant. (a) For what separation dis- tance d between the faster cars does the shock wave remain stationary? If the separation is twice that amount, what are the (b) speed and (c) direction (upstream or downstream) of the shock wave? •••13 ILW You drive on Interstate 10 from San Antonio to Houston, half the time at 55 km/h and the other half at 90 km/h. On the way back you travel half the distance at 55 km/h and the other half at 90 km/h. What is your average speed (a) from San Antonio to Houston, (b) from Houston back to San Antonio, and (c) for the entire trip? (d) What is your average velocity for the entire trip? (e) Sketch x versus t for (a), assuming the motion is all in the positive x direc- tion. Indicate how the average velocity can be found on the sketch. Module 2-2 Instantaneous Velocity and Speed •14 An electron moving along the x axis has a position given by x " 16te$t m, where t is in seconds. How far is the electron from the origin when it momentarily stops? •15 (a) If a particle’s position is given by x " 4 $ 12t # 3t 2 (where t is in seconds and x is in meters), what is its velocity at t " 1 s? (b) Is it moving in the positive or negative direction of x just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (e) Is there ever an instant when the velocity is zero? If so, give the time t; if not, answer no. (f) Is there a time after t " 3 s when the particle is moving in the negative direction of x? If so, give the time t; if not, answer no. •16 The position function x(t) of a particle moving along an x axis is x " 4.0 $ 6.0t 2, with x in meters and t in seconds. (a) At what time and (b) where does the particle (momentarily) stop? At what (c) negative time and (d) positive time does the particle pass through the origin? (e) Graph x versus t for the range $5 s to #5 s. (f) To shift the curve rightward on the graph, should we include the term #20t or the term $20t in x(t)? (g) Does that inclusion increase or decrease the value of x at which the particle momentarily stops? ••17 The position of a particle moving along the x axis is given in centimeters by x " 9.75 # 1.50t 3, where t is in seconds. Calculate (a) the average velocity during the time interval t " 2.00 s to t " 3.00 s; (b) the instantaneous velocity at t " 2.00 s; (c) the instantaneous ve- locity at t " 3.00 s; (d) the instantaneous velocity at t " 2.50 s; and (e) the instantaneous velocity when the particle is midway between its positions at t " 2.00 s and t " 3.00 s. (f) Graph x versus t and in- dicate your answers graphically. Module 2-3 Acceleration •18 The position of a particle moving along an x axis is given by x " 12t 2 $ 2t 3, where x is in meters and t is in seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle at t " 3.0 s. (d) What is the maximum positive coordinate reached by the particle and (e) at what time is it reached? (f) What is the maxi- mum positive velocity reached by the particle and (g) at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving (other than at t " 0)? (i) Determine the av- erage velocity of the particle between t " 0 and t " 3 s. •19 SSM At a certain time a particle had a speed of 18 m/s in the positive x direction, and 2.4 s later its speed was 30 m/s in the opposite direction. What is the average acceleration of the particle during this 2.4 s interval? •20 (a) If the position of a particle is given by x " 20t $ 5t 3, where x is in meters and t is in seconds, when, if ever, is the parti- cle’s velocity zero? (b) When is its acceleration a zero? (c) For what time range (positive or negative) is a negative? (d) Positive? (e) Graph x(t), v(t), and a(t). ••21 From t " 0 to t " 5.00 min, a man stands still, and from t " 5.00 min to t " 10.0 min, he walks briskly in a straight line at a constant speed of 2.20 m/s. What are (a) his average velocity vavg and (b) his average acceleration aavg in the time interval 2.00 min to 8.00 min? What are (c) vavg and (d) aavg in the time interval 3.00 min to 9.00 min? (e) Sketch x versus t and v versus t, and indicate how the answers to (a) through (d) can be obtained from the graphs. ••22 The position of a particle moving along the x axis depends on the time according to the equation x " ct 2 $ bt 3, where x is in me- ters and t in seconds.What are the units of (a) constant c and (b) con- stant b? Let their numerical values be 3.0 and 2.0, respectively. (c) At what time does the particle reach its maximum positive x position? From t " 0.0 s to t " 4.0 s, (d) what distance does the particle move and (e) what is its displacement? Find its velocity at times (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s. Find its acceleration at times (j) 1.0 s, (k) 2.0 s, (l) 3.0 s, and (m) 4.0 s. Module 2-4 Constant Acceleration •23 SSM An electron with an initial velocity v0 " 1.50 ' 10 5 m/s enters a region of length L " 1.00 cm where it is