Ch 5 solutions

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern