Solution Manual For Fundamentals Of Physics Extended, 9th 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. Since the trip consists of two parts, let the displacements during first and second parts of the motion be Δx1 and Δx2, and the corresponding time intervals be Δt1 and Δt2, respectively. Now, because the problem is one-dimensional and both displacements are in the same direction, the total displacement is Δx = Δx1 + Δx2, and the total time for the trip is Δt = Δt1 + Δt2. Using the definition of average velocity given in Eq. 2-2, we have Δx Δx1 + Δx2 = . vavg = Δt Δt1 + Δt2 To find the average speed, we note that during a time Δt 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 = | Δx | = vΔt . 21 22 CHAPTER 2 (a) During the first part of the motion, the displacement is Δx1 = 40 km and the time interval is (40 km) t1 = = 133 . h. (30 km / h) Similarly, during the second part the displacement is Δx2 = 40 km and the time interval is (40 km) t2 = = 0.67 h. (60 km / h) The total displacement is Δx = Δx1 + Δx2 = 40 km + 40 km = 80 km, and the total time elapsed is Δt = Δt1 + Δt2 = 2.00 h. Consequently, the average velocity is vavg = Δx (80 km) = = 40 km/h. Δt (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 is shown below; it consists of two contiguous line segments, the first having a slope of 30 km/h and connecting the origin to (Δt1, Δx1) = (1.33 h, 40 km) and the second having a slope of 60 km/h and connecting (Δt1, Δx1) to (Δt, Δx) = (2.00 h, 80 km). 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 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. Using x = 3t – 4t2 + t3 with SI units understood is efficient (and is the approach we 23 will use), but if we wished to make the units explicit we would write x = (3 m/s)t – (4 m/s2)t2 + (1 m/s3)t3. We will quote our answers to one or two significant figures, and not try to follow the significant figure rules rigorously. (a) Plugging in t = 1 s yields x = 3 – 4 + 1 = 0. (b) With t = 2 s we get x = 3(2) – 4(2)2+(2)3 = –2 m. (c) With t = 3 s we have x = 0 m. (d) Plugging in t = 4 s gives x = 12 m. For later reference, we also note that the position at t = 0 is x = 0. (e) The position at t = 0 is subtracted from the position at t = 4 s to find the displacement Δx = 12 m. (f) The position at t = 2 s is subtracted from the position at t = 4 s to give the displacement Δx = 14 m. Eq. 2-2, then, leads to vavg = Δx 14 m = = 7 m/s. Δt 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). 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 24 CHAPTER 2 Δt = Δ x 200 m = = 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 Δt = 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 Δt L / vs L 1.75 m (b) The amount of time required to reach a depth of D = 5.0 m is t= 5.0 m D = = 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 ⇒ L1 L2 = . t1 t2 From this we obtain ⎛t ⎞ ⎛ 148.15 ⎞ L2 − L1 = ⎜ 2 − 1 ⎟ L1 = ⎜ − 1 ⎟ L1 = 0.00135 L1 ≈ 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 25 vc + vw = d t1 and vc − vw = d 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 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 (96.0 m) = 24.0 m, d= 25.0 m/s − 5.00 m/s v − vs 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 Δx = 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 = Δx 12.0 m = = 2.50 m/s. t 4.80 s 26 CHAPTER 2 (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 savg = 2D D D 72.5 km/h + 68.3 km/h = 70 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 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: 27 14. Using the general property v= d dx exp(bx ) = b exp(bx ) , we write FG H IJ K FG IJ . 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. (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. (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: b g 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. 28 CHAPTER 2 (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 vavg = Δx 50.25 cm − 2175 . cm = Δt 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. (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. 29 (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. (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. 