electrically accelerated (Fig. 2-26). It emerges with v = 5.70 ' 10 m/s. What is its acceleration, assumed constant? •24 Catapulting mushrooms. Certain mushrooms launch their spores by a catapult mecha- nism. As water condenses from the air onto a spore that is attached to the mushroom, a drop grows on one side of the spore and a film grows on the other side. The spore is bent over by the drop’s weight, but when the film reaches the drop, the drop’s water suddenly spreads into the film and the spore springs upward so rapidly that it is slung off into the air. Typically, the spore reaches a speed of 1.6 m/s in a 5.0 mm launch; its speed is then reduced to zero in 1.0 mm by the air. Using those data and assuming constant accelerations, find the acceleration in terms of g during (a) the launch and (b) the speed reduction. •25 An electric vehicle starts from rest and accelerates at a rate of 2.0 m/s2 in a straight line until it reaches a speed of 20 m/s. The vehicle then slows at a constant rate of 1.0 m/s2 until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle travel from start to stop? •26 A muon (an elementary particle) enters a region with a speed of 5.00 ' 10 6 m/s and then is slowed at the rate of 1.25 ' 1014 m/s2. (a) How far does the muon take to stop? (b) Graph x versus t and v versus t for the muon. •27 An electron has a constant acceleration of #3.2 m/s2. At a certain instant its velocity is #9.6 m/s. What is its velocity (a) 2.5 s earlier and (b) 2.5 s later? •28 On a dry road, a car with good tires may be able to brake with a constant deceleration of 4.92 m/s2. (a) How long does such a car, initially traveling at 24.6 m/s, take to stop? (b) How far does it travel in this time? (c) Graph x versus t and v versus t for the deceleration. •29 ILW A certain elevator cab has a total run of 190 m and a max- imum speed of 305 m/min, and it accelerates from rest and then back to rest at 1.22 m/s2. (a) How far does the cab move while ac- celerating to full speed from rest? (b) How long does it take to make the nonstop 190 m run, starting and ending at rest? •30 The brakes on your car can slow you at a rate of 5.2 m/s2. (a) If you are going 137 km/h and suddenly see a state trooper, what is the minimum time in which you can get your car under the 90 km/h speed limit? (The answer reveals the futility of braking to keep your high speed from being detected with a radar or laser gun.) (b) Graph x versus t and v versus t for such a slowing. •31 SSM Suppose a rocket ship in deep space moves with con- stant acceleration equal to 9.8 m/s2, which gives the illusion of nor- mal gravity during the flight. (a) If it starts from rest, how long will it take to acquire a speed one-tenth that of light, which travels at 3.0 ' 108 m/s? (b) How far will it travel in so doing? •32 A world’s land speed record was set by Colonel John P. Stapp when in March 1954 he rode a rocket-propelled sled that moved along a track at 1020 km/h. He and the sled were brought to a stop in 1.4 s. (See Fig. 2-7.) In terms of g, what acceleration did he experience while stopping? •33 SSM ILW A car traveling 56.0 km/h is 24.0 m from a barrier when the driver slams on the brakes. The car hits the barrier 2.00 s later. (a) What is the magnitude of the car’s constant acceleration before impact? (b) How fast is the car traveling at impact? ••34 In Fig. 2-27, a red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis. At time t " 0, the red car is at xr " 0 and the green car is at xg " 220 m. If the red car has a constant velocity of 20 km/h, the cars pass each other at x " 44.5 m, and if it has a constant velocity of 40 km/h, they pass each other at x " 76.6 m. What are (a) the initial velocity and (b) the constant acceleration of the green car? ••35 Figure 2-27 shows a red car and a green car that move toward each other. Figure 2-28 is a graph of their motion, showing the positions xg0 " 270 m and xr0 " $35.0 m at time t " 0. The green car has a con- stant speed of 20.0 m/s and the red car begins from rest. What is the ac- celeration magnitude of the red car? ••36 A car moves along an x axis through a distance of 900 m, starting at rest (at x " 0) and ending at rest (at x " 900 m). Through the first 14 of that distance, its acceleration is #2.25 m/s2. Through the rest of that distance, its acceleration is $0.750 m/s2. What are (a) its travel time through the 900 m and (b) its maxi- mum speed? (c) Graph position x, velocity v, and acceleration a versus time t for the trip. ••37 Figure 2-29 depicts the motion x (m) of a particle moving along an x axis with a constant acceleration. The figure’s vertical scaling is set by xs " 6.0 m. What are the (a) magnitude and (b) di- rection of the particle’s acceleration? ••38 (a) If the maximum acceleration that is tolerable for passengers in a 0 subway train is 1.34 m/s and subway stations