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 - Page 233: 9

Answer

(a) After 2 seconds, the velocity of the rock is $7.56~m/s$ (upward) (b) The velocity on the way up is $6.245~m/s$ The velocity on the way down is $-6.245~m/s$

Work Step by Step

(a) $h = 15t-1.86t^2$ $v = 15-3.72t$ After 2 seconds: $v = 15-3.72(2) = 7.56$ After 2 seconds, the velocity of the rock is $7.56~m/s$ (upward) (b) $h = 15t-1.86t^2 = 25$ $-1.86t^2+15t - 25 = 0$ $1.86t^2-15t + 25 = 0$ We can use the quadratic formula to find the times $t$ when the rock is 25 meters high: $t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $t = \frac{-(-15) \pm \sqrt{(-15)^2-4(1.86)(25)}}{(2)(1.86)}$ $t = \frac{15 \pm \sqrt{225-186}}{3.72}$ $t = 2.3535~~$ or $~~t = 5.711$ We can find the velocity when the rock is 25 meters high on the way up: $v = 15-3.72(2.3535) = 6.245~m/s$ We can find the velocity when the rock is 25 meters high on the way down: $v = 15-3.72(5.711) = -6.245~m/s$ The velocity on the way up is $6.245~m/s$ The velocity on the way down is $-6.245~m/s$
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