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: 30 CHAPTER 2 Δv v(t2 ) − v(t1 ) = . t2 − t1 Δt Let v1 = +18 m/s at t1 = 0 and v2 = –30 m/s at t2 = 2.4 s. Using Eq. 2-7 we find aavg = aavg = v(t2 ) − v(t1 ) (−30 m/s) − (+1 m/s) = = − 20 m/s 2 . t2 − t1 2.4 s − 0 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. 20. We use the functional notation x(t), v(t) and a(t) and find the latter two quantities by differentiating: b g dxtbt g = − 15t + 20 and abt g = dvdtbt g = − 30t vt = 2 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 . s. particle is (momentarily) stopped is t = 20 / 15 = 12 (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. 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 31 direction of motion during 5 min ≤ t ≤ 10 min is taken to be the positive x direction. We also use the fact that Δx = vΔt ' when the velocity is constant during a time interval Δt' . (a) The entire interval considered is Δt = 8 – 2 = 6 min, which is equivalent to 360 s, whereas the sub-interval in which he is moving is only Δt' = 8 − 5 = 3min = 180 s. His position at t = 2 min is x = 0 and his position at t = 8 min is x = vΔ 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 Δt = 9 – 3 = 6 min (360 s again), whereas the sub-interval in which he is moving is Δ t ′ = 9 − 5 = 4 min = 240 s ). His position at t = 3 min is x = 0 and his position at t = 9 min is x = vΔ t ′ = (2.2)(240) = 528 m . Therefore, 528 m − 0 . m / s. vavg = = 147 360 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) = (3.0 m / s 2 )(4.0 s) 2 − (2.0 m / s 3 )(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 Δx = x2 – x1, where x1 = 0 and x2 = –80 m. Thus, Δx = −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/s 2 )(1.0 s) − (6.0 m/s3 )(1.0 s) 2 = 0. (g) Similarly, v(2 s) = (6.0 m/s 2 )(2.0 s) − (6.0 m/s3 )(2.0 s) 2 = − 12m/s . (h) v(3 s) = (6.0 m/s 2 )(3.0 s) − (6.0 m/s3 )(3.0 s)2 = − 36 m/s . (i) v(4 s) = (6.0 m/s 2 )(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/s 2 − (12.0 m/s3 )(1.0 s) = − 6.0 m/s 2 . (k) a (2 s) = 6.0 m/s 2 − (12.0 m/s3 )(2.0 s) = − 18 m/s 2 . 33 (l) a (3 s) = 6.0 m/s 2 − (12.0 m/s3 )(3.0 s) = −30 m/s 2 . (m) a (4 s) = 6.0 m/s 2 − (12.0 m/s3 )(4.0 s) = − 42 m/s 2 . 23. Since the problem involves constant acceleration, the motion of the electron 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) The acceleration can be found by solving Eq. (2-16). 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/s 2 . 2x 2(0.010 m) 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 Δx = 5.0 μ m, the acceleration of the spores during the launch is v 2 − v02 (1.6 m/s) 2 = = 2.56 × 105 m/s 2 = 2.6 × 104 g a= −6 2x 2(5.0 × 10 m) (b) During the speed-reduction stage, the acceleration is v 2 − v02 0 − (1.6 m/s) 2 a= = = −1.28 × 103 m/s 2 = −1.3 × 102 g −3 2x 2(1.0 ×10 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 34 CHAPTER 2 and find x= v 2 − v02 (20 m/s) 2 − (0) 2 = = 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 v 2 = v02 + 2a ( x − x0 ) , we find x= − 1 v02 1 (5.00 × 106 ) 2 =− = 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 x = v0 t + 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 c h v = (9.6 m / s) + 3.2 m / s2 ( −2.5 s) = 16 . m / s. 35 (b) Now, t = +2.5 s and we find c h v = (9.6 m / s) + 3.2 m / s2 (2.5 s) = 18 m / s. 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. 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 Δx as the distance moved during t1, and use Eq. 2-16: (5.08 m/s) 2 = 10.59 m ≈ 10.6 m. v = v + 2a1Δx ⇒ Δx = 2(1.22 m/s 2 ) 2 2 0 (b) Using Eq. 2-11, we have t1 = 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 36 CHAPTER 2 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/s2 ) 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. The constant acceleration stated in the problem permits the use of the equations in Table 2-1. (a) We solve v = v0 + at for the time: t= v − v0 101 (3.0 × 10 8 m / s) = 31 = . × 10 6 s 9.8 m / s 2 a which is equivalent to 1.2 months. (b) We evaluate x = x0 + v0 t + 12 at 2 , with x0 = 0. The result is x= 1 9.8 m/s 2 ) (3.1×106 s) 2 = 4.6 ×1013 m . ( 2 Note that in solving parts (a) and (b), we did not use the equation v 2 = v02 + 2a( x − x0 ) . This equation can be employed for consistency check. The final velocity based on this 37 equation is v = v02 + 2a ( x − x0 ) = 0 + 2(9.8 m/s 2 )(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 / kmIJ 1020 km / hg G b H 3600 s / h K Δv Δt = . s 14 = 202.4 m / s 2 . 