are located 806 m apart, what Figure 2-29 Problem 37. is the maximum speed a subway train can attain between stations? (b) What is the travel time between stations? (c) If a subway train stops for 20 s at each station, what is the maximum average speed of the train, from one start-up to the next? (d) Graph x, v, and a versus t for the interval from one start-up to the next. ••39 Cars A and B move in the same direction in adjacent lanes.The position x of car A is given in Fig. 2-30, from time t " 0 to t " 7.0 s. The figure’s vertical scaling is set by xs " 32.0 m.At t " 0, car B is at x =0, with a velocity of 12 m/s and t (s) a negative constant accelera- Figure 2-30 Problem 39. tion aB. (a) What must aB be such that the cars are (momen- tarily) side by side (momentarily at the same value of x) at t " 4.0 s? (b) For that value of aB, how many times are the cars side by side? (c) Sketch the position x of car B versus time t on Fig. 2-30. How many times will the cars be side by side if the magnitude of accelera- tion aB is (d) more than and (e) less than the answer to part (a)? ••40 You are driving toward a traffic signal when it turns yel- low. Your speed is the legal speed limit of v0 " 55 km/h; your best deceleration rate has the magnitude a " 5.18 m/s2.Your best reaction time to begin braking is T " 0.75 s. To avoid having the front of your car enter the intersection after the light turns red, should you brake to a stop or continue to move at 55 km/h if the distance to the intersection and the duration of the yellow light are (a) 40 m and 2.8 s, and (b) 32 m and 1.8 s? Give an answer of brake, continue, either (if either strategy works), or neither (if neither strategy works and the yellow duration is inappropriate). ••41 As two trains move along a track, their conductors suddenly notice that they are headed toward each other. Figure 2-31 gives their velocities v as functions of time t as the conductors slow the trains. The Figure 2-31 Problem 41. figure’s vertical scaling is set by vs " 40.0 m/s. The slowing processes begin when the trains are 200 m apart.What is their separa- tion when both trains have stopped? •••42 You are arguing over a cell phone while trailing an unmarked police car by 25 m; both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.0 s (long enough for you to look at the phone and yell, “I won’t do that!”). At the beginning of that 2.0 s, the po- lice officer begins braking suddenly at 5.0 m/s2. (a) What is the sep- aration between the two cars when your attention finally returns? Suppose that you take another 0.40 s to realize your danger and begin braking. (b) If you too brake at 5.0 m/s2, what is your speed when you hit the police car? •••43 When a high-speed passenger train traveling at 161 km/h rounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance D " 676 m ahead (Fig. 2-32). The locomotive is moving at 29.0 km/h. The engineer of the high-speed train imme- diately applies the brakes. (a) What must be the magnitude of the resulting constant deceleration if a collision is to be just avoided? (b) Assume that the engineer is at x " 0 when, at t " 0, he first spots the locomotive. Sketch x(t) curves for the locomotive and high-speed train for the cases in which a collision is just avoided and is not quite avoided. •44 When startled, an armadillo will leap upward. Suppose it rises 0.544 m in the first 0.200 s. (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of 0.544 m? (c) How much higher does it go? •45 SSM WWW (a) With what speed must a ball be thrown verti- cally from ground level to rise to a maximum height of 50 m? (b) How long will it be in the air? (c) Sketch graphs of y, v, and a versus t for the ball. On the first two graphs, indicate the time at which 50 m is reached. •46 Raindrops fall 1700 m from a cloud to the ground. (a) If they were not slowed by air resistance, how fast would the drops be moving when they struck the ground? (b) Would it be safe to walk outside during a rainstorm? •47 SSM At a construction site a pipe wrench struck the ground with a speed of 24 m/s. (a) From what height was it inadvertently dropped? (b) How long was it falling? (c) Sketch graphs of y, v, and a versus t for the wrench. •48 A hoodlum throws a stone vertically downward with an ini- tial speed of 12.0 m/s from the roof of a building, 30.0 m above the ground. (a) How long does it take the stone to reach the ground? (b) What is the speed of the stone at impact? •49 SSM A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground? ••50 At time t " 0, apple 1 is dropped from a bridge onto a road- way beneath the bridge; somewhat later, apple 2 is thrown down from the same height. Figure 2-33 gives the vertical positions y of the apples versus t during the falling, until both apples have hit the roadway. The scaling is set by ts " 2.0 s. With approximately what