Solution Manual for Fundamentals of Physics Extended, 10th Edition

Chapter 2 1. The speed (assumed constant) is v = (90 km/h)(1000 m/km)  (3600 s/h) = 25 m/s. Thus, in 0.50 s, the car travels a distance d = vt = (25 m/s)(0.50 s)  13 m. 2. (a) Using the fact that time = distance/velocity while the velocity is constant, we find 73.2 m  73.2 m vavg  73.2 m 73.2 m  1.74 m/s. 1.22 m/s  3.05 m (b) Using the fact that distance = vt while the velocity v is constant, we find vavg  (122 . m / s)(60 s)  (3.05 m / s)(60 s)  2.14 m / s. 120 s (c) The graphs are shown below (with meters and seconds understood). The first consists of two (solid) line segments, the first having a slope of 1.22 and the second having a slope of 3.05. The slope of the dashed line represents the average velocity (in both graphs). The second graph also consists of two (solid) line segments, having the same slopes as before — the main difference (compared to the first graph) being that the stage involving higher-speed motion lasts much longer. 3. THINK This one-dimensional kinematics problem consists of two parts, and we are asked to solve for the average velocity and average speed of the car. EXPRESS Since the trip consists of two parts, let the displacements during first and second parts of the motion be x1 and x2, and the corresponding time intervals be t1 and t2, respectively. Now, because the problem is one-dimensional and both displacements are in the same direction, the total displacement is simply x = x1 + x2, and the total time for the trip is t = t1 + t2. Using the definition of average velocity given in Eq. 2-2, we have x x1  x2 vavg   . t t1  t2 23 CHAPTER 2 24 To find the average speed, we note that during a time t if the velocity remains a positive constant, then the speed is equal to the magnitude of velocity, and the distance is equal to the magnitude of displacement, with d  | x |  vt. ANALYZE (a) During the first part of the motion, the displacement is x1 = 40 km and the time taken is (40 km) t1   133 . h. (30 km / h) Similarly, during the second part of the trip the displacement is x2 = 40 km and the time interval is (40 km) t2   0.67 h. (60 km / h) The total displacement is x = x1 + x2 = 40 km + 40 km = 80 km, and the total time elapsed is t = t1 + t2 = 2.00 h. Consequently, the average velocity is vavg  x (80 km)   40 km/h. t (2.0 h) (b) In this case, the average speed is the same as the magnitude of the average velocity: savg  40 km/h. (c) The graph of the entire trip, shown below, consists of two contiguous line segments, the first having a slope of 30 km/h and connecting the origin to (t1, x1) = (1.33 h, 40 km) and the second having a slope of 60 km/h and connecting (t1, x1) to (t, x) = (2.00 h, 80 km). From the graphical point of view, the slope of the dashed line drawn from the origin to (t, x) represents the average velocity. LEARN The average velocity is a vector quantity that depends only on the net displacement (also a vector) between the starting and ending points. 4. Average speed, as opposed to average velocity, relates to the total distance, as opposed to the net displacement. The distance D up the hill is, of course, the same as the distance down the hill, and since the speed is constant (during each stage of the 25 motion) we have speed = D/t. Thus, the average speed is Dup  Ddown t up  tdown  2D D D  vup vdown which, after canceling D and plugging in vup = 40 km/h and vdown = 60 km/h, yields 48 km/h for the average speed. 5. THINK In this one-dimensional kinematics problem, we’re given the position function x(t), and asked to calculate the position and velocity of the object at a later time. EXPRESS The position function is given as x(t) = (3 m/s)t – (4 m/s2)t2 + (1 m/s3)t3. The position of the object at some instant t0 is simply given by x(t0). For the time interval t1  t  t2 , the displacement is x  x(t2 )  x(t1 ) . Similarly, using Eq. 2-2, the average velocity for this time interval is x x(t2 )  x(t1 ) vavg   . t t2  t1 ANALYZE (a) Plugging in t = 1 s into x(t) yields x(1 s) = (3 m/s)(1 s) – (4 m/s2)(1 s)2 + (1 m/s3)(1 s)3 = 0. (b) With t = 2 s we get x(2 s) = (3 m/s)(2 s) – (4 m/s2) (2 s)2 + (1 m/s3)(2 s)3 = –2 m. (c) With t = 3 s we have x (3 s) = (3 m/s)(3 s) – (4 m/s2) (3 s)2 + (1 m/s3)(3 s)3 = 0 m. (d) Similarly, plugging in t = 4 s gives x(4 s) = (3 m/s)(4 s) – (4 m/s2)(4 s)2 + (1 m/s3) (4 s)3 = 12 m. (e) The position at t = 0 is x = 0. Thus, the displacement between t = 0 and t = 4 s is x  x(4 s)  x(0)  12 m  0  12 m. (f) The position at t = 2 s is subtracted from the position at t = 4 s to give the displacement: x  x(4 s)  x(2 s)  12 m  (2 m)  14 m . Thus, the average velocity is x 14 m vavg    7 m/s. t 2s (g) The position of the object for the interval 0  t  4 is plotted below. The straight line drawn from the point at (t, x) = (2 s, –2 m) to (4 s, 12 m) would represent the average velocity, answer for part (f). CHAPTER 2 26 LEARN Our graphical representation illustrates once again that the average velocity for a time interval depends only on the net displacement between the starting and ending points. 6. Huber’s speed is v0 = (200 m)/(6.509 s) =30.72 m/s = 110.6 km/h, where we have used the conversion factor 1 m/s = 3.6 km/h. Since Whittingham beat Huber by 19.0 km/h, his speed is v1 = (110.6 km/h + 19.0 km/h) = 129.6 km/h, or 36 m/s (1 km/h = 0.2778 m/s). Thus, using Eq. 2-2, the time through a distance of 200 m for Whittingham is  x 200 m t    5.554 s. v1 36 m/s 7. Recognizing that the gap between the trains is closing at a constant rate of 60 km/h, the total time that elapses before they crash is t = (60 km)/(60 km/h) = 1.0 h. During this time, the bird travels a distance of x = vt = (60 km/h)(1.0 h) = 60 km. 8. The amount of time it takes for each person to move a distance L with speed vs is t  L / vs . With each additional person, the depth increases by one body depth d (a) The rate of increase of the layer of people is R dv (0.25 m)(3.50 m/s) d d   s   0.50 m/s t L / vs L 1.75 m (b) The amount of time required to reach a depth of D  5.0 m is D 5.0 m t   10 s R 0.50 m/s 9. Converting to seconds, the running times are t1 = 147.95 s and t2 = 148.15 s, respectively. If the runners were equally fast, then savg1  savg 2 From this we obtain  L1 L2  . t1 t2 27 t   148.15  L2  L1   2  1  L1    1  L1  0.00135L1  1.4 m  147.95   t1  where we set L1  1000 m in the last step. Thus, if L1 and L2 are no different than about 1.4 m, then runner 1 is indeed faster than runner 2. However, if L1 is shorter than L2 by more than 1.4 m, then runner 2 would actually be faster. 10. Let vw be the speed of the wind and vc be the speed of the car. (a) Suppose during time interval t1 , the car moves in the same direction as the wind. Then the effective speed of the car is given by veff ,1  vc  vw , and the distance traveled is d  veff ,1t1  (vc  vw )t1 . On the other hand, for the return trip during time interval t2, the car moves in the opposite direction of the wind and the effective speed would be veff ,2  vc  vw . The distance traveled is d  veff ,2t2  (vc  vw )t2 . The two expressions can be rewritten as d d vc  vw  and vc  vw  t1 t2 1d d  Adding the two equations and dividing by two, we obtain vc     . Thus, 2  t1 t2  method 1 gives the car’s speed vc a in windless situation. (b) If method 2 is used, the result would be d 2d vc    (t1  t2 ) / 2 t1  t2 2d d d  vc  vw vc  vw   v 2  vc2  vw2   vc 1   w   . vc   vc   The fractional difference is 2 vc  vc  vw      (0.0240)2  5.76 104 . vc  vc  11. The values used in the problem statement make it easy to see that the first part of the trip (at 100 km/h) takes 1 hour, and the second part (at 40 km/h) also takes 1 hour. Expressed in decimal form, the time left is 1.25 hour, and the distance that remains is 160 km. Thus, a speed v = (160 km)/(1.25 h) = 128 km/h is needed. 12. (a) Let the fast and the slow cars be separated by a distance d at t = 0. If during the time interval t  L / vs  (12.0 m) /(5.0 m/s)  2.40 s in which the slow car has moved a distance of L  12.0 m , the fast car moves a distance of vt  d  L to join the line of slow cars, then the shock wave would remain stationary. The condition implies a separation of d  vt  L  (25 m/s)(2.4 s) 12.0 m  48.0 m. (b) Let the initial separation at t  0 be d  96.0 m. At a later time t, the slow and CHAPTER 2 28 the fast cars have traveled x  vs t and the fast car joins the line by moving a distance d  x . From t x dx  , vs v we get x vs 5.00 m/s d (96.0 m)  24.0 m, v  vs 25.0 m/s  5.00 m/s which in turn gives t  (24.0 m) /(5.00 m/s)  4.80 s. Since the rear of the slow-car pack has moved a distance of x  x  L  24.0 m 12.0 m  12.0 m downstream, the speed of the rear of the slow-car pack, or equivalently, the speed of the shock wave, is vshock  x 12.0 m   2.50 m/s. t 4.80 s (c) Since x  L , the direction of the shock wave is downstream. 