True if at a0 then v t a0 t v0 true 2 22 at

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Unformatted text preview: = k k = k=1 πk 4n n f (x∗ )∆x = k = lim n→+∞ k=1 πk π π and ∆x = for 0 ≤ x ≤ . 4n 4n 4 Thus π /4 π /4 sec2 x dx = tan x 0 = 1. 0 73. Let f be continuous on a closed interval [a, b] and let F be an antiderivative of f on [a, b]. By Theorem 5.7.2, b b F (b) − F (a) = F (x∗ ) for some x∗ in (a, b). By Theorem 5.6.1, f (x) dx = F (b) − F (a), i.e. f (x) dx = b−a a a F (x∗ )(b − a) = f (x∗ )(b − a). Exercise Set 5.7 3 1. (a) displ = s(3) − s(0) = 3 dt = 3; dist = 0 dt = 3. 0 3 (b) displ = s(3) − s(0) = − 0 3 dt = −3; dist = 3 (c) displ = s(3) − s(0) = 1 3 0 |v (t)|dt = (t − t2 /2) 0 0 |v (t)| dt = 3. 2 v (t)dt = 0 + (t2 /2 − t) 2 1 0 2 3 (1 − t)dt + − (t2 /2 − 3t) 3 2 (t − 3)dt = (t − t2 /2) = 3/2. 2 0 3 + (t2 /2 − 3t) 2 = −1/2; dist = 144 Chapter 5 3 (d) displ = s(3) − s(0) = 2 1 0 2 tdt + 0 5/2 1 3 (5 − 2t)dt = t2 /2 dt + 1 2 1 3 (5 − 2t)dt + dt + tdt + 0 5 /2 2 1 1 v (t)dt = (2t − 5)dt = t2 /2 0 5/2 2 +t 0 1 2 +t + (5t − t2 ) 1 3 + (5t − t2 ) 3 + (t2 − 5t) 2 = 3/2; dist = 2 = 2. 5/2 t 3. (a) v (t) = 20 + 0 a(u)du; add areas of the small blocks to get v (4) ≈ 20 + 1.4 + 3.0 + 4.7 + 6.2 = 35.3 m/s. 6 (b) v (6) = v (4) + 4 a(u)du ≈ 35.3 + 7.5 + 8.6 = 51.4 m/s. 5. (a) s(t) = t3 − t2 + C ; 1 = s(0) = C , so s(t) = t3 − t2 + 1. 1 (b) v (t) = − cos 3t + C1 ; 3 = v (0) = −1 + C1 , C1 = 4, so v (t) = − cos 3t + 4. Then s(t) = − sin 3t + 4t + C2 ; 3 1 3 = s(0) = C2 , so s(t) = − sin 3t + 4t + 3. 3 32 3 t + t + C ; 4 = s(2) = 6 + 2 + C, C = −4 and s(t) = t2 + t − 4. 2 2 7. (a) s(t) = (b) v (t) = −t−1 + C1 , 0 = v (1) = −1 + C1 , C1 = 1 and v (t) = −t−1 + 1 so s(t) = − ln t + t + C2 , 2 = s(1) = 1 + C2 , C2 = 1 and s(t) = − ln t + t + 1. π /2 π /2 9. (a) displacement = s(π/2) − s(0) = sin tdt = − cos t 0 0 0 2π 2π (b) displacement = s(2π )−s(π/2) = π /2 = 1 m; distance = cos tdt = sin t π/2 π/2 | sin t|dt = 1 m. 2π = −1 m; distance = π/2 3π/2 | cos t|dt = − cos tdt+ π/2 2π cos tdt = 3 m. 3π/2 3 11. (a) v (t) = t3 − 3t2 + 2t = t(t − 1)(t − 2), displacement = 1 2 v (t)dt + 0 1 3 −v (t)dt + 3 (b) displacement = 0 0 3 (t3 − 3t2 + 2t)dt = 9/4 m; distance = v (t)dt = 11/4 m. 2 √ √ ( t − 2)dt = 2 3 − 6 m; distance = 3 0 2 13. v = 3t − 1, displacement = 0 3 |v (t)|dt = − 2 (3t − 1) dt = 4 m; distance = 0 |3t − 1| dt = 0 √ v (t)dt = 6 − 2 3 m. 13 m. 3 √ 2√ 2√ 1/ 3t + 1 dt = 3t + 1 + C ; v (0) = 4/3 so C = 2/3, v = 3t + 1 + 2/3, displacement 3 3 15. v = 5 = 1 2√ 2 3t + 1 + 3 3 17. (a) s = a= 0 dt = 296 m; distance = 27 5 1 2√ 2 3t + 1 + 3 3 dt = 296 m. 27 1 2 1 2 2 1 2 sin πt dt = − cos πt + C , s = 0 when t = 0 which gives C = so s = − cos πt + . 