speed is apple 2 thrown down? ••51 As a runaway scientific balloon ascends at 19.6 m/s, one of its instrument packages breaks free of a harness and free-falls. Figure 2-34 gives the vertical velocity of the package versus time, from before it breaks free to when it reaches the ground. (a) What maximum height above the break-free point does it Figure 2-34 Problem 51. rise? (b) How high is the break-free point above the ground? ••52 A bolt is dropped from a bridge under construction, falling 90 m to the valley below the bridge. (a) In how much time does it pass through the last 20% of its fall? What is its speed (b) when it begins that last 20% of its fall and (c) when it reaches the valley beneath the bridge? ••53 SSM ILW A key falls from a bridge that is 45 m above the water. It falls directly into a model boat, moving with constant velocity, that is 12 m from the point of impact when the key is re- leased. What is the speed of the boat? ••54 A stone is dropped into a river from a bridge 43.9 m above the water. Another stone is thrown vertically down 1.00 s after the first is dropped. The stones strike the water at the same time. (a) What is the initial speed of the second stone? (b) Plot velocity versus time on a graph for each stone, taking zero time as the instant the first stone is released. ••55 SSM A ball of moist clay falls 15.0 m to the ground. It is in contact with the ground for 20.0 ms before stopping. (a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground? (Treat the ball as a particle.) (b) Is the average acceleration up or down? ••56 Figure 2-35 shows the speed v versus height y of a ball tossed directly upward, along a y axis. Distance d is 0.40 m. The speed at height yA is vA. The speed at height yB is 13vA. What is speed vA? ••57 To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It re- bounds to a height of 2.00 m. If the ball is in contact with the floor for 12.0 ms, (a) what is the magnitude of its average acceleration during that contact and (b) is the average acceleration up or down? ••58 An object falls a distance h from rest. If it travels 0.50h in the last 1.00 s, find (a) the time and (b) the height of its fall. (c) Explain the physically unacceptable solution of the quadratic equation in t that you obtain. ••59 Water drips from the nozzle of a shower onto the floor 200 cm below. The drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. When the first drop strikes the floor, how far below the nozzle are the (a) second and (b) third drops? ••60 A rock is thrown vertically upward from ground level at time t " 0.At t " 1.5 s it passes the top of a tall tower, and 1.0 s later it reaches its maximum height.What is the height of the tower? •••61 A steel ball is dropped from a building’s roof and passes a window, taking 0.125 s to fall from the top to the bottom of the window, a distance of 1.20 m. It then falls to a sidewalk and bounces back past the window, moving from bottom to top in 0.125 s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is 2.00 s. How tall is the building? •••62 A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend (a) in the top 15.0 cm of this jump and (b) in the bottom 15.0 cm? (The player seems to hang in the air at the top.) •••63 A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.50 s, and the top-to-bottom height of the window is 2.00 m. How high above the window top does the flowerpot go? •••64 A ball is shot vertically up- ward from the surface of another planet. A plot of y versus t for the ball is shown in Fig. 2-36, where y is the height of the ball above its start- ing point and t " 0 at the instant the ball is shot. The figure’s vertical scal- ing is set by ys " 30.0 m. What are the magnitudes of (a) the free-fall accel- eration on the planet and (b) the ini- tial velocity of the ball? Module 2-6 Graphical Integration in Motion Analysis •65 Figure 2-15a gives the acceleration of a volunteer’s head and torso during a rear-end collision. At maximum head ac- celeration, what is the speed of (a) the head and (b) the torso? ••66 In a forward punch in karate, the fist begins at rest at the waist and is brought rapidly forward until the arm is fully ex- tended. The speed v(t) of the fist is given in Fig. 2-37 for someone skilled in karate. The vertical scaling is set by vs " 8.0 m/s. How far has the fist moved at (a) time t " 50 ms and (b) when the speed of the fist is maximum? ••67 When a soccer ball is kicked to- ward a player and the player deflects the ball by “heading” it, the accelera- tion of the head during the collision can be significant. Figure 2-38 gives the measured acceleration a(t) of a soccer player’s head for a bare head and a helmeted head, starting from rest. The scaling on the vertical axis is set by as " 200 m/s2. At time t " 7.0 ms, what is the difference in the speed acquired by the bare