13. (a) Denoting the travel time and distance from San Antonio to Houston as T and D, respectively, the average speed is savg1  D (55 km/h)(T/2)  (90 km/h)(T / 2)   72.5 km/h T T which should be rounded to 73 km/h. (b) Using the fact that time = distance/speed while the speed is constant, we find savg2  D D  D/2  68.3 km/h /2 T 55 km/h  90Dkm/h which should be rounded to 68 km/h. (c) The total distance traveled (2D) must not be confused with the net displacement (zero). We obtain for the two-way trip 2D savg   70 km/h. D D  72.5 km/h 68.3 km/h (d) Since the net displacement vanishes, the average velocity for the trip in its entirety is zero. (e) In asking for a sketch, the problem is allowing the student to arbitrarily set the distance D (the intent is not to make the student go to an atlas to look it up); the student can just as easily arbitrarily set T instead of D, as will be clear in the following discussion. We briefly describe the graph (with kilometers-per-hour understood for the slopes): two contiguous line segments, the first having a slope of 55 and connecting the origin to (t1, x1) = (T/2, 55T/2) and the second having a slope of 90 and connecting (t1, x1) to (T, D) where D = (55 + 90)T/2. The average velocity, from the 29 graphical point of view, is the slope of a line drawn from the origin to (T, D). The graph (not drawn to scale) is depicted below: 14. Using the general property v d dx exp(bx)  b exp(bx) , we write F I G H J K F IJ . G H K dx d (19t ) de  t   e  t  (19t )  dt dt dt If a concern develops about the appearance of an argument of the exponential (–t) apparently having units, then an explicit factor of 1/T where T = 1 second can be inserted and carried through the computation (which does not change our answer). The result of this differentiation is v  16(1  t )e  t with t and v in SI units (s and m/s, respectively). We see that this function is zero when t = 1 s. Now that we know when it stops, we find out where it stops by plugging our result t = 1 into the given function x = 16te–t with x in meters. Therefore, we find x = 5.9 m. 15. We use Eq. 2-4 to solve the problem. (a) The velocity of the particle is v dx d  (4  12t  3t 2 )  12  6t . dt dt Thus, at t = 1 s, the velocity is v = (–12 + (6)(1)) = –6 m/s. (b) Since v  0, it is moving in the –x direction at t = 1 s. (c) At t = 1 s, the speed is |v| = 6 m/s. (d) For 0  t  2 s, |v| decreases until it vanishes. For 2  t  3 s, |v| increases from zero to the value it had in part (c). Then, |v| is larger than that value for t  3 s. (e) Yes, since v smoothly changes from negative values (consider the t = 1 result) to positive (note that as t  + , we have v  + ). One can check that v = 0 when t  2 s. CHAPTER 2 30 (f) No. In fact, from v = –12 + 6t, we know that v  0 for t  2 s. 16. We use the functional notation x(t), v(t), and a(t) in this solution, where the latter two quantities are obtained by differentiation: bg dxdtbt g  12t and abt g dvdtbt g  12 vt  with SI units understood. (a) From v(t) = 0 we find it is (momentarily) at rest at t = 0. (b) We obtain x(0) = 4.0 m. (c) and (d) Requiring x(t) = 0 in the expression x(t) = 4.0 – 6.0t2 leads to t = 0.82 s for the times when the particle can be found passing through the origin. (e) We show both the asked-for graph (on the left) as well as the ―shifted‖ graph that is relevant to part (f). In both cases, the time axis is given by –3  t  3 (SI units understood). (f) We arrived at the graph on the right (shown above) by adding 20t to the x(t) expression. (g) Examining where the slopes of the graphs become zero, it is clear that the shift causes the v = 0 point to correspond to a larger value of x (the top of the second curve shown in part (e) is higher than that of the first). 17. We use Eq. 2-2 for average velocity and Eq. 2-4 for instantaneous velocity, and work with distances in centimeters and times in seconds. (a) We plug into the given equation for x for t = 2.00 s and t = 3.00 s and obtain x2 = 21.75 cm and x3 = 50.25 cm, respectively. The average velocity during the time interval 2.00  t  3.00 s is x 50.25 cm  2175 . cm vavg   t 3.00 s  2.00 s which yields vavg = 28.5 cm/s. 2 (b) The instantaneous velocity is v  dx dt  4.5t , which, at time t = 2.00 s, yields v = (4.5)(2.00)2 = 18.0 cm/s. 31 (c) At t = 3.00 s, the instantaneous velocity is v = (4.5)(3.00)2 = 40.5 cm/s. (d) At t = 2.50 s, the instantaneous velocity is v = (4.5)(2.50)2 = 28.1 cm/s. (e) Let tm stand for the moment when the particle is midway between x2 and x3 (that is, when the particle is at xm = (x2 + x3)/2 = 36 cm). Therefore, xm  9.75  15 . tm3  tm  2.596 in seconds. Thus, the instantaneous speed at this time is v = 4.5(2.596)2 = 30.3 cm/s. (f) The answer to part (a) is given by the slope of the straight line between t = 2 and t = 3 in this x-vs-t plot. The answers to parts (b), (c), (d), and (e) correspond to the slopes of tangent lines (not shown but easily imagined) to the curve at the appropriate points. 18. (a) Taking derivatives of x(t) = 12t2 – 2t3 we obtain the velocity and the acceleration functions: v(t) = 24t – 6t2 and a(t) = 24 – 12t with length in meters and time in seconds. Plugging in the value t = 3 yields x(3)  54 m . (b) Similarly, plugging in the value t = 3 yields v(3) = 18 m/s. (c) For t = 3, a(3) = –12 m/s2. (d) At the maximum x, we must have v = 0; eliminating the t = 0 root, the velocity equation reveals t = 24/6 = 4 s for the time of maximum x. Plugging t = 4 into the equation for x leads to x = 64 m for the largest x value reached by the particle. (e) From (d), we see that the x reaches its maximum at t = 4.0 s. (f) A maximum v requires a = 0, which occurs when t = 24/12 = 2.0 s. This, inserted into the velocity equation, gives vmax = 24 m/s. (g) From (f), we see that the maximum of v occurs at t = 24/12 = 2.0 s. (h) In part (e), the particle was (momentarily) motionless at t = 4 s. The acceleration at that time is readily found to be 24 – 12(4) = –24 m/s2. CHAPTER 2 32 (i) The average velocity is defined by Eq. 2-2, so we see that the values of x at t = 0 and t = 3 s are needed; these are, respectively, x = 0 and x = 54 m (found in part (a)). Thus, 54  0 vavg = = 18 m/s. 3 0 19. THINK In this one-dimensional kinematics problem, we’re given the speed of a particle at two instants and asked to calculate its average acceleration. EXPRESS We represent the initial direction of motion as the +x direction. The average acceleration over a time interval t1  t  t2 is given by Eq. 2-7: aavg  v v(t2 )  v(t1 )  . t t2  t1 ANALYZE Let v1 = +18 m/s at t1  0 and v2 = –30 m/s at t2 = 2.4 s. Using Eq. 2-7 we find v(t )  v(t1 ) (30 m/s)  (1 m/s) aavg  2    20 m/s 2 . t2  t1 2.4 s  0 LEARN The average acceleration has magnitude 20 m/s2 and is in the opposite direction to the particle’s initial velocity. This makes sense because the velocity of the particle is decreasing over the time interval. With t1  0 , the velocity of the particle as a function of time can be written as v  v0  at  (18 m/s)  (20 m/s2 )t . 20. We use the functional notation x(t), v(t) and a(t) and find the latter two quantities by differentiating: dx t dv t vt    15t 2  20 and a t    30t t dt bg bg bg bg with SI units understood. These expressions are used in the parts that follow. (a) From 0   15t 2  20 , we see that the only positive value of t for which the particle is (momentarily) stopped is t  20 / 15  12 . s. (b) From 0 = – 30t, we find a(0) = 0 (that is, it vanishes at t = 0). (c) It is clear that a(t) = – 30t is negative for t > 0. (d) The acceleration a(t) = – 30t is positive for t < 0. (e) The graphs are shown below. SI units are understood. 33 21. We use Eq. 2-2 (average velocity) and Eq. 2-7 (average acceleration). Regarding our coordinate choices, the initial position of the man is taken as the origin and his direction of motion during 5 min  t  10 min is taken to be the positive x direction. We also use the fact that x  vt ' when the velocity is constant during a time interval t' . (a) The entire interval considered is t = 8 – 2 = 6 min, which is equivalent to 360 s, whereas the sub-interval in which he is moving is only t'  8  5  3min  180 s. His position at t = 2 min is x = 0 and his position at t = 8 min is x  v t   (2.2)(180)  396 m . Therefore, 396 m  0 vavg   110 . m / s. 360 s (b) The man is at rest at t = 2 min and has velocity v = +2.2 m/s at t = 8 min. Thus, keeping the answer to 3 significant figures, aavg  2.2 m / s  0  0.00611 m / s2 . 