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. The problem statement (see part (a)) indicates that a = constant, which allows us to use Table 2-1. (a) We take x0 = 0, and solve x = v0t + 21 at2 (Eq. 2-15) for the acceleration: a = 2(x – v0t)/t2. Substituting x = 24.0 m, v0 = 56.0 km/h = 15.55 m/s and t = 2.00 s, we find a= 2( x − v0t ) 2 ( 24.0m − ( 15.55m/s ) ( 2.00s ) ) = = − 3.56m/s 2 , 2 t2 ( 2.00s ) or | a | = 3.56 m/s 2 . The negative sign indicates that the acceleration is opposite to the direction of motion of the car. The car is slowing down. (b) We evaluate v = v0 + at as follows: c v = 1555 . m / s − 356 . m / s2 h b2.00 sg = 8.43 m / s which can also be converted to 30.3 km/h. 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 38 CHAPTER 2 We simultaneously solve these equations and obtain the following results: (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/s 2 . 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 Δx1 = 1 where a2 = −0.75 m/s2 and Δx2 = Δx1 = vo1 t1 + 2 a1 t12 Δx2 = v2 t2 − 2 a2 t22 900 4 m 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 + 2a1Δx1 = 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: 39 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 / s2 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. b g (a) Equation 2-16 leads to v12 = v02 + 2a1 x1 ⇒ v1 = 2 (1.34 m/s 2 ) ( 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 41 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. (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 (Δx) 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: 1 xpolice – 10.6 = 18.56(t − t0) – 2 (5)(t − t0)2 1 xyou = 30.56(t − t0) – 2 (5)(t − t0)2 . 42 CHAPTER 2 Subtracting equations, we find 10.6 = (30.56 – 18.56)(t − t0) ⇒ 0.883 s = t − t0. 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 vA (which is also the final velocity of the train, if the rear-end collision is barely avoided). We note that the distance Δx consists of the original gap between them, D, as well as the forward distance traveled during this time by the locomotive vA t . Therefore, vt + v A Δ x D + vA t D = = = + vA . 2 t t t We now use Eq. 2-11 to eliminate time from the equation. Thus, vt + v A D = + vA v A − vt / a 2 b g which leads to F v + v − v IJ FG v − v IJ = − 1 bv − v g . a=G H 2 K H D K 2D Hence, 1 FG 29 km − 161 kmIJ = −12888 km / h a=− h h K 2(0.676 km) H t A A A 2 t A t 2 2 which we convert as follows: F 1000 mIJ FG 1 h IJ = −0.994 m / s a = c −12888 km / h h G H 1 km K H 3600 sK 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 Δy replacing Δx) because this is constant acceleration motion. The ground level 43 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 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/s 2 ) (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. In this problem a ball is being thrown vertically upward. Its subsequent motion is under the influence of gravity. 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 Δy replacing Δx) 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 = v 2 = 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 . (a) At the highest point v = 0 and v0 = 2 gy . Since y = 50 m we find v0 = 2 gy = 2(9.8 m/s 2 )(50 m) = 31.3 m/s. (b) Using the result from (a) for v0, we find the total flight time to be 44 CHAPTER 2 T= 2v0 2(31.3 m/s) = = 6.39 s ≈ 6.4 s . g 9.8 m/s 2 (c) SI units are understood in the x and v graphs shown. The acceleration graph is a horizontal line at –9.8 m/s2. 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 Δy replacing Δx). (a) Using Eq. 2-16 and taking the negative root (since the final velocity is downward), we have v = − v02 − 2 g Δy = − 0 − 2(9.8 m/s 2 )(−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. 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 Δy replacing Δx). (a) Starting the clock at the moment the wrench is dropped (v0 = 0), then v 2 = v02 − 2 g Δy leads to Δy = − (−24 m/s) 2 = −29.4 m 2(9.8 m/s 2 ) so that it fell through a height of 29.4 m. 45 (b) Solving v = v0 – gt for time, we find: 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. As the wrench falls, with a = − g 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 gΔy = 27.1 m / s where the positive root has been chosen in order to give speed (which is the magnitude of the velocity vector). 