head and the speed acquired by the helmeted head? ••68 A salamander of the genus Hydromantes captures prey by launching its tongue as a projectile: The skeletal part of the tongue is shot for- ward, unfolding the rest of the tongue, until the outer portion lands on the prey, sticking to it. Figure 2-39 shows the acceleration mag- nitude a versus time t for the acceleration phase of the launch in a typical situation. The indicated accelerations are a2 " 400 m/s2 and a1 " 100 m/s2. What is the outward speed of the tongue at the end of the acceleration phase? ••69 ILW How far does the run- ner whose velocity–time graph is shown in Fig. 2-40 travel in 16 s? The figure’s vertical scaling is set by vs " 8.0 m/s. •••70 Two particles move along an x axis. The position of particle 1 is given by x " 6.00t2 # 3.00t # 2.00 (in meters and seconds); the ac- celeration of particle 2 is given by a " $8.00t (in meters per second squared and seconds) and, at t " 0, its velocity is 20 m/s. When the velocities of the particles match, what is their velocity? Additional Problems ------------------- 71 In an arcade video game, a spot is programmed to move across the screen according to x " 9.00t $ 0.750t 3, where x is dis- tance in centimeters measured from the left edge of the screen and t is time in seconds. When the spot reaches a screen edge, at either x " 0 or x " 15.0 cm, t is reset to 0 and the spot starts moving again according to x(t). (a) At what time after starting is the spot instan- taneously at rest? (b) At what value of x does this occur? (c) What is the spot’s acceleration (including sign) when this occurs? (d) Is it moving right or left just prior to coming to rest? (e) Just after? (f) At what time t . 0 does it first reach an edge of the screen? 72 A rock is shot vertically upward from the edge of the top of a tall building. The rock reaches its maximum height above the top of the building 1.60 s after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground 6.00 s after it is launched. In SI units: (a) with what upward velocity is the rock shot, (b) what maximum height above the top of the building is reached by the rock, and (c) how tall is the building? 73 At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.2 m/s2. At the same instant a truck, traveling with a constant speed of 9.5 m/s, overtakes and passes the automobile. (a) How far beyond the traffic signal will the automobile overtake the truck? (b) How fast will the automo- bile be traveling at that instant? 74 A pilot flies horizontally at 1300 km/h, at height h " 35 m above initially level ground. However, at time t " 0, the pilot be- gins to fly over ground sloping upward at angle u " 4.3° (Fig. 2-41). If the pilot does not change the airplane’s heading, at what time t does the plane strike the ground? 75 To stop a car, first you require a certain reaction time to be- gin braking; then the car slows at a constant rate. Suppose that the total distance moved by your car during these two phases is 56.7 m when its initial speed is 80.5 km/h, and 24.4 m when its initial speed is 48.3 km/h. What are (a) your reaction time and (b) the magni- tude of the acceleration? 76 Figure 2-42 shows part of a street where traffic flow is to be controlled to allow a platoon of cars to move smoothly along the street. Suppose that the platoon leaders have just reached intersection 2, where the green appeared when they were distance d from the intersection. They continue to travel at a cer- tain speed vp (the speed limit) to reach intersection 3, where the green appears when they are distance d from it. The intersections are separated by distances D23 and D12. (a) What should be the time delay of the onset of green at intersection 3 relative to that at intersection 2 to keep the platoon moving smoothly? Suppose, instead, that the platoon had been stopped by a red light at intersection 1. When the green comes on there, the leaders require a certain time tr to respond to the change and an additional time to accelerate at some rate a to the cruising speed vp. (b) If the green at intersection 2 is to appear when the leaders are distance d from that intersection, how long after the light at intersection 1 turns green should the light at intersection 2 turn green? 77 SSM A hot rod can accelerate from 0 to 60 km/h in 5.4 s. (a) What is its average acceleration, in m/s2, during this time? (b) How far will it travel during the 5.4 s, assuming its acceleration is con- stant? (c) From rest, how much time would it require to go a distance of 0.25 km if its acceleration could be maintained at the value in (a)? 