(a) During the first part of the motion, the displacement is x 1 = 40 km and the time interval is t 1 (30 km / h) (80 km) 40 km/h. t. (2.

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Claude Lucas
5 years ago
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1 Chapter 3. Since the trip consists of two parts, let the displacements during first and second parts of the motion be x and x, and the corresponding time interals be t and t, respectiely. Now, because the problem is one-dimensional and both displacements are in the same direction, the total displacement is x = x + x, and the total time for the trip is t = t + t. Using the definition of aerage elocity gien in Eq. -, we hae x x x ag. t t t To find the aerage speed, we note that during a time t if the elocity remains a positie constant, then the speed is equal to the magnitude of elocity, and the distance is equal to the magnitude of displacement, with d x t. (a) uring the first part of the motion, the displacement is x = 4 km and the time interal is ( 4 km) t 33. h. (3 km / h) Similarly, during the second part the displacement is x = 4 km and the time interal is ( 4 km) t. 67 h. (6 km / h) The total displacement is x = x + x = 4 km + 4 km = 8 km, and the total time elapsed is t = t + t =. h. Consequently, the aerage elocity is ag x (8 km) 4 km/h. t (. h) (b) In this case, the aerage speed is the same as the magnitude of the aerage elocity: sag 4 km/h. (c) The graph of the entire trip is shown below; it consists of two contiguous line segments, the first haing a slope of 3 km/h and connecting the origin to (t, x ) = (.33 h, 4 km) and the second haing a slope of 6 km/h and connecting (t, x ) to (t, x) = (. h, 8 km).

2 CHAPTER 4. Aerage speed, as opposed to aerage elocity, relates to the total distance, as opposed to the net displacement. The distance 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 hae speed = /t. Thus, the aerage speed is t up up t down down up down which, after canceling and plugging in up = 4 km/h and down = 6 km/h, yields 48 km/h for the aerage speed. 7. Recognizing that the gap between the trains is closing at a constant rate of 6 km/h, the total time that elapses before they crash is t = (6 km)/(6 km/h) =. h. uring this time, the bird traels a distance of x = t = (6 km/h)(. h) = 6 km.. The alues used in the problem statement make it easy to see that the first part of the trip (at km/h) takes hour, and the second part (at 4 km/h) also takes hour. Expressed in decimal form, the time left is.5 hour, and the distance that remains is 6 km. Thus, a speed = (6 km)/(.5 h) = 8 km/h is needed. 7. We use Eq. - for aerage elocity and Eq. -4 for instantaneous elocity, and work with distances in centimeters and times in seconds. (a) We plug into the gien equation for x for t =. s and t = 3. s and obtain x =.75 cm and x 3 = 5.5 cm, respectiely. The aerage elocity during the time interal. t 3. s is ag x t 5. 5 cm 75. cm 3. s. s which yields ag = 8.5 cm/s. dx (b) The instantaneous elocity is 4. 5t, which, at time t =. s, yields = (4.5)(.) = 8. cm/s. (c) At t = 3. s, the instantaneous elocity is = (4.5)(3.) = 4.5 cm/s.

3 3 (d) At t =.5 s, the instantaneous elocity is = (4.5)(.5) = 8. cm/s. (e) Let t m stand for the moment when the particle is midway between x and x 3 (that is, when the particle is at x m = (x + x 3 )/ = 36 cm). Therefore, 3 x t t. 596 m m m in seconds. Thus, the instantaneous speed at this time is = 4.5(.596) = 3.3 cm/s. (f) The answer to part (a) is gien by the slope of the straight line between t = and t = 3 in this x-s-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 cure at the appropriate points.. We use the functional notation x(t), (t) and a(t) and find the latter two quantities by differentiating: dx t d t btg b g t abtg b g 5 and 3t t with SI units understood. These expressions are used in the parts that follow. (a) From 5t, we see that the only positie alue of t for which the particle is (momentarily) stopped is t / 5. s. (b) From = 3t, we find a() = (that is, it anishes at t = ). (c) It is clear that a(t) = 3t is negatie for t >. (d) The acceleration a(t) = 3t is positie for t <. (e) The graphs are shown below. SI units are understood.