49. 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 Δy replacing Δx) because this is constant acceleration motion. We are placing the coordinate origin on the ground. We note that the initial velocity of the package is 46 CHAPTER 2 the same as the velocity of the balloon, v0 = +12 m/s, and that its initial coordinate is y0 = +80 m. (a) We solve y = y0 + v0t − 21 gt 2 for time, with y = 0, 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 = 5.4 s (b) If we wish to avoid using the result from part (a), we could use Eq. 2-16, but if that is not a concern, then a variety of formulas from Table 2-1 can be used. For instance, Eq. 2-11 leads to v = v0 − gt = 12 m/s − (9.8 m/s 2 )(5.447 s) = −41.38 m/s Its final speed is about 41 m/s. 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 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 Δy = (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 < 8 portion of the graph: 1 Δy = (19.6 m/s)(6.0 s) + 2 (–9.8 m/s2)(6.0 s)2 ≈ –59 m , or | Δy |= 59 m . 47 52. The full extent of the bolt’s fall is given by 1 y – y0 = –2 g t2 where y – y0 = –90 m (if upward is chosen as the positive y direction). Thus the time for the full fall is found to be t = 4.29 s. The first 80% of its free-fall distance is given by –72 = –g τ2/2, which requires time τ = 3.83 s. (a) Thus, the final 20% of its fall takes t – τ = 0.45 s. (b) We can find that speed using v = −gτ. Therefore, |v| = 38 m/s, approximately. (c) Similarly, vfinal = − g t ⇒ |vfinal| = 42 m/s. 53. The speed of the boat is constant, given by vb = d/t. Here, d is the distance of the boat from the bridge when the key is dropped (12 m) and t is the time the key takes in falling. To calculate t, we put the origin of the coordinate system at the point where the key is dropped and take the y axis to be positive in the downward direction. Taking the time to be zero at the instant the key is dropped, we compute the time t when y = 45 m. Since the initial velocity of the key is zero, the coordinate of the key is given by y = 12 gt 2 . Thus, t= 2y 2(45 m) = = 3.03 s . g 9.8 m / s2 Therefore, the speed of the boat is vb = 12 m = 4.0 m / s . 3.03 s 54. (a) 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 Eq. 2-15 (with Δy replacing Δx) because this is constant acceleration motion. We use primed variables (except t) with the first stone, which has zero initial velocity, and unprimed variables with the second stone (with initial downward velocity –v0, so that v0 is being used for the initial speed). SI units are used throughout. Δy′ = 0 ( t ) − 1 2 gt 2 Δy = ( −v0 )( t − 1) − 1 2 g ( t − 1) 2 Since the problem indicates Δy’ = Δy = –43.9 m, we solve the first equation for t (finding t = 2.99 s) and use this result to solve the second equation for the initial speed of the second stone: 48 CHAPTER 2 −43.9 m = ( −v0 ) (1.99 s ) − 1 2 9.8 m/s 2 ) (1.99 s ) ( 2 which leads to v0 = 12.3 m/s. (b) The velocity of the stones are given by v′y = d (Δy′) = − gt , dt vy = d ( Δy ) = −v0 − g (t − 1) dt The plot is shown below: 55. During contact with the ground its average acceleration is given by Δv aavg = Δt where Δv is the change in its velocity during contact with the ground and Δt = 20.0 ×10−3 s is the duration of contact. Thus, we must first find the velocity of the ball just before it hits the ground (y = 0). (a) Now, to find the velocity just before contact, we take t = 0 to be when it is dropped. Using Eq. (2-16) with y0 = 15.0 m , we obtain v = − v02 − 2 g ( y − y0 ) = − 0 − 2(9.8 m/s 2 )(0 − 15 m) = −17.15 m/s where the negative sign is chosen since the ball is traveling downward at the moment of contact. Consequently, the average acceleration during contact with the ground is aavg = Δv 0 − (−17.1 m/s) = = 857 m/s 2 . −3 Δt 20.0 × 10 s 49 (b) The fact that the result is positive indicates that this acceleration vector points upward. In a later chapter, this will be directly related to the magnitude and direction of the force exerted by the ground on the ball during the collision. 56. We use Eq. 2-16, vB2 = vA2 + 2a(yB – yA), 1 with a = –9.8 m/s2, yB – yA = 0.40 m, and vB = 3 vA. It is then straightforward to solve: vA = 3.0 m/s, approximately. 