Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.7 - Rates of Change in the Natural and Social Sciences - 3.7 Exercises: 8

Answer

a. $100 ft$ b. $16\frac{ft}{s}$ and $-16\frac{ft}{s}$

Work Step by Step

a. $v(t) = \frac{d(s)}{dx}$ $v'(t) = \frac{d(80t)}{dt} - \frac{16t^{2}}{dt}$ $v'(t) = 80 - 32t$ $v'(t) = 0$ $0 = 80 - 32t$ Solve for $t$: $32t = 80$ $t = \frac{80}{32}$ $t = 2.5 s$ $s = 80t - 16t^{2}$ $s = 80(2.5) - 16(2.5)^2$ $s = 200 - 16(6.25)$ $s = 200 - 100$ $s = 100 ft$ b. $s = 80t - 16t^{2}$ $96 = 80t -16t^{2}$ $80t -16t^{2}-96 = 0$ Simplify by $-16$: $-16(t^{2} - 5t +6) = 0$ $t^{2} - 5t + 6 = 0$ Now factorize: $(t-3)(t-2) = 0$ $t = 3$ and $t=2$ $v(t) = \frac{ds}{dt}$ $v(t) = 80 - 32t$ $v(2) = 80 - 32(2)$ $v(2) = 80 - 64$ $v(2) = 16\frac{ft}{s}$ $v(3) = 80 - 32(3)$ $v(3) = 80 - 96$ $v(3) = -16\frac{ft}{s}$
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