2 π 2 π π 2 π dv π 1 = cos πt. When t = 1 : s = 2/π , v = 1, |v | = 1, a = 0. dt 2 2 |v (t)|dt = 38 Page number 262 black Exercise Set 5.7 145 3 3 Chapter 5 t dt = − t2 + C1 , v = 0 when t = 0 which gives C1 = 0 so v = − t2 . 2 2 1303 13 13 s=− t2 dt = − t + C2 , s = 1 when t = 0 which gives C2 = 1 so s = − t + 1. When t = 1 : s = 1/2, 2 2 2 v = −3/2, |v | = 3/2, a = −3. (b) v = −3 6 0 19. By inspection the velocity is positive for t > 0, and during the first second the ant is at most 5/2 cm from the T 6 v (t) dt = starting position. For T > 1 the displacement of the ant during the time interval [0, T ] is given by –130 T 5/2 + 1 √ (6 t −(t1/t) dt = 5/2 + (4t3/2 − ln t) a) 0 T 1 = −3/2 + 4T 3/2 − ln T , and the displacement equals 4 cm if 4T 3/2 − ln T = 11/2, T ≈ 1.272 s. vember 10, 2008 16:01 ”ISM ET chapter 5” Sheet number 38 Page number 262 black 20 3 20 3 t2 + 3t − 5;21. s(t) = (20t2 − 110t +120) dt = t − 55t2 +120t + C . But s = 0 when t = 0, so C = 0 and s = t − 55t2 +120t. 3 3 70 262 Moreover, a(t) = d v (t) = 40t − 110. Chapter 5 dt 180 180 25 0 25 130 6 0 –30 t) 0 a (t ) 0 6 0 6 –40 true –130 v (t ) s (t ) a (t ) 23. True; if a(t) = a0 then v (t) = a0 t + v0 . true 2 22. a(t) = 4t − 30, v (t) = 2t2 − 30t + 3, s(t) = t3 − 15t2 + 3t − 5; 3 25. False; consider v (t) = t on [−1, 1]. 1100 500 70 27. (a) The displacement is positive on (0, 5). 5.5 0 (0.75, 2.97) 25 0 –1100 0 –110 s (t ) 5 25 0 25 –30 v (t ) a (t ) –1.5 23. 263 24. true 9 Page number true; if a(t) = a0 then v (t) = a0 t + v0 black 5 (b) The displacement is − sin 5 + 5 cos 5 ≈ 4.877. 2 1.5 263 25. (a) The displacementtis positive on (0, 5). 29. false; consider v (t) = on [−1, 1] 26. true 0.5 5.5 27. (a) positive on (0, 0.74) and (2.97, 5), negative on (0.75, 2.97) 0 0 0 0 1 5 0 –1.5 (b) For 0 < T < 5 the displacement is disp = T /2 − sin(T ) + T cos(T ) 5 0 5 0 –0.3 (b) For 0 < T < 1 the displacement is 146 ￿ ￿ 12 1 disp = T− ln(10T + 1) 2 ￿ ￿200 3 1 1 1 3 (b) The displacement is 1 +26e−5 . + − 0.1 2 ln 10 − 2 T + T + T 200 2 4 20 400 , 1) 0, t < 4 ￿ 31. (a) a(t) = 0 0, t < 4 −10, t > 4 31. (a) a(t) = −10, t...
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This document was uploaded on 02/02/2014.

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