57. The average acceleration during contact with the floor is aavg = (v2 – v1) / Δt, where v1 is its velocity just before striking the floor, v2 is its velocity just as it leaves the floor, and Δt is the duration of contact with the floor (12 × 10–3 s). (a) Taking the y axis to be positively upward and placing the origin at the point where the ball is dropped, we first find the velocity just before striking the floor, using v12 = v02 − 2 gy . With v0 = 0 and y = – 4.00 m, the result is v1 = − −2 gy = − −2(9.8 m/s 2 ) (−4.00 m) = −8.85 m/s where the negative root is chosen because the ball is traveling downward. To find the velocity just after hitting the floor (as it ascends without air friction to a height of 2.00 m), we use v 2 = v22 − 2 g ( y − y0 ) with v = 0, y = –2.00 m (it ends up two meters below its initial drop height), and y0 = – 4.00 m. Therefore, v2 = 2 g ( y − y0 ) = 2(9.8 m/s 2 ) (−2.00 m + 4.00 m) = 6.26 m/s . Consequently, the average acceleration is aavg = v2 − v1 6.26 m/s − (− 8.85 m/s) = = 1.26 × 103 m/s 2 . Δt 12.0 × 10−3 s (b) The positive nature of the result indicates that the acceleration vector points upward. In a later chapter, this will be directly related to the magnitude and direction of the force exerted by the ground on the ball during the collision. 58. We choose down as the +y direction and set the coordinate origin at the point where it was dropped (which is when we start the clock). We denote the 1.00 s duration mentioned in the problem as t – t' where t is the value of time when it lands and t' is one second prior to that. The corresponding distance is y – y' = 0.50h, where y denotes the location of the ground. In these terms, y is the same as h, so we have h –y' = 0.50h or 0.50h = y' . (a) We find t' and t from Eq. 2-15 (with v0 = 0): 50 CHAPTER 2 1 2 y′ y′ = gt ′2 ⇒ t ′ = g 2 y= 1 2 2y . gt ⇒ t = g 2 Plugging in y = h and y' = 0.50h, and dividing these two equations, we obtain b g 2 0.50h / g t′ = = 0.50 . 2h / g t Letting t' = t – 1.00 (SI units understood) and cross-multiplying, we find t − 100 . = t 0.50 ⇒ t = 100 . 1 − 0.50 which yields t = 3.41 s. (b) Plugging this result into y = 12 gt 2 we find h = 57 m. (c) In our approach, we did not use the quadratic formula, but we did “choose a root” when we assumed (in the last calculation in part (a)) that 0.50 = +0.707 instead of –0.707. If we had instead let 0.50 = –0.707 then our answer for t would have been roughly 0.6 s, which would imply that t' = t – 1 would equal a negative number (indicating a time before it was dropped), which certainly does not fit with the physical situation described in the problem. 59. 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 Δy replacing Δx) because this is constant acceleration motion. The ground level is taken to correspond to the origin of the y-axis. (a) The time drop 1 leaves the nozzle is taken as t = 0 and its time of landing on the floor t1 can be computed from Eq. 2-15, with v0 = 0 and y1 = –2.00 m. 1 −2 y −2(−2.00 m) y1 = − gt12 ⇒ t1 = = = 0.639 s . 2 g 9.8 m/s 2 At that moment, the fourth drop begins to fall, and from the regularity of the dripping we conclude that drop 2 leaves the nozzle at t = 0.639/3 = 0.213 s and drop 3 leaves the nozzle at t = 2(0.213 s) = 0.426 s. Therefore, the time in free fall (up to the moment drop 1 lands) for drop 2 is t2 = t1 – 0.213 s = 0.426 s. Its position at the moment drop 1 strikes the floor is 1 1 y2 = − gt22 = − (9.8 m/s 2 )(0.426 s) 2 = −0.889 m, 2 2 or about 89 cm below the nozzle. 51 (b) The time in free fall (up to the moment drop 1 lands) for drop 3 is t3 = t1 –0.426 s = 0.213 s. Its position at the moment drop 1 strikes the floor is 1 1 y3 = − gt32 = − (9.8 m/s 2 )(0.213 s) 2 = −0.222 m, 2 2 or about 22 cm below the nozzle. 60. To find the “launch” velocity of the rock, we apply Eq. 2-11 to the maximum height (where the speed is momentarily zero) v = v0 − gt ⇒ 0 = v0 − ( 9.8 m/s 2 ) ( 2.5 s ) so that v0 = 24.5 m/s (with +y up). Now we use Eq. 2-15 to find the height of the tower (taking y0 = 0 at the ground level) y − y0 = v0t + 1 2 1 2 at ⇒ y − 0 = ( 24.5 m/s )(1.5 s ) − ( 9.8 m/s 2 ) (1.5 s ) . 