360 s (c) Now, the entire interval considered is t = 9 – 3 = 6 min (360 s again), whereas the sub-interval in which he is moving is t   9  5  4min  240 s ). His position at t  3 min is x = 0 and his position at t = 9 min is x  vt   (2.2)(240)  528 m . Therefore, 528 m  0 vavg   147 . m / s. 360 s (d) The man is at rest at t = 3 min and has velocity v = +2.2 m/s at t = 9 min. Consequently, aavg = 2.2/360 = 0.00611 m/s2 just as in part (b). (e) The horizontal line near the bottom of this x-vs-t graph represents the man standing at x = 0 for 0  t < 300 s and the linearly rising line for 300  t  600 s represents his constant-velocity motion. The lines represent the answers to part (a) and (c) in the sense that their slopes yield those results. The graph of v-vs-t is not shown here, but would consist of two horizontal ―steps‖ (one at v = 0 for 0  t x0. Since we seek the maximum, we reject the first root (t = 0) and accept the second (t = 1s). (d) In the first 4 s the particle moves from the origin to x = 1.0 m, turns around, and goes back to x(4 s)  (30 . m / s2 )(4.0 s) 2  (2.0 m / s3 )(4.0 s) 3   80 m . The total path length it travels is 1.0 m + 1.0 m + 80 m = 82 m. (e) Its displacement is x = x2 – x1, where x1 = 0 and x2 = –80 m. Thus, x  80 m . The velocity is given by v = 2ct – 3bt2 = (6.0 m/s2)t – (6.0 m/s3)t2. (f) Plugging in t = 1 s, we obtain v(1 s)  (6.0 m/s2 )(1.0 s)  (6.0 m/s3 )(1.0 s)2  0. (g) Similarly, v(2 s)  (6.0 m/s2 )(2.0 s)  (6.0 m/s3 )(2.0 s) 2  12m/s . (h) v(3 s)  (6.0 m/s2 )(3.0 s)  (6.0 m/s3 )(3.0 s)2  36 m/s . (i) v(4 s)  (6.0 m/s2 )(4.0 s)  (6.0 m/s3 )(4.0 s)2  72 m/s . The acceleration is given by a = dv/dt = 2c – 6b = 6.0 m/s2 – (12.0 m/s3)t. (j) Plugging in t = 1 s, we obtain a(1 s)  6.0 m/s2  (12.0 m/s3 )(1.0 s)  6.0 m/s2 . (k) a(2 s)  6.0 m/s2  (12.0 m/s3 )(2.0 s)  18 m/s 2 . 35 (l) a(3 s)  6.0 m/s2  (12.0 m/s3 )(3.0 s)  30 m/s2 . (m) a(4 s)  6.0 m/s2  (12.0 m/s3 )(4.0 s)   42 m/s 2 . 23. THINK The electron undergoes a constant acceleration. Given the final speed of the electron and the distance it has traveled, we can calculate its acceleration. EXPRESS Since the problem involves constant acceleration, the motion of the electron can be readily analyzed using the equations given in Table 2-1: v  v0  at (2  11) 1 x  x0  v0t  at 2 2 (2  15) v 2  v02  2a( x  x0 ) (2  16) The acceleration can be found by solving Eq. 2-16. ANALYZE With v0  1.50 105 m/s , v  5.70 106 m/s , x0 = 0 and x = 0.010 m, we find the average acceleration to be a v 2  v02 (5.7 106 m/s)2  (1.5 105 m/s)2   1.62 1015 m/s2 . 2x 2(0.010 m) LEARN It is always a good idea to apply other equations in Table 2-1 not used for solving the problem as a consistency check. For example, since we now know the value of the acceleration, using Eq. 2-11, the time it takes for the electron to reach its final speed would be v  v0 5.70 106 m/s  1.5 105 m/s t   3.426 109 s a 1.62 1015 m/s 2 Substituting the value of t into Eq. 2-15, the distance the electron travels is 1 1 x  x0  v0t  at 2  0  (1.5 105 m/s)(3.426 109 s)  (1.62 1015 m/s 2 )(3.426 10 9 s)2 2 2  0.01 m This is what was given in the problem statement. So we know the problem has been solved correctly. 24. In this problem we are given the initial and final speeds, and the displacement, and are asked to find the acceleration. We use the constant-acceleration equation given in Eq. 2-16, v2 = v20 + 2a(x – x0). (a) Given that v0  0 , v  1.6 m/s, and x  5.0 m, the acceleration of the spores during the launch is CHAPTER 2 36 a v 2  v02 (1.6 m/s)2   2.56 105 m/s 2  2.6 104 g 6 2x 2(5.0 10 m) (b) During the speed-reduction stage, the acceleration is a v 2  v02 0  (1.6 m/s)2   1.28 103 m/s 2  1.3 102 g 2x 2(1.0 103 m) The negative sign means that the spores are decelerating. 25. We separate the motion into two parts, and take the direction of motion to be positive. In part 1, the vehicle accelerates from rest to its highest speed; we are given v0 = 0; v = 20 m/s and a = 2.0 m/s2. In part 2, the vehicle decelerates from its highest speed to a halt; we are given v0 = 20 m/s; v = 0 and a = –1.0 m/s2 (negative because the acceleration vector points opposite to the direction of motion). (a) From Table 2-1, we find t1 (the duration of part 1) from v = v0 + at. In this way, 20  0  2.0t1 yields t1 = 10 s. We obtain the duration t2 of part 2 from the same equation. Thus, 0 = 20 + (–1.0)t2 leads to t2 = 20 s, and the total is t = t1 + t2 = 30 s. (b) For part 1, taking x0 = 0, we use the equation v2 = v20 + 2a(x – x0) from Table 2-1 and find v 2  v02 (20 m/s) 2  (0) 2 x   100 m . 2a 2(2.0 m/s 2 ) This position is then the initial position for part 2, so that when the same equation is used in part 2 we obtain v 2  v02 (0)2  (20 m/s) 2 x  100 m   . 2a 2(1.0 m/s 2 ) Thus, the final position is x = 300 m. That this is also the total distance traveled should be evident (the vehicle did not “backtrack” or reverse its direction of motion). 26. The constant-acceleration condition permits the use of Table 2-1. (a) Setting v = 0 and x0 = 0 in v2  v02  2a( x  x0 ) , we find 1 v02 1 (5.00  106 )2 x    0.100 m . 2 a 2 1.25  1014 Since the muon is slowing, the initial velocity and the acceleration must have opposite signs. (b) Below are the time plots of the position x and velocity v of the muon from the moment it enters the field to the time it stops. The computation in part (a) made no reference to t, so that other equations from Table 2-1 (such as v  v0  at and 37 x  v0t  12 at 2 ) are used in making these plots. 27. We use v = v0 + at, with t = 0 as the instant when the velocity equals +9.6 m/s. (a) Since we wish to calculate the velocity for a time before t = 0, we set t = –2.5 s. Thus, Eq. 2-11 gives v  (9.6 m / s)  3.2 m / s2 ( 2.5 s)  16 . m / s. c h (b) Now, t = +2.5 s and we find v  (9.6 m / s)  c 3.2 m / s h (2.5 s)  18 m / s. 2 28. We take +x in the direction of motion, so v0 = +24.6 m/s and a = – 4.92 m/s2. We also take x0 = 0. (a) The time to come to a halt is found using Eq. 2-11: 0  v0  at  t  24.6 m/s  5.00 s .  4.92 m/s 2 (b) Although several of the equations in Table 2-1 will yield the result, we choose Eq. 2-16 (since it does not depend on our answer to part (a)). 0  v02  2ax  x   (24.6 m/s) 2  61.5 m . 2   4.92 m/s 2  (c) Using these results, we plot v0t  12 at 2 (the x graph, shown next, on the left) and v0 + at (the v graph, on the right) over 0  t  5 s, with SI units understood. CHAPTER 2 38 29. We assume the periods of acceleration (duration t1) and deceleration (duration t2) are periods of constant a so that Table 2-1 can be used. Taking the direction of motion to be +x then a1 = +1.22 m/s2 and a2 = –1.22 m/s2. We use SI units so the velocity at t = t1 is v = 305/60 = 5.08 m/s. (a) We denote x as the distance moved during t1, and use Eq. 2-16: v 2  v02  2a1x  x  (b) Using Eq. 2-11, we have t1  (5.08 m/s) 2  10.59 m  10.6 m. 2(1.22 m/s2 ) v  v0 5.08 m/s   4.17 s. a1 1.22 m/s 2 The deceleration time t2 turns out to be the same so that t1 + t2 = 8.33 s. The distances traveled during t1 and t2 are the same so that they total to 2(10.59 m) = 21.18 m. This implies that for a distance of 190 m – 21.18 m = 168.82 m, the elevator is traveling at constant velocity. This time of constant velocity motion is t3  168.82 m  33.21 s. 5.08 m / s Therefore, the total time is 8.33 s + 33.21 s  41.5 s. 30. We choose the positive direction to be that of the initial velocity of the car (implying that a < 0 since it is slowing down). We assume the acceleration is constant and use Table 2-1. (a) Substituting v0 = 137 km/h = 38.1 m/s, v = 90 km/h = 25 m/s, and a = –5.2 m/s2 into v = v0 + at, we obtain t 25 m / s  38 m / s  2.5 s . 5.2 m / s2 (b) We take the car to be at x = 0 when the brakes are applied (at time t = 0). Thus, the coordinate of the car as a function of time is given by x   38 m/s  t  1 5.2 m/s 2  t 2  2 in SI units. This function is plotted from t = 0 to t = 2.5 s on the graph to the right. We have not shown the v-vs-t graph here; it is a descending straight line from v0 to v. 31. THINK The rocket ship undergoes a constant acceleration from rest, and we want to know the time elapsed and the distance traveled when the rocket reaches a certain speed. 