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Unformatted text preview: 5 0 -1200 27. 25 -200 s(t) 0 25 -30 v( t ) (a) from the graph the velocity is positive, so the displacement is always increasing and is therefore positive a(t) v 0.5 0.4 0.3 0.2 t 1 (b) 28. 2 3 4 5 s(t) = t/2 + (t + 1)e−t (a) If t0 &lt; 1 then the area between the velocity curve and the t-axis, between t = 0 and t = t0 , will always be negative, so the displacement will be negative. v 0.1 t 0.2 0.4 0.6 0.8 -0.1 -0.2 (b) 29. (a) 1 t2 − 2 200 s(t) = a(t) = 0, −10, ln(t + 0.1) − t&lt;4 t&gt;4 t2 t 1 + − ln 10 4 20 200 (b) 25, t &lt; 4 65 − 10t, t &gt; 4 v (t) = a v 2 4 12 t 20 2 -5 4 6 -20 -40 -10 25t, 65t − 5t2 − 80, (c) x(t) = (d) x(6.5) = 131.25 t&lt;4 , so x(8) = 120, x(12) = −20 t&gt;4 8 10 12 t 1 Exercise Set 7.7 238 2 v 2 − v0 2(s − s0 ) (a) 2 From exercise 30 part (a), in Section 3 of Chapter 6, v 2 = v0 − 2g (s − s0 ), so a = −g = (b) From exercise 30 part (b), in Section 3 of Chapter 6, s = s0 + 1 (v0 + v )t, so t = 2 (c) From exercise 30 part (c), in Section 3 of Chapter 6, s = s0 + vt + 1 gt2 = s0 + vt − 1 at2 2 2 31. (a) a = −1 mi/h/s = −22/15 ft/s2 32. Take t = 0 when deceleration begins, then a = −10 so v = −10t + C1 , but v = 88 when t = 0 which gives C1 = 88 thus v = −10t + 88, t ≥ 0 30. (a) (b) 2(s − s0 ) v0 + v a = 30 km/h/min = 1/7200 km/s2 v = 45 mi/h = 66 ft/s, 66 = −10t + 88, t = 2.2 s (b) v = 0 (the car is stopped) when t = 8.8 s s= v dt = (−10t + 88)dt = −5t2 + 88t + C2 , and taking s = 0 when t = 0, C2 = 0 so s = −5t2 + 88t. At t = 8.8, s = 387.2. The car travels 387.2 ft before coming to a stop. 33. a = a0 ft/s2 , v = a0 t + v0 = a0 t + 132 ft/s, s = a0 t2 /2 + 132t + s0 = a0 t2 /2 + 132t ft; s = 200 ft 121 20 when t = , so when v = 88 ft/s. Solve 88 = a0 t + 132 and 200 = a0 t2 /2 + 132t to get a0 = − 5 11 121 t + 132. s = −12.1t2 + 132t, v = − 5 70 121 242 ft/s2 ft/s when t = s (b) v = 55 mi/h = (a) a0 = − 5 3 33 60 (c) v = 0 when t = s 11 34. dv/dt = 3, v = 3t + C1 , but v = v0 when t = 0 so C1 = v0 , v = 3t + v0 . From ds/dt = v = 3t + v0 we get s = 3t2 /2 + v0 t + C2 and, with s = 0 when t = 0, C2 = 0 so s = 3t2 /2 + v0 t. s = 40 when t = 4 thus 40 = 3(4)2 /2 + v0 (4), v0 = 4 m/s 35. Suppose s = s0 = 0, v = v0 = 0 at t = t0 = 0; s = s1 = 120, v = v1 at t = t1 ; and s = s2 , v = v2 = 12 at t = t2 . From Exercise 30(a), 2 2 v1 − v0 , v 2 = 2as1 = 5.2(120) = 624. Applying the formula again, 2(s1 − s0 ) 1 2 v 2 − v1 2 , v 2 = v1 − 3(s2 − s1 ), so −1.5 = a = 2 2(s2 − s1 ) 2 2 2 s2 = s1 − (v2 − v1 )/3 = 120 − (144 − 624)/3 = 280 m. 2.6 = a = 36. a(t) = 4, 0, t&lt;2 , so, with v0 = 0, v (t) = t&gt;2 4t, t &lt; 2 and, since s0 = 0, s(t) = 8, t &gt; 2 2t2 , t &lt; 2 8t − 8, t &gt; 2 s = 100 when 8t − 8 = 100, t = 108/8 = 13.5 s 37. The truck’s velocity is vT = 50 and its position is sT = 50t + 5000. The car’s acceleration is aC = 2, so vC = 2t, sC = t2 (initial position and initial velocity of the car are both zero). sT = sC when 50t + 5000 = t2 , t2 − 50t − 5000 = (t + 50)(t − 100) = 0, t = 100 s and sC = sT = t2 = 10, 000 ft 38. Let t = 0 correspond to the time when the leader is 100 m from the ﬁnish line; let s = 0 correspond to the ﬁnish line. Then vC = 12, sC = 12t − 115; aL = 0.5 for√ &gt; 0, vL = 0.5t + 8, sL = 0.25t2 + 8t − 100. t sC = 0 at t = 115/12 ≈ 9.58 s, and sL = 0 at t = −16 + 4 41 ≈ 9.61, so the challenger wins. 39. s = 0 and v = 112 when t = 0 so v (t) = −32t + 112, s(t) = −16t2 + 112t (a) v (3) = 16 ft/s, v (5) = −48 ft/s (b) v = 0 when the projectile is at its maximum height so −32t + 112 = 0, t = 7/2 s, s(7/2) = −16(7/2)2 + 112(7/2) = 196 ft. 239 Chapter 7 (c) s = 0 when it reaches the ground so −16t2 + 112t = 0, −16t(t − 7) = 0, t = 0, 7 of which t = 7 is when it is at ground level on its way down. v (7) = −112, |v | = 112 ft/s. 40. s = 112 when t = 0 so s(t) = −16t2 + v0 t + 112. But s = 0 when t = 2 thus −16(2)2 + v0 (2) + 112 = 0, v0 = −24 ft/s. 41. (a) s(t) = 0 when it hits the ground, s(t) = −16t2 + 16t = −16t(t − 1) = 0 when t = 1 s. (b) The projectile moves upward until it gets to its highest point where v (t) = 0, v (t) = −32t + 16 = 0 when t = 1/2 s. √ 555/4 s 42. s(t) = 0 when the rock hits the ground, s(t) = −16t2 + 555 = 0 when t = √ √ √ (b) v (t) = −32t, v ( 555/4) = −8 555, the speed at impact is 8 555 ft/s 43. (a) s(t) = 0 when the package hits the ground, √ s(t) = −16t2 + 20t + 200 = 0 when t = (5 + 5 33)/8 s √ √ √ (b) v (t) = −32t + 20, v [(5 + 5 33)/8] = −20 33, the speed at impact is 20 33 ft/s 44. (a) s(t) = 0 when the stone hits the ground, s(t) = −16t2 − 96t + 112 = −16(t2 + 6t − 7) = −16(t + 7)(t − 1) = 0 when t = 1 s (b) v (t) = −32t − 96, v (1) = −128, the speed at impact is 128 ft/s 45. (a) s(t) = −4.9t2 + 49t + 150 and v (t) = −9.8t + 49 (a) the projectile reaches its maximum height when v (t) = 0, −9.8t + 49 = 0, t = 5 s (b) s(5) = −4.9(5)2 + 49(5) + 150 = 272.5 m (c) the projectile reaches its starting point when s(t) = 150, −4.9t2 + 49t + 150 = 150,...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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