2 2 Thus, we obtain y = 26 m. 61. We choose down as the +y direction and place the coordinate origin at the top of the building (which has height H). During its fall, the ball passes (with velocity v1) the top of the window (which is at y1) at time t1, and passes the bottom (which is at y2) at time t2. We are told y2 – y1 = 1.20 m and t2 – t1 = 0.125 s. Using Eq. 2-15 we have b g 21 g bt − t g y2 − y1 = v1 t2 − t1 + 2 2 1 which immediately yields v1 = 1.20 m − 12 ( 9.8 m/s 2 ) ( 0.125 s ) 2 0.125 s = 8.99 m/s. From this, Eq. 2-16 (with v0 = 0) reveals the value of y1: v12 = 2 gy1 ⇒ y1 = (8.99 m/s) 2 = 4.12 m. 2(9.8 m/s 2 ) It reaches the ground (y3 = H) at t3. Because of the symmetry expressed in the problem (“upward flight is a reverse of the fall’’) we know that t3 – t2 = 2.00/2 = 1.00 s. And this means t3 – t1 = 1.00 s + 0.125 s = 1.125 s. Now Eq. 2-15 produces 1 y3 − y1 = v1 (t3 − t1 ) + g (t3 − t1 ) 2 2 1 y3 − 4.12 m = (8.99 m/s) (1.125 s) + (9.8 m/s 2 ) (1.125 s) 2 2 52 CHAPTER 2 which yields y3 = H = 20.4 m. 62. The height reached by the player is y = 0.76 m (where we have taken the origin of the y axis at the floor and +y to be upward). (a) The initial velocity v0 of the player is v0 = 2 gy = 2(9.8 m/s 2 ) (0.76 m) = 3.86 m/s . This is a consequence of Eq. 2-16 where velocity v vanishes. As the player reaches y1 = 0.76 m – 0.15 m = 0.61 m, his speed v1 satisfies v02 − v12 = 2 gy1 , which yields v1 = v02 − 2 gy1 = (3.86 m/s) 2 − 2(9.80 m/s 2 ) (0.61 m) = 1.71 m/s . The time t1 that the player spends ascending in the top Δy1 = 0.15 m of the jump can now be found from Eq. 2-17: Δy1 = 2 ( 0.15 m ) 1 = 0.175 s ( v1 + v ) t1 ⇒ t1 = 2 1.71 m/s + 0 which means that the total time spent in that top 15 cm (both ascending and descending) is 2(0.175 s) = 0.35 s = 350 ms. (b) The time t2 when the player reaches a height of 0.15 m is found from Eq. 2-15: 0.15 m = v0t2 − 1 2 1 gt2 = (3.86 m/s)t2 − (9.8 m/s 2 )t22 , 2 2 which yields (using the quadratic formula, taking the smaller of the two positive roots) t2 = 0.041 s = 41 ms, which implies that the total time spent in that bottom 15 cm (both ascending and descending) is 2(41 ms) = 82 ms. 63. The time t the pot spends passing in front of the window of length L = 2.0 m is 0.25 s each way. We use v for its velocity as it passes the top of the window (going up). Then, with a = –g = –9.8 m/s2 (taking down to be the –y direction), Eq. 2-18 yields L 1 1 L = vt − gt 2 ⇒ v = − gt . t 2 2 The distance H the pot goes above the top of the window is therefore (using Eq. 2-16 with the final velocity being zero to indicate the highest point) ( 2.00 m / 0.25 s − (9.80 m/s )(0.25 s) / 2 ) = 2.34 m. v 2 ( L / t − gt / 2 ) H= = = 2g 2g 2(9.80 m/s 2 ) 2 2 2 53 64. The graph shows y = 25 m to be the highest point (where the speed momentarily vanishes). The neglect of “air friction” (or whatever passes for that on the distant planet) is certainly reasonable due to the symmetry of the graph. (a) To find the acceleration due to gravity gp on that planet, we use Eq. 2-15 (with +y up) 1 1 2 y − y0 = vt + g p t 2 ⇒ 25 m − 0 = ( 0 )( 2.5 s ) + g p ( 2.5 s ) 2 2 so that gp = 8.0 m/s2. (b) That same (max) point on the graph can be used to find the initial velocity. y − y0 = 1 ( v0 + v ) t 2 ⇒ 25 m − 0 = 1 ( v0 + 0 ) ( 2.5 s ) 2 Therefore, v0 = 20 m/s. 65. The key idea here is that the speed of the head (and the torso as well) at any given time can be calculated by finding the area on the graph of the head’s acceleration versus time, as shown in Eq. 2-26: ⎛ area between the acceleration curve ⎞ v1 − v0 = ⎜ ⎟ ⎝ and the time axis, from t0 to t1 ⎠ (a) From Fig. 2.14a, we see that the head begins to accelerate from rest (v0 = 0) at t0 = 110 ms and reaches a maximum value of 90 m/s2 at t1 = 160 ms. The area of this region is 1 area = (160 − 110) × 10−3s ⋅ ( 90 m/s 2 ) = 2.25 m/s 2 which is equal to v1, the speed at t1. (b) To compute the speed of the torso at t1=160 ms, we divide the area into 4 regions: From 0 to 40 ms, region A has zero area. From 40 ms to 100 ms, region B has the shape of a triangle with area 1 area B = (0.0600 s)(50.0 m/s 2 ) = 1.50 m/s . 