H 0 h 256 ft a the 0 when 6 4 thus t1642 b first

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Unformatted text preview: > 4 a Chapter 5 1 (b) v (t) = 4 12 20 –0.3 –5 25, t < 4 65 − 10t, t > 4 v t 2 ￿ 24 6 t 8 10 12 –20 –10 3 T+ 400 –40 ￿ 25t, t4 25, t < < 4 , so x(8) = 120, x(12) = −20 (c) (b) t)v= ) = x( (t ￿ 65t 6525210t, <,t > > 4 − 5t , −t80 4 t 4 − (b) v (t) = 65 − 10t, t > 4 (d) x(6.5) = 131.25 v 32. Take t 20 0 when deceleration begins, then a = −11 so v = −11t + C1 , but v = 88 when t = 0 = which gives C1 = 88 thus vt = −11t + 88, t ≥ 0 2 4 6 8 10 12 (a) v = 45 mi/h = 66 ft/s, 66 = −11t + 88, t = 2 s –20 (b) v = 0 (the car is stopped) when t = 8 s ￿ –40 ￿ 11 s = v dt = (−11t + 88)dt = − t2 + 88t + C2 , and taking s = 0 when t = 0, C2 = 0 so 2 11 2 s= t 88t travels 352 ft before stop. t 352. so x(8) = 120, x(12)x− )2= + 25t,. At t = 8, s = < 4 ,Thexcar = 120, x(12) = −20. coming to a (6.5) = 131.25. (c) =t −20 ( so (8) (d) x 65t − 5t2 − 80, t > 4 33. a = a0 ft/s2 , v 2 a0 t + v0 = a0 t + 132 ft/s, s = a0 t2 /2 +2132t + s0 = a0 t2 /2 +2132t ft; s = 200 ft = 33. a = a0 ft/s , v = a0 t + v0 = a0 t + 132 ft/s, s = a0 t /2 + 132t + s0 = a0 t /2 + 132t ft; s = 200 ft when v = 88 121 20 t , when v = −11t 88 Solve + 1320 t when t = 200 = + 132 to get a = a = when when 20= so s = −12.1t2 + 132t, hen a = −11 so v = Solve + C=, a0 t88 = a88 + 132 and a0 t2 /2 a0 t2 /2t + 132t to0 get −0121 − ft/s. 88 ft/s. 1 but v = and 200 = 0 5 t = 11 , 11 5 8, t ≥ 0 121 v=− t + 132. t + 88, t = 2 s 5 242 70 60 121 =8s ft/s2 . (b) v = 55 mi/h = ft/s when t = s. (c) v = 0 when t = s. (a) a0 = − 5 3 33 11 11 2 t + 88t + C2 , and taking s = 0 when t = 0, C2 = 0 so 2 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 . 2 v 2 − v0 2. The car travels 352formulas (10) and to a stop. get that in the case of constant acceleration, a = From ft before coming (11), we . This implies that 2(s − s0 ) 2 2 2 2 v1 − v0 v 2 − v1 2 2.6 = a = , v1 = 2as1 = 5.2(120) = 624. Applying the formula again, −1.5 = a = , v2 = 2 2 ft/s, s = a0 t /2 + 132t + s0 2(s1 0 t s0 ) + 132t ft; s = 200 ft = a − /2 2(s2 − s1 ) 2 2 2 20 v1 − 3(s − s1 ), so s2 = 121 (v2 − v1 )/3 = 120 − (144 − 624)/3 = 280 m. s− 2 , and 200 = a0 t2 /2 + 1322 to get a0 = − 1 when t = t 5 11 37. The truck’s velocity is vT = 50 and its position is sT = 50t + 2500. The car’s acceleration is aC = 4 ft/s2 , so vC = 4t, sC = 2t2 (initial position and initial velocity of the car are both zero). sT = sC when 50t + 2500 = 2t2 , 2t2 − 50t − 2500 = 2(t + 25)(t − 50) = 0, t = 50 s and sC = sT = 2t2 = 5000 ft. 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. (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. Exercise Set 5.8 265 (a) the projectile reaches its maximum height when v (t) = 0, −9.8t + 49 = 0, t = 5 s Exercise Set 5.8 (b) s(5) = −4.9(5)2 + 49(5) + 150 = 272.5 m 147 41. (a) s(the= 0 when it hits the ground, s(point−16t2 +(16t = −16t(t.9t2 += 0 when t= 150, (c) t) projectile reaches its starting t) = when s t) = 150, −4 − 1) 49t + 150 = 1 s. −4.9t(t − 10) = 0, t = 10 s (b) The projectile moves49 = −49until it gets to its highest point where v (t) = 0, v (t) = −32t + 16 = 0 when (d) v (10) = −9.8(10) + upward m/s t = 1/2 s. (e) s(t) = 0 when the projectile hits the ground, −4.9t2 + 49t + 150 = 0 when (use the quadratic 1 43. s(t) = formula) − ≈gt2 .46 60t − 4.9t2 m and v (t) = v0 − gt = 60 − 9.8t m/s. s0 + v0 t t 2 12 = s (a)) v (t(12.46) = −9.= 60./9.8+ 49.12 −73.1, the speed at impact is about 73.1 m/s (f v ) = 0 when t 8(12 46) ≈ 6 ≈ s. (b) s(60/9.8) ≈ 183.67 m. 44. take s = 0 at the water level and let h be the height of the bridge, then s = h and v = 0 when t = 0 so s(t) = −16t2 + h (c) Another 6.12 s; solve for t in s(t) = 0 to get this result, or use the symmetry of the parabola s = 60t − 4.9t2 about s = line t = t .12 in the−-s plane. h = 0, h = 256 ft (a) the 0 when 6= 4 thus t16(4)2 + (b) First, find how long it takes for the stone to hit the water (find t for s = 0) : −16t2 + h = 0, √ (d) Also 60 m/s, as seen from the symmetry of the parabola (or compute v (6.12)). t = h/4. Next, find how long it takes the sound to travel to the bridge: this time is h/1080 because the speed is constant at 1080 ft/s. Finally, use the fact that the total of these two √ 45. s(t) = −4.9t2 + 49t + 150 and v (t) = −9.8t + 49. √ √ h h times must be 4 s: + = 4, h + 270 h v t) = + 8t + h = 0, t = s. (a) The model rocket reaches its maximum height when =(4320,0,h−9.270 49− 4320 = 50, and by 1080 4 ￿ √ −270 ± (270)2 + 4(4320) 2 formula the quadratic h= , reject the n...
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This document was uploaded on 02/02/2014.

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