39 EXPRESS Since the problem involves constant acceleration, the motion of the rocket can be readily analyzed using the equations in Table 2-1: v  v0  at (2  11) 1 x  x0  v0t  at 2 2 v 2  v02  2a( x  x0 ) (2  15) (2  16) ANALYZE (a) Given that a  9.8 m/s2 , v0  0 and v  0.1c  3.0 107 m/s , we can solve v  v0  at for the time: t v  v0 3.0 107 m/s  0   3.1106 s 2 a 9.8 m/s which is about 1.2 months. So it takes 1.2 months for the rocket to reach a speed of 0.1c starting from rest with a constant acceleration of 9.8 m/s2. (b) To calculate the distance traveled during this time interval, we evaluate x  x0  v0t  12 at 2 , with x0 = 0 and v0 = 0 . The result is 1 x   9.8 m/s2  (3.1106 s)2  4.6 1013 m. 2 LEARN In solving parts (a) and (b), we did not use Eq. (2-16): v2  v02  2a( x  x0 ) . This equation can be used to check our answers. The final velocity based on this equation is v  v02  2a( x  x0 )  0  2(9.8 m/s2 )(4.6 1013 m  0)  3.0 107 m/s , which is what was given in the problem statement. So we know the problems have been solved correctly. 32. The acceleration is found from Eq. 2-11 (or, suitably interpreted, Eq. 2-7). a F 1000 m / km I 1020 km / hg b G H3600 s / h J K v t  14 . s  202.4 m / s2 . In terms of the gravitational acceleration g, this is expressed as a multiple of 9.8 m/s2 as follows:  202.4 m/s 2  a g  21g . 2   9.8 m/s  33. THINK The car undergoes a constant negative acceleration to avoid impacting a barrier. Given its initial speed, we want to know the distance it has traveled and the time elapsed prior to the impact. CHAPTER 2 40 EXPRESS Since the problem involves constant acceleration, the motion of the car can be readily analyzed using the equations in Table 2-1: v  v0  at (2  11) 1 x  x0  v0t  at 2 2 (2  15) v 2  v02  2a( x  x0 ) (2  16) We take x0 = 0 and v0 = 56.0 km/h = 15.55 m/s to be the initial position and speed of the car. Solving Eq. 2-15 with t = 2.00 s gives the acceleration a. Once a is known, the speed of the car upon impact can be found by using Eq. 2-11. ANALYZE (a) Using Eq. 2-15, we find the acceleration to be a 2( x  v0t ) 2 (24.0 m)  (15.55 m/s)(2.00 s)     3.56m/s 2 , 2 2 t (2.00 s) or | a | 3.56 m/s2 . The negative sign indicates that the acceleration is opposite to the direction of motion of the car; the car is slowing down. (b) The speed of the car at the instant of impact is v  v0  at  15.55 m/s  (3.56 m/s2 )(2.00 s)  8.43 m/s which can also be converted to 30.3 km/h. LEARN In solving parts (a) and (b), we did not use Eq. 1-16. This equation can be used as a consistency check. The final velocity based on this equation is v  v02  2a( x  x0 )  (15.55 m/s)2  2(3.56 m/s 2 )(24 m  0)  8.43 m/s , which is what was calculated in (b). This indicates that the problems have been solved correctly. 34. Let d be the 220 m distance between the cars at t = 0, and v1 be the 20 km/h = 50/9 m/s speed (corresponding to a passing point of x1 = 44.5 m) and v2 be the 40 km/h =100/9 m/s speed (corresponding to a passing point of x2 = 76.6 m) of the red car. We have two equations (based on Eq. 2-17): 1 where t1 = x1  v1 1 where t2 = x2  v2 d – x1 = vo t1 + 2 a t12 d – x2 = vo t2 + 2 a t22 We simultaneously solve these equations and obtain the following results: 41 (a) The initial velocity of the green car is vo =  13.9 m/s. or roughly  50 km/h (the negative sign means that it’s along the –x direction). (b) The corresponding acceleration of the car is a =  2.0 m/s2 (the negative sign means that it’s along the –x direction). 35. The positions of the cars as a function of time are given by 1 1 xr (t )  xr 0  ar t 2  (35.0 m)  ar t 2 2 2 xg (t )  xg 0  vg t  (270 m)  (20 m/s)t where we have substituted the velocity and not the speed for the green car. The two cars pass each other at t  12.0 s when the graphed lines cross. This implies that 1 (270 m)  (20 m/s)(12.0 s)  30 m  (35.0 m)  ar (12.0 s) 2 2 which can be solved to give ar  0.90 m/s2 . 36. (a) Equation 2-15 is used for part 1 of the trip and Eq. 2-18 is used for part 2: 1 where a1 = 2.25 m/s2 and x1 = 1 where a2 = 0.75 m/s2 and x2 = x1 = vo1 t1 + 2 a1 t12 x2 = v2 t2  2 a2 t22 900 m 4 3(900) m 4 In addition, vo1 = v2 = 0. Solving these equations for the times and adding the results gives t = t1 + t2 = 56.6 s. (b) Equation 2-16 is used for part 1 of the trip:  900  2 2 v2 = (vo1)2 + 2a1x1 = 0 + 2(2.25)   = 1013 m /s  4  which leads to v = 31.8 m/s for the maximum speed. 37. (a) From the figure, we see that x0 = –2.0 m. From Table 2-1, we can apply x – x0 = v0t + 1 2 at2 with t = 1.0 s, and then again with t = 2.0 s. This yields two equations for the two unknowns, v0 and a: CHAPTER 2 42 1 2 0.0   2.0 m   v0 1.0 s   a 1.0 s  2 1 2 6.0 m   2.0 m   v0  2.0 s   a  2.0 s  . 2 Solving these simultaneous equations yields the results v0 = 0 and a = 4.0 m/s2. (b) The fact that the answer is positive tells us that the acceleration vector points in the +x direction. 38. We assume the train accelerates from rest ( v0  0 and x0  0 ) at a1  134 . m / s until the midway point and then decelerates at a2  134 . m / s2 until it comes to a stop v2  0 at the next station. The velocity at the midpoint is v1, which occurs at x1 = 806/2 = 403m. 2 b g (a) Equation 2-16 leads to v12  v02  2a1 x1  v1  2 1.34 m/s2   403 m   32.9 m/s. (b) The time t1 for the accelerating stage is (using Eq. 2-15) x1  v0t1  2  403 m  1 2 a1t1  t1   24.53 s . 2 1.34 m/s 2 Since the time interval for the decelerating stage turns out to be the same, we double this result and obtain t = 49.1 s for the travel time between stations. (c) With a ―dead time‖ of 20 s, we have T = t + 20 = 69.1 s for the total time between start-ups. Thus, Eq. 2-2 gives 806 m vavg   117 . m/ s . 69.1 s (d) The graphs for x, v and a as a function of t are shown below. The third graph, a(t), consists of three horizontal ―steps‖ — one at 1.34 m/s2 during 0 < t < 24.53 s, and the next at –1.34 m/s2 during 24.53 s < t < 49.1 s and the last at zero during the ―dead time‖ 49.1 s < t 5/2 then there are no (real) solutions to the equation; the cars are never side by side. (e) Here we have 102  2(20)(aB) > 0 at two different times.  two real roots. The cars are side by side 40. We take the direction of motion as +x, so a = –5.18 m/s2, and we use SI units, so v0 = 55(1000/3600) = 15.28 m/s. CHAPTER 2 44 (a) The velocity is constant during the reaction time T, so the distance traveled during it is dr = v0T – (15.28 m/s) (0.75 s) = 11.46 m. We use Eq. 2-16 (with v = 0) to find the distance db traveled during braking: v 2  v02  2adb  db   (15.28 m/s)2 2  5.18 m/s 2  which yields db = 22.53 m. Thus, the total distance is dr + db = 34.0 m, which means that the driver is able to stop in time. And if the driver were to continue at v0, the car would enter the intersection in t = (40 m)/(15.28 m/s) = 2.6 s, which is (barely) enough time to enter the intersection before the light turns, which many people would consider an acceptable situation. (b) In this case, the total distance to stop (found in part (a) to be 34 m) is greater than the distance to the intersection, so the driver cannot stop without the front end of the car being a couple of meters into the intersection. And the time to reach it at constant speed is 32/15.28 = 2.1 s, which is too long (the light turns in 1.8 s). The driver is caught between a rock and a hard place. 41. The displacement (x) for each train is the ―area‖ in the graph (since the displacement is the integral of the velocity). Each area is triangular, and the area of a triangle is 1/2(base) × (height). Thus, the (absolute value of the) displacement for one train (1/2)(40 m/s)(5 s) = 100 m, and that of the other train is (1/2)(30 m/s)(4 s) = 60 m. The initial ―gap‖ between the trains was 200 m, and according to our displacement computations, the gap has narrowed by 160 m. Thus, the answer is 200 – 160 = 40 m. 42. (a) Note that 110 km/h is equivalent to 30.56 m/s. During a two-second interval, you travel 61.11 m. The decelerating police car travels (using Eq. 2-15) 51.11 m. In light of the fact that the initial ―gap‖ between cars was 25 m, this means the gap has narrowed by 10.0 m – that is, to a distance of 15.0 m between cars. (b) First, we add 0.4 s to the considerations of part (a). During a 2.4 s interval, you travel 73.33 m. The decelerating police car travels (using Eq. 2-15) 58.93 m during that time. The initial distance between cars of 25 m has therefore narrowed by 14.4 m. Thus, at the start of your braking (call it t0) the gap between the cars is 10.6 m. The speed of the police car at t0 is 30.56 – 5(2.4) = 18.56 m/s. Collision occurs at time t when xyou = xpolice (we choose coordinates such that your position is x = 0 and the police car’s position is x = 10.6 m at t0). Eq. 2-15 becomes, for each car: xpolice – 10.6 = 18.56(t  t0) – 2 (5)(t  t0)2 1 xyou = 30.56(t  t0) – 2 (5)(t  t0)2 1 . Subtracting equations, we find 10.6 = (30.56 – 18.56)(t  t0)  0.883 s = t  t0. 