15 f x has an absolute minimum at x 2 and m e2 4 a 3

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Unformatted text preview: 1 − 2| + |607 − (−1)| = 611 ft (a) v (t) = −(3π/2) sin(πt/2), a(t) = −(3π 2 /4) cos(πt/2) (b) s(1) = 0 ft, v (1) = −3π/2 ft/s, speed = 3π/2 ft/s, a(1) = 0 ft/s2 (c) v = 0 at t = 0, 2, 4 (d) v changes sign at t = 0, 2, 4 and a changes sign at t = 1, 3, 5, so the particle is speeding up for 0 < t < 1, 2 < t < 3 and 4 < t < 5, and it is slowing down for 1 < t < 2 and 3 < t < 4 (e) total distance = |s(2) − s(0)| + |s(4) − s(2)| + |s(5) − s(4)| = | − 3 − 3| + |3 − (−3)| + |0 − 3| = 15 ft (a) v (t) = (b) s(1) = 1/5 ft, v (1) = 3/25 ft/s, speed = 3/25 ft/s, a(1) = −22/125 ft/s2 (c) 14. v (t) = 3t2 − 12t, a(t) = 6t − 12 (d) 13. (a) (c) 12. speeding up at t = 3; slowing down at t = 1, 5; neither at t = 2, 4 (b) 11. to the right at t = 1, stopped at t = 2, otherwise to the left v = 0 at t = 2 2t(t2 − 12) 4 − t2 , a(t) = (t2 + 4)2 (t2 + 4)3 (d) √ √ a changes sign at t = 2√ 3, so the particle is speeding up for 2 < t < 2 3 and it is slowing down for 0 < t < 2 and for 2 3 < t (e) total distance = |s(2) − s(0)| + |s(5) − s(2)| = 15. v (t) = 1 19 1 5 −0 + − = ft 4 29 4 58 2t(t2 − 15) 5 − t2 , a(t) = (t2 + 5)2 (t2 + 5)3 0.2 0.25 0.01 0 0 0 10 20 20 0 s (t) -0.05 v(t) -0.15 a(t) 195 Chapter 6 (a) (c) 16. √ 5 √ √ 5/10 at t = 5 √ √ √ a changes sign at t = 15, so the particle is speeding up for 5 < t < 15 and slowing down √ √ for 0 < t < 5 and 15 < t v = 0 at t = (b) s= v (t) = (1 − t)e−t , a(t) = (t − 2)e−t 1 0.4 0.1 0 0 0 6 8 10 0 -0.2 s (t) -2 v( t ) (a) 18. 19. v = 0 at t = 1 (c) 17. (b) a (t) a changes sign at t = 2, so the particle is speeding up for 1 < t < 2 and slowing down for 0 < t < 1 and 2 < t s = −3t + 2 v = −3 a=0 Constant speed s = t3 − 6t2 + 9t + 1 v = 3(t − 1)(t − 3) a = 6(t − 2) Speeding up t=2 (Stopped) t = 3 1 3 Slowing down s 5 Speeding up Slowing down (Stopped) t = 4 t=3 t=0 9 t+1 (t + 4)(t − 2) v= (t + 1)2 18 a= (t + 1)3 s=t+ Slowing down t = 2 (Stopped) 16 18 20 s Speeding up (Stopped) t = 2 5 0 ≤ t ≤ 2π t > 2π s= cos t, 1, − sin t, 0, 0 ≤ t ≤ 2π t > 2π a= 22. t = 1 (Stopped) t=0 s = t3 − 9t2 + 24t v = 3(t − 2)(t − 4) a = 6(t − 3) v= 21. s t=0 2 0 20. s = 1/e at t = 1 − cos t, 0, 0 ≤ t < 2π t > 2π Slowing down Slowing down t=π t = 3π/2 t = π/2 -1 0 9 s (Stopped permanently) t = 2π t=0 1 Speeding up s √ 15t2 − 6t − 2 5t2 − 6t + 2 3 + 39 √ is always positive, a(t) = v (t) = has a positive root at t = 15 2t3/2 t Slowing down 0 3 + 39 15 Speeding up s Exercise Set 6.3 23. (a) 196 v = 10t − 22, speed = |v | = |10t − 22|. d|v |/dt does not exist at t = 2.2 which is the only critical point. If t = 1, 2.2, 3 then |v | = 12, 0, 8. The maximum speed is 12 ft/s. (b) the distance from the origin is |s| = |5t2 − 22t| = |t(5t − 22)|, but t(5t − 22) < 0 for 1 ≤ t ≤ 3 so |s| = −(5t2 − 22t) = 22t − 5t2 , d|s|/dt = 22 − 10t, thus the only critical point is t = 2.2. d2 |s|/dt2 < 0 so the particle is farthest from the origin when t = 2.2. Its position is s = 5(2.2)2 − 22(2.2) = −24.2. 24. 200t d|v | 600(4 − t2 ) 200t , speed = |v | = 2 for t ≥ 0. = 0 when t = 2, which is the =2 (t2 + 12)2 (t + 12)2 dt (t + 12)3 only critical point in (0, +∞). By the first derivative test there is a relative maximum, and hence an absolute maximum, at t = 2. The maximum speed is 25/16 ft/s to the left. v=− 25. s(t) = s0 − 1 gt2 = s0 − 4.9t2 m, v = −9.8t m/s, a = −9.8 m/s2 2 (a) |s(1.5) − s(0)| = 11.025 m (b) v (1.5) = −14.7 m/s (c) |v (t)| = 12 when t = 12/9.8 = 1.2245 s (d) s(t) − s0 = −100 when 4.9t2 = 100, t = 4.5175 s √ 800 =5 2 16 27. (a) s(t) = s0 − 1 gt2 = 800 − 16t2 ft, s(t) = 0 when t = 2 (b) 26. √ √ v (t) = −32t and v (5 2) = −160 2 ≈ 226.27 ft/s = 154.28 mi/h s(t) = s0 + v0 t − 1 gt2 = 60t − 4.9t2 m and v (t) = v0 − gt = 60 − 9.8t m/s 2 (a) v (t) = 0 when t = 60/9.8 ≈ 6.12 s (b) s(60/9.8) ≈ 183.67 m (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 the line t = 6.12 in the t-s plane (d) also 60 m/s, as seen from the symmetry of the parabola (or compute v (6.12)) 28. (a) they are the same (b) s(t) = v0 t − 1 gt2 and v (t) = v0 − gt; s(t) = 0 when t = 0, 2v0 /g ; 2 v (0) = v0 and v (2v0 /g ) = v0 − g (2v0 /g ) = −v0 so the speed is the same at launch (t = 0) and at return (t = 2v0 /g ). 29. If g = 32 ft/s2 , s0 = 7 and v0 is unknown, then s(t) = 7 + v0 t − 16t2 and v (t) = v0 32t; s = smax when 2 v = 0, or√ = v0 /32; and smax = 208 yields 208 = s(v0 /32) = 7 + v0 (v0 /32) − 16(v0 /32)2 = 7 + v0 /64, t so v0 = 8 201 ≈ 113.42 ft/s. 30. (a) 2 2 2 Use (6) and then (5) to get v 2 = v0 − 2v0 gt + g 2 t2 = v0 − 2g (v0 t − 1 gt2 ) = v0 − 2g (s − s0 ). 2 (b) Add v0 to both sides of (6): 2v0 − gt = v0 + v , v0 − 1 gt = 1 (v0 + v...
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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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