78 A red train traveling at 72 km/h and a green train traveling at 144 km/h are headed toward each other along a straight, level track. When they are 950 m apart, each engineer sees the other’s train and applies the brakes. The brakes slow each train at the rate of 1.0 m/s2. Is there a collision? If so, answer yes and give the speed of the red train and the speed of the green train at impact, respec- tively. If not, answer no and give the separation between the trains when they stop. 79 At time t " 0, a rock climber accidentally allows a piton to fall freely from a high point on the rock wall to the valley below him. Then, after a short delay, his climbing part- ner, who is 10 m higher on the wall, throws a piton down- ward. The positions y of the pitons versus t during the falling are given in Fig. 2-43. With what speed is the second piton thrown? 80 A train started from rest and moved with constant accelera- tion. At one time it was traveling 30 m/s, and 160 m farther on it was traveling 50 m/s. Calculate (a) the acceleration, (b) the time re- quired to travel the 160 m mentioned, (c) the time required to at- tain the speed of 30 m/s, and (d) the distance moved from rest to the time the train had a speed of 30 m/s. (e) Graph x versus t and v versus t for the train, from rest. 81 SSM A particle’s acceleration along an x axis is a " 5.0t, with t in seconds and a in meters per second squared. At t " 2.0 s, its velocity is #17 m/s. What is its velocity at t " 4.0 s? 82 Figure 2-44 gives the ac- celeration a versus time t for a particle moving along an x axis. The a-axis scale is set by as " 12.0 m/s2. At t " $2.0 s, the particle’s velocity is 7.0 m/s. What is its velocity at t " 6.0 s? 83 Figure 2-45 shows a simple device for measuring your reaction time. It consists of a cardboard strip marked with a scale and two large dots. A friend holds the strip vertically, with thumb and forefinger at the dot on the right in Fig. 2-45. You then posi- tion your thumb and forefinger at the other dot (on the left in Fig. 2-45), being careful not to touch the strip. Your friend re- leases the strip, and you try to pinch it as soon as possible after you see it begin to fall. The mark at the place where you pinch the strip gives your reaction time. (a) How far from the lower dot should you place the 50.0 ms mark? How much higher should you place the marks for (b) 100, (c) 150, (d) 200, and (e) 250 ms? (For example, should the 100 ms marker be 2 times as far from the dot as the 50 ms marker? If so, give an answer of 2 times. Can you find any pattern in the answers?) 84 A rocket-driven sled running on a straight, level track is used to investigate the effects of large accelerations on humans. One such sled can attain a speed of 1600 km/h in 1.8 s, starting from rest. Find (a) the acceleration (assumed constant) in terms of g and (b) the distance traveled. 85 A mining cart is pulled up a hill at 20 km/h and then pulled back down the hill at 35 km/h through its original level. (The time required for the cart’s reversal at the top of its climb is negligible.) What is the average speed of the cart for its round trip, from its original level back to its original level? 86 A motorcyclist who is moving along an x axis directed to- ward the east has an acceleration given by a " (6.1 $ 1.2t) m/s2 for 0 0 t 0 6.0 s. At t " 0, the velocity and position of the cyclist are 2.7 m/s and 7.3 m. (a) What is the maximum speed achieved by the cyclist? (b) What total distance does the cyclist travel be- tween t " 0 and 6.0 s? 87 SSM When the legal speed limit for the New York Thruway was increased from 55 mi/h to 65 mi/h, how much time was saved by a motorist who drove the 700 km between the Buffalo entrance and the New York City exit at the legal speed limit? 88 A car moving with constant acceleration covered the distance between two points 60.0 m apart in 6.00 s. Its speed as it passed the second point was 15.0 m/s. (a) What was the speed at the first point? (b) What was the magnitude of the acceleration? (c) At what prior distance from the first point was the car at rest? (d) Graph x versus t and v versus t for the car, from rest (t " 0). 89 SSM A certain juggler usually tosses balls vertically to a height H. To what height must they be tossed if they are to spend twice as much time in the air? 90 A particle starts from the ori- gin at t " 0 and moves along the positive x axis. A graph of the veloc- ity of the particle as a function of the time is shown in Fig. 2-46; the v-axis scale is set by vs " 4.0 m/s. (a) What is the coordinate of the particle at t " 5.0 s? (b) What is the velocity of the particle at t " 5.0 s? (c) What is the acceleration of the particle at t " 5.0 s? (d) What is the average ve- locity of the particle between t " 1.0 s and t " 5.0 s? (e) What is the average acceleration of the particle between t " 1.0 s and t " 5.0 s? 91 A rock is dropped from a 100-m-high cliff. How long does it take to fall (a) the first 50 m and (b) the second 50 m? 92 Two subway stops are separated by 1100 m. If a subway train accelerates at #1.2 m/s2 from rest through the first half of the dis- tance and decelerates at $1.2 m/s2 through the second half, what are (a) its travel time and (b) its maximum speed? (c) Graph x, v, and a versus t for the trip. 93 A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B, 3.00 m higher than A, with speed 1 2 v. Calculate (a) the speed v and (b) the maximum height reached by the stone above point B. 94 A rock is dropped (from rest) from the top of a 60-m-tall building. How far above the ground is the rock 1.2 s before it reaches the ground? 