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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. dv /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 , ds/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
dv 
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 ts 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.
 Spring '14
 The Land

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