45 At that time your speed is 30.56 + a(t  t0) = 30.56 – 5(0.883)  26 m/s (or 94 km/h). 43. In this solution we elect to wait until the last step to convert to SI units. Constant acceleration is indicated, so use of Table 2-1 is permitted. We start with Eq. 2-17 and denote the train’s initial velocity as vt and the locomotive’s velocity as v (which is also the final velocity of the train, if the rear-end collision is barely avoided). We note that the distance x consists of the original gap between them, D, as well as the forward distance traveled during this time by the locomotive v t . Therefore, vt  v x D  v t D     v . 2 t t t We now use Eq. 2-11 to eliminate time from the equation. Thus, vt  v D   v 2 v  vt / a b g which leads to v v I 1 Fv  v  v IJF aG  G J H 2 KHD K 2 D bv  v g. Hence, 1 km km I F a 29  161 J  12888 km / h G 2(0.676 km) H h h K t   2 t   t 2 2 which we convert as follows: 1000 m I F1 h I F ac 12888 km / h h G H1 km J KG H3600 sJ K 0.994 m / s 2 2 2 so that its magnitude is |a| = 0.994 m/s2. A graph is shown here for the case where a collision is just avoided (x along the vertical axis is in meters and t along the horizontal axis is in seconds). The top (straight) line shows the motion of the locomotive and the bottom curve shows the motion of the passenger train. The other case (where the collision is not quite avoided) would be similar except that the slope of the bottom curve would be greater than that of the top line at the point where they meet. 44. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the motion. We are allowed to use Table 2-1 (with y replacing x) because this is constant acceleration motion. The ground level is taken to correspond to the origin of the y axis. (a) Using y  v0t  21 gt 2 , with y = 0.544 m and t = 0.200 s, we find CHAPTER 2 46 v0  y  gt 2 / 2 0.544 m  (9.8 m/s 2 ) (0.200 s) 2 / 2   3.70 m/s . t 0.200 s (b) The velocity at y = 0.544 m is v  v0  gt  3.70 m/s  (9.8 m/s2 ) (0.200 s)  1.74 m/s . (c) Using v 2  v02  2 gy (with different values for y and v than before), we solve for the value of y corresponding to maximum height (where v = 0). y v02 (3.7 m/s)2   0.698 m. 2 g 2(9.8 m/s 2 ) Thus, the armadillo goes 0.698 – 0.544 = 0.154 m higher. 45. THINK As the ball travels vertically upward, its motion is under the influence of gravitational acceleration. The kinematics is one-dimensional. EXPRESS We neglect air resistance for the duration of the motion (between ―launching‖ and ―landing‖), so a = –g = –9.8 m/s2 (we take downward to be the –y direction). We use the equations in Table 2-1 (with y replacing x) because this is a = constant motion: v  v0  gt (2  11) 1 y  y0  v0t  gt 2 2 v 2  v02  2 g ( y  y0 ) (2  15) (2  16) We set y0 = 0. Upon reaching the maximum height y, the speed of the ball is momentarily zero (v = 0). Therefore, we can relate its initial speed v0 to y via the equation 0  v2  v02  2 gy. The time it takes for the ball to reach maximum height is given by v  v0  gt  0 , or t  v0 / g . Therefore, for the entire trip (from the time it leaves the ground until the time it returns to the ground), the total flight time is T  2t  2v0 / g. ANALYZE (a) At the highest point v = 0 and v0  2 gy . With y = 50 m, we find the initial speed of the ball to be v0  2 gy  2(9.8 m/s2 )(50 m)  31.3 m/s. (b) Using the result from (a) for v0, the total flight time of the ball is T 2v0 2(31.3 m/s)   6.39 s g 9.8 m/s2 47 (c) The plots of y, v and a as a function of time are shown below. The acceleration graph is a horizontal line at –9.8 m/s2. At t = 3.19 s, y = 50 m. LEARN In calculating the total flight time of the ball, we could have used Eq. 2-15. At t  T  0 , the ball returns to its original position ( y  0 ). Therefore, 2v 1 y  v0T  gT 2  0  T  0 2 g 46. Neglect of air resistance justifies setting a = –g = –9.8 m/s2 (where down is our –y direction) for the duration of the fall. This is constant acceleration motion, and we may use Table 2-1 (with y replacing x). (a) Using Eq. 2-16 and taking the negative root (since the final velocity is downward), we have v   v02  2 g y   0  2(9.8 m/s2 )(1700 m)  183 m/s . Its magnitude is therefore 183 m/s. (b) No, but it is hard to make a convincing case without more analysis. We estimate the mass of a raindrop to be about a gram or less, so that its mass and speed (from part (a)) would be less than that of a typical bullet, which is good news. But the fact that one is dealing with many raindrops leads us to suspect that this scenario poses an unhealthy situation. If we factor in air resistance, the final speed is smaller, of course, and we return to the relatively healthy situation with which we are familiar. 47. THINK The wrench is in free fall with an acceleration a = –g = –9.8 m/s2. EXPRESS We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the fall. This is constant acceleration motion, which justifies the use of Table 2-1 (with y replacing x): v  v0  gt (2  11) 1 y  y0  v0t  gt 2 2 (2  15) v 2  v02  2 g ( y  y0 ) (2  16) Since the wrench had an initial speed v0 = 0, knowing its speed of impact allows us to apply Eq. 2-16 to calculate the height from which it was dropped. CHAPTER 2 48 ANALYZE (a) Using v 2  v02  2ay , we find the initial height to be y  v02  v 2 0  (24 m/s)2   29.4 m. 2a 2(9.8 m/s 2 ) So that it fell through a height of 29.4 m. (b) Solving v = v0 – gt for time, we obtain a flight time of t v0  v 0  (24 m/s)   2.45 s. g 9.8 m/s 2 (c) SI units are used in the graphs, and the initial position is taken as the coordinate origin. The acceleration graph is a horizontal line at –9.8 m/s2. LEARN As the wrench falls, with a   g  0 , its speed increases but its velocity becomes more negative, as indicated by the second graph above. 48. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the fall. This is constant acceleration motion, which justifies the use of Table 2-1 (with y replacing x). (a) Noting that y = y – y0 = –30 m, we apply Eq. 2-15 and the quadratic formula (Appendix E) to compute t: v0  v02  2 gy 1 2 y  v0t  gt  t  2 g which (with v0 = –12 m/s since it is downward) leads, upon choosing the positive root (so that t > 0), to the result: t 12 m/s  (12 m/s)2  2(9.8 m/s 2 )(30 m)  1.54 s. 9.8 m/s 2 (b) Enough information is now known that any of the equations in Table 2-1 can be used to obtain v; however, the one equation that does not use our result from part (a) is Eq. 2-16: v  v02  2 gy  27.1 m / s 49 where the positive root has been chosen in order to give speed (which is the magnitude of the velocity vector). 49. THINK In this problem a package is dropped from a hot-air balloon which is ascending vertically upward. We analyze the motion of the package under the influence of gravity. EXPRESS We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the motion. This allows us to use Table 2-1 (with y replacing x): v  v0  gt (2  11) 1 y  y0  v0t  gt 2 2 (2  15) v 2  v02  2 g ( y  y0 ) (2  16) We place the coordinate origin on the ground and note that the initial velocity of the package is the same as the velocity of the balloon, v0 = +12 m/s and that its initial coordinate is y0 = +80 m. The time it takes for the package to hit the ground can be found by solving Eq. 2-15 with y = 0. ANALYZE (a) We solve 0  y  y0  v0t  12 gt 2 for time using the quadratic formula (choosing the positive root to yield a positive value for t): t v0  v02  2 gy0 g  12 m/s  (12 m/s) 2  2  9.8 m/s 2  80 m  9.8 m/s 2  5.45 s . (b) The speed of the package when it hits the ground can be calculated using Eq. 2-11. The result is v  v0  gt  12 m/s  (9.8 m/s2 )(5.447 s)  41.38 m/s . Its final speed is 41.38 m/s. LEARN Our answers can be readily verified by using Eq. 2-16 which was not used in either (a) or (b). The equation leads to v   v02  2 g ( y  y0 )   (12 m/s)2  2(9.8 m/s2 )(0  80 m)  41.38 m/s which agrees with that calculated in (b). 1 50. The y coordinate of Apple 1 obeys y – yo1 = – 2 g t2 where y = 0 when t = 2.0 s. This allows us to solve for yo1, and we find yo1 = 19.6 m. The graph for the coordinate of Apple 2 (which is thrown apparently at t = 1.0 s with CHAPTER 2 50 velocity v2) is 1 y – yo2 = v2(t – 1.0) – 2 g (t – 1.0)2 where yo2 = yo1 = 19.6 m and where y = 0 when t = 2.25 s. Thus, we obtain |v2| = 9.6 m/s, approximately. 