95 SSM An iceboat has a constant velocity toward the east when a sudden gust of wind causes the iceboat to have a constant accel- eration toward the east for a period of 3.0 s. A plot of x versus t is shown in Fig. 2-47, where t " 0 is taken to be the instant the wind starts to blow and the positive x axis is toward the east. (a) What is the acceleration of the iceboat during the 3.0 s interval? (b) What is the velocity of the iceboat at the end of the 3.0 s interval? (c) If the acceleration remains constant for an additional 3.0 s, how far does the iceboat travel during this second 3.0 s interval? 96 A lead ball is dropped in a lake from a diving board 5.20 m above the water. It hits the water with a certain velocity and then sinks to the bottom with this same constant velocity. It reaches the bottom 4.80 s after it is dropped. (a) How deep is the lake? What are the (b) magnitude and (c) direction (up or down) of the aver- age velocity of the ball for the entire fall? Suppose that all the wa- ter is drained from the lake. The ball is now thrown from the diving board so that it again reaches the bottom in 4.80 s. What are the (d) magnitude and (e) direction of the initial velocity of the ball? 97 The single cable supporting an unoccupied construction ele- vator breaks when the elevator is at rest at the top of a 120-m-high building. (a) With what speed does the elevator strike the ground? (b) How long is it falling? (c) What is its speed when it passes the halfway point on the way down? (d) How long has it been falling when it passes the halfway point? 98 Two diamonds begin a free fall from rest from the same height, 1.0 s apart. How long after the first diamond begins to fall will the two diamonds be 10 m apart? 99 A ball is thrown vertically downward from the top of a 36.6- m-tall building. The ball passes the top of a window that is 12.2 m above the ground 2.00 s after being thrown. What is the speed of the ball as it passes the top of the window? 100 A parachutist bails out and freely falls 50 m. Then the para- chute opens, and thereafter she decelerates at 2.0 m/s2. She reaches the ground with a speed of 3.0 m/s. (a) How long is the parachutist in the air? (b) At what height does the fall begin? 101 A ball is thrown down vertically with an initial speed of v0 from a height of h. (a) What is its speed just before it strikes the ground? (b) How long does the ball take to reach the ground? What would be the answers to (c) part a and (d) part b if the ball were thrown upward from the same height and with the same ini- tial speed? Before solving any equations, decide whether the an- swers to (c) and (d) should be greater than, less than, or the same as in (a) and (b). 102 The sport with the fastest moving ball is jai alai, where measured speeds have reached 303 km/h. If a professional jai alai player faces a ball at that speed and involuntarily blinks, he blacks out the scene for 100 ms. How far does the ball move dur- ing the blackout? 103 If a baseball pitcher throws a fastball at a horizontal speed of 160 km/h, how long does the ball take to reach home plate 18.4 m away? 104 A proton moves along the x axis according to the equation x " 50t # 10t 2, where x is in meters and t is in seconds. Calculate (a) the average velocity of the proton during the first 3.0 s of its motion, (b) the instantaneous velocity of the proton at t " 3.0 s, and (c) the instantaneous acceleration of the proton at t " 3.0 s. (d) Graph x versus t and indicate how the answer to (a) can be obtained from the plot. (e) Indicate the answer to (b) on the graph. (f) Plot v versus t and indicate on it the answer to (c). 105 A motorcycle is moving at 30 m/s when the rider applies the brakes, giving the motorcycle a constant deceleration. During the 3.0 s interval immediately after braking begins, the speed decreases to 15 m/s. What distance does the motorcycle travel from the instant braking begins until the motorcycle stops? 106 A shuffleboard disk is accelerated at a constant rate from rest to a speed of 6.0 m/s over a 1.8 m distance by a player using a cue. At this point the disk loses contact with the cue and slows at a constant rate of 2.5 m/s2 until it stops. (a) How much time elapses from when the disk begins to accelerate until it stops? (b) What total distance does the disk travel? 