2 From 100 to 120 ms, region C has the shape of a rectangle with area area C = (0.0200 s) (50.0 m/s 2 ) = 1.00 m/s. From 110 to 160 ms, region D has the shape of a trapezoid with area 1 (0.0400 s) (50.0 + 20.0) m/s 2 = 1.40 m/s. 2 Substituting these values into Eq. 2-26, with v0 = 0 then gives area D = 54 CHAPTER 2 v1 − 0 = 0 + 1.50 m/s + 1.00 m/s + 1.40 m/s = 3.90 m/s, or v1 = 3.90 m/s. 66. The key idea here is that the position of an object at any given time can be calculated by finding the area on the graph of the object’s velocity versus time, as shown in Eq. 2-25: ⎛ area between the velocity curve ⎞ x1 − x0 = ⎜ ⎟. ⎝ and the time axis, from t0 to t1 ⎠ (a) To compute the position of the fist at t = 50 ms, we divide the area in Fig. 2-34 into two regions. From 0 to 10 ms, region A has the shape of a triangle with area area A = 1 (0.010 s) (2 m/s) = 0.01 m. 2 From 10 to 50 ms, region B has the shape of a trapezoid with area area B = 1 (0.040 s) (2 + 4) m/s = 0.12 m. 2 Substituting these values into Eq. 2-25 with x0 = 0 then gives x1 − 0 = 0 + 0.01 m + 0.12 m = 0.13 m, or x1 = 0.13 m. (b) The speed of the fist reaches a maximum at t1 = 120 ms. From 50 to 90 ms, region C has the shape of a trapezoid with area area C = 1 (0.040 s) (4 + 5) m/s = 0.18 m. 2 From 90 to 120 ms, region D has the shape of a trapezoid with area area D = 1 (0.030 s) (5 + 7.5) m/s = 0.19 m. 2 Substituting these values into Eq. 2-25, with x0 = 0 then gives x1 − 0 = 0 + 0.01 m + 0.12 m + 0.18 m + 0.19 m = 0.50 m, or x1 = 0.50 m. 67. The problem is solved using Eq. 2-26: ⎛ area between the acceleration curve ⎞ v1 − v0 = ⎜ ⎟ ⎝ and the time axis, from t0 to t1 ⎠ 55 To compute the speed of the unhelmeted, bare head at t1 = 7.0 ms, we divide the area under the a vs. t graph into 4 regions: From 0 to 2 ms, region A has the shape of a triangle with area 1 area A = (0.0020 s) (120 m/s 2 ) = 0.12 m/s. 2 From 2 ms to 4 ms, region B has the shape of a trapezoid with area area B = 1 (0.0020 s) (120 + 140) m/s2 = 0.26 m/s. 2 From 4 to 6 ms, region C has the shape of a trapezoid with area area C = 1 (0.0020 s) (140 + 200) m/s2 = 0.34 m/s. 2 From 6 to 7 ms, region D has the shape of a triangle with area 1 area D = (0.0010 s) (200 m/s 2 ) = 0.10 m/s. 2 Substituting these values into Eq. 2-26, with v0=0 then gives vunhelmeted = 0.12 m/s + 0.26 m/s + 0.34 m/s + 0.10 m/s = 0.82 m/s. Carrying out similar calculations for the helmeted head, we have the following results: From 0 to 3 ms, region A has the shape of a triangle with area 1 (0.0030 s) (40 m/s 2 ) = 0.060 m/s. 2 From 3 ms to 4 ms, region B has the shape of a rectangle with area area A = area B = (0.0010 s) (40 m/s 2 ) = 0.040 m/s. From 4 to 6 ms, region C has the shape of a trapezoid with area area C = 1 (0.0020 s) (40 + 80) m/s2 = 0.12 m/s. 2 From 6 to 7 ms, region D has the shape of a triangle with area 1 area D = (0.0010 s) (80 m/s 2 ) = 0.040 m/s. 2 Substituting these values into Eq. 2-26, with v0 = 0 then gives vhelmeted = 0.060 m/s + 0.040 m/s + 0.12 m/s + 0.040 m/s = 0.26 m/s. 56 CHAPTER 2 Thus, the difference in the speed is Δv = vunhelmeted − vhelmeted = 0.82 m/s − 0.26 m/s = 0.56 m/s. 68. This problem can be solved by noting that velocity can be determined by the graphical integration of acceleration versus time. The speed of the tongue of the salamander is simply equal to the area under the acceleration curve: 1 −2 1 1 (10 s)(100 m/s 2 ) + (10−2 s)(100 m/s 2 + 400 m/s 2 ) + (10−2 s)(400 m/s 2 ) 2 2 2 = 5.0 m/s. v = area = z 69. Since v = dx / dt (Eq. 2-4), then Δx = v dt , which corresponds to the area under the v vs t graph. Dividing the total area A into rectangular (base × height) and triangular 21 base × height areas, we have b g A = A0 <t <2 + A2 < t <10 + A10 < t <12 + A12 < t 0 for times less than t = 2 s, then the spot had been moving rightward. (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 2 Δx = at = vt truck 2 car which leads to FG H IJ b g K 58 CHAPTER 2 t= Therefore, 2v 2 ( 9.5 m/s ) = = 8.6 s . a 2.2 m/s 2 Δx = 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/s 2 ) ( 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 Δx = h 35 m = = 4655 . m ≈ 0.465 km. tan θ tan 4.3° This corresponds to a time of flight found from Eq. 2-2 (with v = vavg since it is constant) Δx 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 = v0 t r + v0 t b + 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 1 v02 v2 1 v02 x = v0 t r − 0 + = v0 t r − . 2 a 2 a a We write this equation for each of the initial velocities: x1 = v01tr − 2 1 v01 2 a x2 = v02 tr − 2 1 v02 . 