51. (a) With upward chosen as the +y direction, we use Eq. 2-11 to find the initial velocity of the package: v = vo + at  0 = vo – (9.8 m/s2)(2.0 s) which leads to vo = 19.6 m/s. Now we use Eq. 2-15: 1 y = (19.6 m/s)(2.0 s) + 2 (–9.8 m/s2)(2.0 s)2  20 m . We note that the ―2.0 s‖ in this second computation refers to the time interval 2 < t < 4 in the graph (whereas the ―2.0 s‖ in the first computation referred to the 0 < t < 2 time interval shown in the graph). (b) In our computation for part (b), the time interval (―6.0 s‖) refers to the 2 < t 0 for times less than t = 2 s, then the spot had been moving rightward. 61 (e) As implied by our answer to part (c), it moves leftward for times immediately after t = 2 s. In fact, the expression found in part (a) guarantees that for all t > 2, v 2 s cannot be the right edge; it is the left edge (x = 0). Solving the expression given in the problem statement (with x = 0) for positive t yields the answer: the spot reaches the left edge at t = 12 s  3.46 s. 72. We adopt the convention frequently used in the text: that “up” is the positive y direction. (a) At the highest point in the trajectory v = 0. Thus, with t = 1.60 s, the equation v = v0 – gt yields v0 = 15.7 m/s. 1 (b) One equation that is not dependent on our result from part (a) is y – y0 = vt + 2gt2; this readily gives ymax – y0 = 12.5 m for the highest (“max”) point measured relative to where it started (the top of the building). 1 (c) Now we use our result from part (a) and plug into y y0 = v0t + 2gt2 with t = 6.00 s and y = 0 (the ground level). Thus, we have 1 0 – y0 = (15.68 m/s)(6.00 s) – 2 (9.8 m/s2)(6.00 s)2. Therefore, y0 (the height of the building) is equal to 82.3 m. 73. We denote the required time as t, assuming the light turns green when the clock reads zero. By this time, the distances traveled by the two vehicles must be the same. (a) Denoting the acceleration of the automobile as a and the (constant) speed of the truck as v then 1 x  at 2  vt truck 2 car which leads to F IJ bg G H K t Therefore, 2v 2  9.5 m/s    8.6 s . a 2.2 m/s 2 x  vt   9.5 m/s 8.6 s   82 m . (b) The speed of the car at that moment is vcar  at   2.2 m/s2  8.6 s   19 m/s . 74. If the plane (with velocity v) maintains its present course, and if the terrain continues its upward slope of 4.3°, then the plane will strike the ground after traveling CHAPTER 2 62 x  h 35 m   4655 . m  0.465 km. tan  tan 4.3 This corresponds to a time of flight found from Eq. 2-2 (with v = vavg since it is constant) x 0.465 km t   0.000358 h  1.3 s. v 1300 km / h This, then, estimates the time available to the pilot to make his correction. 75. We denote tr as the reaction time and tb as the braking time. The motion during tr is of the constant-velocity (call it v0) type. Then the position of the car is given by x  v0tr  v0tb  1 2 atb 2 where v0 is the initial velocity and a is the acceleration (which we expect to be negative-valued since we are taking the velocity in the positive direction and we know the car is decelerating). After the brakes are applied the velocity of the car is given by v = v0 + atb. Using this equation, with v = 0, we eliminate tb from the first equation and obtain v 2 1 v02  v02 x  v0tr  0   v0tr  . a 2 a 2 a We write this equation for each of the initial velocities: x1  v01tr  2 1 v01 , 2 a x2  v02tr  2 1 v02 . 2 a Solving these equations simultaneously for tr and a we get tr  2 2 v02 x1  v01 x2 v01v02 v02  v01 b g and a 2 2 1 v02 v01  v01v02 . 2 v02 x1  v01 x2 (a) Substituting x1 = 56.7 m, v01 = 80.5 km/h = 22.4 m/s, x2 = 24.4 m and v02 = 48.3 km/h = 13.4 m/s, we find tr  2 2 v02 x1  v01 x2 (13.4 m/s) 2 (56.7 m)  (22.4 m/s) 2 (24.4 m)  v01v02 (v02  v01 ) (22.4 m/s)(13.4 m/s)(13.4 m/s  22.4 m/s)  0.74 s. (b) Similarly, substituting x1 = 56.7 m, v01 = 80.5 km/h = 22.4 m/s, x2 = 24.4 m, and 63 v02 = 48.3 km/h = 13.4 m/s gives 2 2  v01v02 1 v02v01 1 (13.4 m/s)(22.4 m/s) 2  (22.4 m/s)(13.4 m/s) 2 a  2 v02 x1  v01 x2 2 (13.4 m/s)(56.7 m)  (22.4 m/s)(24.4 m)  6.2 m/s 2 . The magnitude of the deceleration is therefore 6.2 m/s2. Although rounded-off values are displayed in the above substitutions, what we have input into our calculators are the ―exact‖ values (such as v02  161 12 m/s). 76. (a) A constant velocity is equal to the ratio of displacement to elapsed time. Thus, for the vehicle to be traveling at a constant speed v p over a distance D23 , the time delay should be t  D23 / v p . (b) The time required for the car to accelerate from rest to a cruising speed v p is t0  v p / a . During this time interval, the distance traveled is x0  at02 / 2  v 2p / 2a. The car then moves at a constant speed v p over a distance D12  x0  d to reach intersection 2, and the time elapsed is t1  ( D12  x0  d ) / v p . Thus, the time delay at intersection 2 should be set to v p D12  (v 2p / 2a)  d D12  x0  d ttotal  tr  t0  t1  tr    tr   a vp a vp vp  tr  1 v p D12  d  2 a vp 77. THINK The speed of the rod changes due to a nonzero acceleration. EXPRESS Since the problem involves constant acceleration, the motion of the rod can be readily analyzed using the equations given in Table 2-1. We take +x to be in the direction of motion, so 1000 m / km v  60 km / h   16.7 m / s 3600 s / h b gF G H IJ K and a > 0. The location where the rod starts from rest (v0 = 0) is taken to be x0 = 0. ANALYZE (a) Using Eq. 2-7, we find the average acceleration to be aavg  v v  v0 16.7 m/s  0    3.09 m/s2 . t t  t0 5.4 s  0 (b) Assuming constant acceleration a  aavg  3.09 m/s2 , the total distance traveled during the 5.4-s time interval is CHAPTER 2 64 1 1 x  x0  v0t  at 2  0  0  (3.09 m/s 2 )(5.4 s)2  45 m 2 2 (c) Using Eq. 2-15, the time required to travel a distance of x = 250 m is: x 2  250 m  1 2 2x at  t    12.73 s 2 a 3.1 m/s 2 LEARN The displacement of the rod as a function of time can be written as 1 x(t )  (3.09 m/s 2 )t 2 . Note that we could have chosen Eq. 2-17 to solve for (b): 2 1 1 x   v0  v  t  16.7 m/s  5.4 s   45 m. 2 2 78. We take the moment of applying brakes to be t = 0. The deceleration is constant so that Table 2-1 can be used. Our primed variables (such as v0  72 km/h = 20 m/s ) refer to one train (moving in the +x direction and located at the origin when t = 0) and unprimed variables refer to the other (moving in the –x direction and located at x0 = +950 m when t = 0). We note that the acceleration vector of the unprimed train points in the positive direction, even though the train is slowing down; its initial velocity is v0 = –144 km/h = –40 m/s. Since the primed train has the lower initial speed, it should stop sooner than the other train would (were it not for the collision). Using Eq 2-16, it should stop (meaning v  0 ) at  v   v0   0  (20 m/s)2  200 m . x  2 2 2 a 2 m/s 2 The speed of the other train, when it reaches that location, is v  v02  2ax   40 m/s   2 1.0 m/s2   200 m  950 m  2  10 m/s using Eq 2-16 again. Specifically, its velocity at that moment would be –10 m/s since it is still traveling in the –x direction when it crashes. If the computation of v had failed (meaning that a negative number would have been inside the square root) then we would have looked at the possibility that there was no collision and examined how far apart they finally were. A concern that can be brought up is whether the primed train collides before it comes to rest; this can be studied by computing the time it stops (Eq. 2-11 yields t = 20 s) and seeing where the unprimed train is at that moment (Eq. 2-18 yields x = 350 m, still a good distance away from contact). 1 79. The y coordinate of Piton 1 obeys y – y01 = – 2 g t2 where y = 0 when t = 3.0 s. This allows us to solve for yo1, and we find y01 = 44.1 m. The graph for the coordinate of Piton 2 (which is thrown apparently at t = 1.0 s with velocity v1) is 65 1 y – y02 = v1(t–1.0) – 2 g (t – 1.0)2 where y02 = y01 + 10 = 54.1 m and where (again) y = 0 when t = 3.0 s. obtain |v1| = 17 m/s, approximately. Thus we 80. We take +x in the direction of motion. We use subscripts 1 and 2 for the data. Thus, v1 = +30 m/s, v2 = +50 m/s, and x2 – x1 = +160 m. (a) Using these subscripts, Eq. 2-16 leads to a v22  v12 (50 m/s) 2  (30 m/s) 2   5.0 m/s 2 . 