107 The head of a rattlesnake can accelerate at 50 m/s2 in striking a victim. If a car could do as well, how long would it take to reach a speed of 100 km/h from rest? 108 A jumbo jet must reach a speed of 360 km/h on the runway for takeoff. What is the lowest constant acceleration needed for takeoff from a 1.80 km runway? 109 An automobile driver increases the speed at a constant rate from 25 km/h to 55 km/h in 0.50 min. A bicycle rider speeds up at a constant rate from rest to 30 km/h in 0.50 min. What are the magni- tudes of (a) the driver’s acceleration and (b) the rider’s acceleration? 110 On average, an eye blink lasts about 100 ms. How far does a MiG-25 “Foxbat” fighter travel during a pilot’s blink if the plane’s average velocity is 3400 km/h? 111 A certain sprinter has a top speed of 11.0 m/s. If the sprinter starts from rest and accelerates at a constant rate, he is able to reach his top speed in a distance of 12.0 m. He is then able to main- tain this top speed for the remainder of a 100 m race. (a) What is his time for the 100 m race? (b) In order to improve his time, the sprinter tries to decrease the distance required for him to reach his top speed. What must this distance be if he is to achieve a time of 10.0 s for the race? 112 The speed of a bullet is measured to be 640 m/s as the bullet emerges from a barrel of length 1.20 m.Assuming constant accelera- tion, find the time that the bullet spends in the barrel after it is fired. 113 The Zero Gravity Research Facility at the NASA Glenn Research Center includes a 145 m drop tower.This is an evacuated ver- tical tower through which, among other possibilities, a 1-m-diameter sphere containing an experimental package can be dropped. (a) How long is the sphere in free fall? (b) What is its speed just as it reaches a catching device at the bottom of the tower? (c) When caught, the sphere experiences an average deceleration of 25g as its speed is reduced to zero.Through what distance does it travel during the deceleration? 114 A car can be braked to a stop from the autobahn-like speed of 200 km/h in 170 m. Assuming the acceleration is constant, find its magnitude in (a) SI units and (b) in terms of g. (c) How much time Tb is required for the braking? Your reaction time Tr is the time you require to perceive an emergency, move your foot to the brake, and begin the braking. If Tr " 400 ms, then (d) what is Tb in terms of Tr, and (e) is most of the full time required to stop spent in reacting or braking? Dark sunglasses delay the visual signals sent from the eyes to the visual cortex in the brain, increasing Tr. (f) In the extreme case in which Tr is increased by 100 ms, how much farther does the car travel during your reaction time? 115 In 1889, at Jubbulpore, India, a tug-of-war was finally won af- ter 2 h 41 min, with the winning team displacing the center of the rope 3.7 m. In centimeters per minute, what was the magnitude of the average velocity of that center point during the contest? 116 Most important in an investigation of an airplane crash by the U.S. National Transportation Safety Board is the data stored on the airplane’s flight-data recorder, commonly called the “black box” in spite of its orange coloring and reflective tape. The recorder is engi- neered to withstand a crash with an average deceleration of magni- tude 3400g during a time interval of 6.50 ms. In such a crash, if the recorder and airplane have zero speed at the end of that time inter- val, what is their speed at the beginning of the interval? 117 From January 26, 1977, to September 18, 1983, George Meegan of Great Britain walked from Ushuaia, at the southern tip of South America, to Prudhoe Bay in Alaska, covering 30 600 km. In meters per second, what was the magnitude of his average velocity during that time period? 118 The wings on a stonefly do not flap, and thus the insect cannot fly. However, when the insect is on a water surface, it can sail across the surface by lifting its wings into a breeze. Suppose that you time stoneflies as they move at constant speed along a straight path of a certain length. On average, the trips each take 7.1 s with the wings set as sails and 25.0 s with the wings tucked in. (a) What is the ratio of the sailing speed vs to the nonsailing speed vns? (b) In terms of vs, what is the difference in the times the insects take to travel the first 2.0 m along the path with and without sailing? 119 The position of a particle as it moves along a y axis is given by y " (2.0 cm) sin (pt/4), with t in seconds and y in centimeters. (a) What is the average veloc- ity of the particle between t " 0 and t " 2.0 s? (b) What is the instan- taneous velocity of the particle at t " 0, 1.0, and 2.0 s? (c) What is the average acceleration of the particle between t " 0 and t " 2.0 s? (d) What is the instantaneous acceleration of the particle at t " 0, 1.0, and 2.0 s?