2 a and Solving these equations simultaneously for tr and a we get 59 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 v02 = 48.3 km/h = 13.4 m/s gives a=− 2 2 − v01v02 1 v02 v01 1 (13.4 m/s)(22.4 m/s) 2 − (22.4 m/s)(13.4 m/s)2 =− 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 Δx0 = at02 / 2 = v 2p / 2a. The car then moves at a constant speed v p over a distance D12 − Δx0 − d to reach intersection 2, and the time elapsed is t1 = ( D12 − Δx0 − d ) / v p . Thus, the time delay at intersection 2 should be set to v p D12 − (v 2p / 2a ) − d D12 − Δx0 − d ttotal = tr + t0 + t1 = tr + + = tr + + a vp a vp vp = tr + 1 v p D12 − d + 2 a vp 77. Since the problem involves constant acceleration, the motion of the rod can be readily analyzed using the equations in Table 2-1. We take +x in the direction of motion, so 60 CHAPTER 2 b v = 60 km / h m / km I g FGH 1000 J = + 16.7 m / s 3600 s / h K and a > 0. The location where it starts from rest (v0 = 0) is taken to be x0 = 0. (a) Using Eq. 2-7, we find the average acceleration to be aavg = Δv v − v0 16.7 m/s − 0 = = = 3.09 m/s 2 ≈ 3.1 m/s 2 . Δt t − t0 5.4 s − 0 (b) Assuming constant acceleration a = aavg = 3.09 m/s 2 , the total distance traveled during the 5.4-s time interval is 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 = = = 13 s . 2 a 3.1 m/s 2 Note that the displacement of the rod as a function of time can be written as 1 x(t ) = (3.09 m/s 2 )t 2 . Also 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 2a′ −2 m/s 2 The speed of the other train, when it reaches that location, is v = v02 + 2aΔx = = 10 m/s ( −40 m/s ) + 2 (1.0 m/s 2 ) ( 200 m − 950 m ) 2 61 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 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/s 2 (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. 62 CHAPTER 2 81. 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 82. The velocity v at t = 6 (SI units and two significant figures understood) is 6 1 vgiven + ∫ adt . A quick way to implement this is to recall the area of a triangle (2 −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 Δy = − 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 84. We take the direction of motion as +x, take x0 = 0 and use SI units, so v = 1600(1000/3600) = 444 m/s. 63 (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 the textbook in §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. 64 CHAPTER 2 87. The time it 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 v1 = 55 mi/h = (55 mi/h) 1609 m/mi = 24.58 m/s 3600 s/h v2 = 65 mi/h = (65 mi/h) 1609 m/mi = 29.05 m/s 3600 s/h With d = 700 km = 7.0 ×105 m , the time difference between the two is ⎛1 ⎛ ⎞ 1⎞ 1 1 Δt = t1 − t2 = d ⎜ − ⎟ = (7.0 × 105 m) ⎜ − ⎟ = 4383 s = 73 min ⎝ 24.58 m/s 29.05 m/s ⎠ ⎝ v1 v2 ⎠ or 1 h and 13 min. 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: x= 1 1 ( 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 x= − v02 (5.00 m/s) 2 = − = − 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, t′ = −v0 −5.00 m/s = = − 3.0s a 1.67 m/s 2 indicates the moment the car was at rest. SI units are understood. 65 89. 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 Δy replacing Δx) because this is constant acceleration motion. When something is thrown straight up and is caught at the level it was thrown from, the time of flight t is half of its time of ascent ta, which is given by Eq. 2-18 with Δy = H and v = 0 (indicating the maximum point). H = vt a + 1 2 gta 2 ⇒ ta = 2H g Writing these in terms of the total time in the air t = 2ta we have H= 1 2 gt 8 ⇒ 2H . g t=2 We consider two throws, one to height H1 for total time t1 and another to height H2 for total time t2, and we set up a ratio: FG IJ H K H2 81 gt22 t = 1 2 = 2 H1 8 gt1 t1 2 from which we conclude that if t2 = 2t1 (as is required by the problem) then H2 = 22H1 = 4H1. 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: 72 CHAPTER 2 Δ y = v0 t − 1 2 gt 2 ⇒ t= v0 + v02 − 2 gΔy g with the positive root again chosen to yield t > 0. With y = 0 and y0 = h, we obtain t= v02 + 2 gh + v0 . g 102. We assume constant velocity motion and use Eq. 2-2 (with vavg = v > 0). Therefore, F GH Δx = vΔt = 303 km h FG 1000 m / kmIJ I c100 × 10 sh = 8.4 m. H 3600 s / h K JK −3