2  x2  x1  2 160 m  (b) We find the time interval corresponding to the displacement x2 – x1 using Eq. 2-17: t2  t1  2  x2  x1  2 160 m    4.0 s . v1  v2 30 m/s  50 m/s (c) Since the train is at rest (v0 = 0) when the clock starts, we find the value of t1 from Eq. 2-11: 30 m/s v1  v0  at1  t1   6.0 s . 5.0 m/s2 (d) The coordinate origin is taken to be the location at which the train was initially at rest (so x0 = 0). Thus, we are asked to find the value of x1. Although any of several equations could be used, we choose Eq. 2-17: x1  1 1  v0  v1  t1   30 m/s  6.0 s   90 m . 2 2 (e) The graphs are shown below, with SI units understood. 81. THINK The particle undergoes a non-constant acceleration along the +x-axis. An integration is required to calculate velocity. EXPRESS With a non-constant acceleration a(t )  dv / dt , the velocity of the CHAPTER 2 66 t1 particle at time t1 is given by Eq. 2-27: v1  v0   a(t )dt , where v0 is the velocity at t0 time t0. In our situation, we have a  5.0t. In addition, we also know that v0  17 m/s at t0  2.0 s. ANALYZE Integrating (from t = 2 s to variable t = 4 s) the acceleration to get the velocity and using the values given in the problem, leads to t t 1 1 v  v0   adt  v0   (5.0t )dt  v0  (5.0)(t 2  t02 ) = 17 + 2 (5.0)(42 – 22) = 47 m/s. t0 t0 2 LEARN The velocity of the particle as a function of t is 1 1 v(t )  v0  (5.0)(t 2  t02 )  17  (5.0)(t 2  4)  7  2.5t 2 2 2 in SI units (m/s). Since the acceleration is linear in t, we expect the velocity to be quadratic in t, and the displacement to be cubic in t. 82. The velocity v at t = 6 (SI units and two significant figures understood) is 6 vgiven   adt . A quick way to implement this is to recall the area of a triangle ( 2 1 2 base × height). The result is v = 7 m/s + 32 m/s = 39 m/s. 83. The object, once it is dropped (v0 = 0) is in free fall (a = –g = –9.8 m/s2 if we take down as the –y direction), and we use Eq. 2-15 repeatedly. (a) The (positive) distance D from the lower dot to the mark corresponding to a certain reaction time t is given by y   D   21 gt 2 , or D = gt2/2. Thus, for t1  50.0 ms , c9.8 m / s hc50.0  10 sh 0.0123 m = 1.23 cm. D  3 2 1 2 2  9.8 m/s   100  10 s   0.049 m = 4D . (b) For t = 100 ms, D  3 2 2 2 2 1 2  9.8 m/s   150  10 s   0.11m = 9D . (c) For t = 150 ms, D  3 2 3 3 2 1 2  9.8 m/s   200  10 s   0.196 m =16D . (d) For t = 200 ms, D  3 2 4 4 2 1 2 c9.8 m / s hc250  10 sh 0.306 m = 25D . (e) For t = 250 ms, D  3 2 4 5 2 2 1 67 84. We take the direction of motion as +x, take x0 = 0 and use SI units, so v = 1600(1000/3600) = 444 m/s. (a) Equation 2-11 gives 444 = a(1.8) or a = 247 m/s2. We express this as a multiple of g by setting up a ratio:  247 m/s 2  a g  25 g. 2   9.8 m/s  (b) Equation 2-17 readily yields x 1 1  v0  v  t   444 m/s 1.8 s   400 m. 2 2 85. Let D be the distance up the hill. Then average speed = total distance traveled total time of travel = 2D D D + 20 km/h 35 km/h  25 km/h . 86. We obtain the velocity by integration of the acceleration: t v  v0   (6.1 1.2t )dt  . 0 Lengths are in meters and times are in seconds. The student is encouraged to look at the discussion in Section 2-7 to better understand the manipulations here. (a) The result of the above calculation is v  v0  6.1 t  0.6 t 2 , where the problem states that v0 = 2.7 m/s. The maximum of this function is found by knowing when its derivative (the acceleration) is zero (a = 0 when t = 6.1/1.2 = 5.1 s) and plugging that value of t into the velocity equation above. Thus, we find v  18 m/s . (b) We integrate again to find x as a function of t: t t 0 0 x  x0   v dt    (v0  6.1t   0.6 t 2 ) dt   v0t  3.05 t 2  0.2 t 3 . With x0 = 7.3 m, we obtain x = 83 m for t = 6. This is the correct answer, but one has the right to worry that it might not be; after all, the problem asks for the total distance traveled (and x  x0 is just the displacement). If the cyclist backtracked, then his total distance would be greater than his displacement. Thus, we might ask, “did he backtrack?” To do so would require that his velocity be (momentarily) zero at some point (as he reversed his direction of motion). We could solve the above quadratic equation for velocity, for a positive value of t where v = 0; if we did, we would find that at t = 10.6 s, a reversal does indeed happen. However, in the time interval we are concerned with in our problem (0 ≤ t ≤ 6 s), there is no reversal and the displacement is the same as the total distance traveled. 87. THINK In this problem we’re given two different speeds, and asked to find the difference in their travel times. CHAPTER 2 68 EXPRESS The time is takes to travel a distance d with a speed v1 is t1  d / v1 . Similarly, with a speed v2 the time would be t2  d / v2 . The two speeds in this problem are 1609 m/mi v1  55 mi/h  (55 mi/h)  24.58 m/s 3600 s/h v2  65 mi/h  (65 mi/h) 1609 m/mi  29.05 m/s 3600 s/h ANALYZE With d  700 km  7.0 105 m , the time difference between the two is 1   1 1 1 t  t1  t2  d     (7.0 105 m)     4383 s v v 24.58 m/s 29.05 m/s    1 2   73 min or about 1.2 h. LEARN The travel time was reduced from 7.9 h to 6.9 h. Driving at higher speed (within the legal limit) reduces travel time. 88. The acceleration is constant and we may use the equations in Table 2-1. (a) Taking the first point as coordinate origin and time to be zero when the car is there, we apply Eq. 2-17: 1 1 x   v  v0  t  15.0 m/s  v0   6.00 s  . 2 2 With x = 60.0 m (which takes the direction of motion as the +x direction) we solve for the initial velocity: v0 = 5.00 m/s. (b) Substituting v = 15.0 m/s, v0 = 5.00 m/s, and t = 6.00 s into a = (v – v0)/t (Eq. 2-11), we find a = 1.67 m/s2. (c) Substituting v = 0 in v 2  v02  2ax and solving for x, we obtain v02 (5.00 m/s)2 x      7.50m , 2a 2 1.67 m/s 2  or | x |  7.50 m . (d) The graphs require computing the time when v = 0, in which case, we use v = v0 + at’ = 0. Thus, v 5.00 m/s t  0    3.0s a 1.67 m/s2 indicates the moment the car was at rest. SI units are understood. 69 89. THINK In this problem we explore the connection between the maximum height an object reaches under the influence of gravity and the total amount of time it stays in air. EXPRESS Neglecting air resistance and setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the motion, we analyze the motion of the ball using Table 2-1 (with y replacing x). We set y0 = 0. Upon reaching the maximum height H, the speed of the ball is momentarily zero (v = 0). Therefore, we can relate its initial speed v0 to H via the equation 0  v2  v02  2 gH  v0  2 gH . The time it takes for the ball to reach maximum height is given by v  v0  gt  0 , or t  v0 / g  2H / g . ANALYZE If we want the ball to spend twice as much time in air as before, i.e., t   2t , then the new maximum height H  it must reach is such that t   2H  / g . Solving for H  we obtain 1 1 1  H   gt 2  g (2t )2  4  gt 2   4 H . 2 2 2  LEARN Since H : t 2 , doubling t means that H must increase fourfold. Note also that for t   2t , the initial speed must be twice the original speed: v0  2v0 . 90. (a) Using the fact that the area of a triangle is 12 (base) (height) (and the fact that the integral corresponds to the area under the curve) we find, from t = 0 through t = 5 s, the integral of v with respect to t is 15 m. Since we are told that x0 = 0 then we conclude that x = 15 m when t = 5.0 s. (b) We see directly from the graph that v = 2.0 m/s when t = 5.0 s. (c) Since a = dv/dt = slope of the graph, we find that the acceleration during the interval 4 < t 0. With y = 0 and y0 = h, this becomes t v02  2 gh  v0 . g (c) If it were thrown upward with that speed from height h then (in the absence of air friction) it would return to height h with that same downward speed and would therefore yield the same final speed (before hitting the ground) as in part (a). An important perspective related to this is treated later in the book (in the context of energy conservation). (d) Having to travel up before it starts its descent certainly requires more time than in part (b). The calculation is quite similar, however, except for now having +v0 in the equation where we had put in –v0 in part (b). The details follow: 1 y  v0t  gt 2 2 v0  v02  2 gy  t g CHAPTER 2 76 with the positive root again chosen to yield t > 0. With y = 0 and y0 = h, we obtain v02  2 gh  v0 . g 102. We assume constant velocity motion and use Eq. 2-2 (with vavg = v > 0). Therefore, t F km F1000 m / kmII c100  10 sh 8.4 m. G H3600 s / h J KJ H h G K x  vt  303 3 103. Assuming the horizontal velocity of the ball is constant, the horizontal displacement is x  vt , where x is the horizontal distance