4 4 CHAPTER 5. We separate the motion into two parts, and take the direction of motion to be positie. In part, the ehicle accelerates from rest to its highest speed; we are gien = ; = m/s and a =. m/s. In part, the ehicle decelerates from its highest speed to a halt; we are gien = m/s; = and a =. m/s (negatie because the acceleration ector points opposite to the direction of motion). (a) From Table -, we find t (the duration of part ) from = + at. In this way,.t yields t = s. We obtain the duration t of part from the same equation. Thus, = + (.)t leads to t = s, and the total is t = t + t = 3 s. (b) For part, taking x =, we use the equation = + a(x x ) from Table - and find ( m/s) () x m. a (. m/s ) This position is then the initial position for part, so that when the same equation is used in part we obtain () ( m/s) x m. a (. m/s ) Thus, the final position is x = 3 m. That this is also the total distance traeled should be eident (the ehicle did not "backtrack" or reerse its direction of motion). 8. We take +x in the direction of motion, so = +4.6 m/s and a = 4.9 m/s. We also take x =. (a) The time to come to a halt is found using Eq. -: 4.6 m/s at t 5. s. 4.9 m/s

5 (b) Although seeral of the equations in Table - will yield the result, we choose Eq. -6 (since it does not depend on our answer to part (a)). (4.6 m/s) ax x 6.5 m. 4.9 m/s (c) Using these results, we plot t 5 at (the x graph, shown next, on the left) and + at (the graph, on the right) oer t 5 s, with SI units understood. 46. Neglect of air resistance justifies setting a = g = 9.8 m/s (where down is our y direction) for the duration of the fall. This is constant acceleration motion, and we may use Table - (with y replacing x). (a) Using Eq. -6 and taking the negatie root (since the final elocity is downward), we hae g y (9.8 m/s )( 7 m) 83 m/s. Its magnitude is therefore 83 m/s. (b) No, but it is hard to make a conincing 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 relatiely healthy situation with which we are familiar. 54. (a) We neglect air resistance, which justifies setting a = g = 9.8 m/s (taking down as the y direction) for the duration of the motion. We are allowed to use Eq. -5 (with y replacing x) because this is constant acceleration motion. We use primed ariables (except t) with the first stone, which has zero initial elocity, and unprimed ariables with the second stone (with initial downward elocity, so that is being used for the initial speed). SI units are used throughout.

6 6 CHAPTER y t gt y t g t Since the problem indicates y = y = 43.9 m, we sole the first equation for t (finding t =.99 s) and use this result to sole the second equation for the initial speed of the second stone: 43.9 m.99 s 9.8 m/s.99 s which leads to =.3 m/s. (b) The elocity of the stones are gien by d( y) d( y) y gt, y g( t ) The plot is shown below: 6. The height reached by the player is y =.76 m (where we hae taken the origin of the y axis at the floor and +y to be upward). (a) The initial elocity of the player is gy (9.8 m/s )(.76 m) 3.86 m/s. This is a consequence of Eq. -6 where elocity anishes. As the player reaches y

7 7 =.76 m.5 m =.6 m, his speed satisfies gy, which yields gy (3.86 m/s) (9.8 m/s )(.6 m).7 m/s. The time t that the player spends ascending in the top y =.5 m of the jump can now be found from Eq. -7:.5 m y t t.75 s.7 m/s which means that the total time spent in that top 5 cm (both ascending and descending) is (.75 s) =.35 s = 35 ms. (b) The time t when the player reaches a height of.5 m is found from Eq. -5:.5 m t gt (3.86 m/s) t (9.8 m/s ) t, which yields (using the quadratic formula, taking the smaller of the two positie roots) t =.4 s = 4 ms, which implies that the total time spent in that bottom 5 cm (both ascending and descending) is (4 ms) = 8 ms. 64. The graph shows y = 5 m to be the highest point (where the speed momentarily anishes). The neglect of air friction (or whateer passes for that on the distant planet) is certainly reasonable due to the symmetry of the graph. (a) To find the acceleration due to graity g p on that planet, we use Eq. -5 (with +y up) y y t g pt 5 m.5 s g p.5 s so that g p = 8. m/s. (b) That same (max) point on the graph can be used to find the initial elocity. Therefore, = m/s. y y t 5 m.5 s

1. The speed (assumed constant) is (90 km/h)(1000 m/km) (3600 s/h) = 25 m/s. Thus, during 0.50 s, the car travels (0.50)(25) 13 m.

1. The speed (assumed constant) is (90 km/h)(1000 m/km) (3600 s/h) = 25 m/s. Thus, during 0.50 s, the car travels (0.50)(25) 13 m. . The speed (assumed constant) is (90 km/h)(000 m/km) (3600 s/h) = 5 m/s. Thus, during 0.50 s, the car travels (0.50)(5) 3 m. . Huber s speed is v 0 =(00 m)/(6.509 s)=30.7 m/s = 0.6 km/h, where we have