traveled, t is the time, and v is the (horizontal) velocity. Converting v to meters per second, we have 160 km/h = 44.4 m/s. Thus x 18.4 m t    0.414 s. v 44.4 m / s The velocity-unit conversion implemented above can be figured ―from basics‖ (1000 m = 1 km, 3600 s = 1 h) or found in Appendix D. 104. In this solution, we make use of the notation x(t) for the value of x at a particular t. Thus, x(t) = 50t + 10t2 with SI units (meters and seconds) understood. (a) The average velocity during the first 3 s is given by vavg  x(3)  x(0) (50)(3)  (10)(3) 2  0   80 m / s. t 3 (b) The instantaneous velocity at time t is given by v = dx/dt = 50 + 20t, in SI units. At t = 3.0 s, v = 50 + (20)(3.0) = 110 m/s. (c) The instantaneous acceleration at time t is given by a = dv/dt = 20 m/s2. It is constant, so the acceleration at any time is 20 m/s2. (d) and (e) The graphs that follow show the coordinate x and velocity v as functions of time, with SI units understood. The dashed line marked (a) in the first graph runs from t = 0, x = 0 to t = 3.0s, x = 240 m. Its slope is the average velocity during the first 3s of motion. The dashed line marked (b) is tangent to the x curve at t = 3.0 s. Its slope is the instantaneous velocity at t = 3.0 s. 77 105. We take +x in the direction of motion, so v0 = +30 m/s, v1 = +15 m/s and a < 0. The acceleration is found from Eq. 2-11: a = (v1 – v0)/t1 where t1 = 3.0 s. This gives a = –5.0 m/s2. The displacement (which in this situation is the same as the distance traveled) to the point it stops (v2 = 0) is, using Eq. 2-16, v22  v02  2ax  x   (30 m/s)2  90 m. 2(5 m/s2 ) 106. The problem consists of two constant-acceleration parts: part 1 with v0 = 0, v = 6.0 m/s, x = 1.8 m, and x0 = 0 (if we take its original position to be the coordinate origin); and, part 2 with v0 = 6.0 m/s, v = 0, and a2 = –2.5 m/s2 (negative because we are taking the positive direction to be the direction of motion). (a) We can use Eq. 2-17 to find the time for the first part 1 x – x0 = 2(v0 + v) t1 1  1.8 m – 0 = 2(0 + 6.0 m/s) t1 so that t1 = 0.6 s. And Eq. 2-11 is used to obtain the time for the second part v  v0  a2t2  0 = 6.0 m/s + (–2.5 m/s2)t2 from which t2 = 2.4 s is computed. Thus, the total time is t1 + t2 = 3.0 s. (b) We already know the distance for part 1. We could find the distance for part 2 from several of the equations, but the one that makes no use of our part (a) results is Eq. 2-16 v 2  v02  2a2 x2  0 = (6.0 m/s)2 + 2(–2.5 m/s2)x2 which leads to x2 = 7.2 m. Therefore, the total distance traveled by the shuffleboard disk is (1.8 + 7.2) m = 9.0 m. 107. The time required is found from Eq. 2-11 (or, suitably interpreted, Eq. 2-7). First, we convert the velocity change to SI units: v  (100 km / h) 1000 m / km I F G H3600 s / h J K 27.8 m / s . Thus, t = v/a = 27.8/50 = 0.556 s. 108. From Table 2-1, v 2  v02  2ax is used to solve for a. Its minimum value is amin  v2  v02 (360 km / h) 2   36000 km / h 2 2 xmax 2(180 . km) which converts to 2.78 m/s2. CHAPTER 2 78 109. (a) For the automobile v = 55 – 25 = 30 km/h, which we convert to SI units: b g 1000 m/ km v (30 km / h) 3600 s/ h a   0.28 m / s2 . t (0.50 min)(60 s / min) (b) The change of velocity for the bicycle, for the same time, is identical to that of the car, so its acceleration is also 0.28 m/s2. 110. Converting to SI units, we have v = 3400(1000/3600) = 944 m/s (presumed constant) and t = 0.10 s. Thus, x = vt = 94 m. 111. This problem consists of two parts: part 1 with constant acceleration (so that the equations in Table 2-1 apply), v0 = 0, v = 11.0 m/s, x = 12.0 m, and x0 = 0 (adopting the starting line as the coordinate origin); and, part 2 with constant velocity (so that x – x0 = vt applies) with v = 11.0 m/s, x0 = 12.0, and x = 100.0 m. (a) We obtain the time for part 1 from Eq. 2-17 x  x0  b g b g 1 1 v0  v t1  12.0  0  0  110 . t1 2 2 so that t1 = 2.2 s, and we find the time for part 2 simply from 88.0 = (11.0)t2  t2 = 8.0 s. Therefore, the total time is t1 + t2 = 10.2 s. (b) Here, the total time is required to be 10.0 s, and we are to locate the point xp where the runner switches from accelerating to proceeding at constant speed. The equations for parts 1 and 2, used above, therefore become x p  0  12  0  11.0 m/s  t1 100.0 m  x p  11.0 m/s 10.0 s  t1  where in the latter equation, we use the fact that t2 = 10.0 – t1. Solving the equations for the two unknowns, we find that t1 = 1.8 s and xp = 10.0 m. 112. The bullet starts at rest (v0 = 0) and after traveling the length of the barrel ( x  1.2 m ) emerges with the given velocity (v = 640 m/s), where the direction of motion is the positive direction. Turning to the constant acceleration equations in Table 2-1, we use x  12 (v0  v) t . Thus, we find t = 0.00375 s (or 3.75 ms). 113. There is no air resistance, which makes it quite accurate to set a = –g = –9.8 m/s2 (where downward is the –y direction) for the duration of the fall. We are allowed to use Table 2-1 (with y replacing x) because this is constant acceleration motion; in fact, when the acceleration changes (during the process of catching the ball) we will again assume constant acceleration conditions; in this case, we have a2 = +25g = 245 m/s2. (a) The time of fall is given by Eq. 2-15 with v0 = 0 and y = 0. Thus, 79 t 2 y0 2(145 m)   5.44 s. g 9.8 m/s2 (b) The final velocity for its free-fall (which becomes the initial velocity during the catching process) is found from Eq. 2-16 (other equations can be used but they would use the result from part (a)) v   v02  2 g( y  y0 )   2 gy0  53.3 m / s where the negative root is chosen since this is a downward velocity. Thus, the speed is | v | 53.3 m/s. (c) For the catching process, the answer to part (b) plays the role of an initial velocity (v0 = –53.3 m/s) and the final velocity must become zero. Using Eq. 2-16, we find v 2  v02 (53.3 m/s)2 y2    5.80 m , 2a2 2(245 m/s 2 ) or | y2 |  5.80 m. The negative value of y2 signifies that the distance traveled while arresting its motion is downward. 114. During Tr the velocity v0 is constant (in the direction we choose as +x) and obeys v0 = Dr/Tr where we note that in SI units the velocity is v0 = 200(1000/3600) = 55.6 m/s. During Tb the acceleration is opposite to the direction of v0 (hence, for us, a Tr, most of the full time required to stop is spent in braking. (f) We are only asked what the increase in distance D is, due to Tr = 0.100 s, so we simply have D  v0 Tr   55.6 m/s  0.100 s   5.56 m . 115. The total time elapsed is t  2 h 41 min  161 min and the center point is displaced by x  3.70 m  370 cm. Thus, the average velocity of the center point is vavg  x 370 cm   2.30 cm/min. t 161 min 116. Using Eq. 2-11, v  v0  at , we find the initial speed to be v0  v  at  0  (3400)(9.8 m/s2 )(6.5 103 s)  216.6 m/s 117. The total number of days walked is (including the first and the last day, and leap year) N  340  365  365  366  365  365  261  2427 Thus, the average speed of the walk is savg  d 3.06 107 m   0.146 m/s. t (2427 days)(86400 s/day) 118. (a) Let d be the distance traveled. The average speed with and without wings set as sails are vs  d / ts and vns  d / tns , respectively. Thus, the ratio of the two speeds is vs d / ts tns 25.0 s     3.52 vns d / tns ts 7.1s (b) The difference in time expressed in terms of vs is t  tns  ts  d d d d d (2.0 m) 5.04 m     2.52  2.52  vns vs (vs / 3.52) vs vs vs vs 119. (a) Differentiating y(t )  (2.0 cm)sin( t / 4) with respect to t, we obtain v y (t )  dy      cm/s  cos( t / 4) dt  2  The average velocity between t = 0 and t = 2.0 s is 81 2 1 1   2  t  v dt  cm/s y   0 cos   dt  0 (2.0 s) (2.0 s)  2   4  /2 1   2 cm  0 cos x dx  1.0 cm/s (2.0 s) vavg  (b) The instantaneous velocities of the particle at t = 0, 1.0 s, and 2.0 s are, respectively,    v y (0)   cm/s  cos(0)  cm/s 2 2   2   v y (1.0 s)   cm/s  cos( / 4)  cm/s 4 2    v y (2.0 s)   cm/s  cos( / 2)  0 2  (c) Differentiating v y (t ) with respect to t, we obtain the following expression for acceleration: dv y   2  a y (t )    cm/s 2  sin( t / 4) dt  8  The average acceleration between t = 0 and t = 2.0 s is 2  2  t  1 1  2 a dt   cm/s 2   sin   dt  y  (2.0 s) 0 (2.0 s)  8  4  0  / 2 1   1      2    cm/s  0 sin x dx    cm/s    cm/s (2.0 s)  2 (2.0 s)  2 4   aavg  (d) The instantaneous accelerations of the particle at t = 0, 1.0 s, and 2.0 s are, respectively,  2  a y (0)    cm/s 2  sin(0)  0  8  2    2 2 a y (1.0 s)    cm/s 2  sin( / 4)   cm/s 2 8 16   2   2 2 a y (2.0 s)    cm/s  sin( / 2)   cm/s 2 8  8 