Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Section 4.2 - The Natural Exponential Function - 4.2 Exercises - Page 342: 25

Answer

(a)$v(0)=0$. (b) $v(5)\approx113.78ft/s$ $v(10)\approx155.64ft/s$ (c) See the image below. (d) The terminal velocity of the diver is $180ft/s$

Work Step by Step

So, we have a function of velocity $v(t)$ in a given time $t$: $v(t)=180(1-e^{-0.2t})$ (a) Initial velocity is the velocity at $t=0$ $v(0)=180(1-e^{-0.2\times0})=180(1-e^0)=180(1-1)=180\times0=0$ Which is logical, because at $t=0$ the sky diver hasn't started a motion yet, so the velocity is 0. (b) $t=5$ $v(5)=180(1-e^{-0.2\times5})=180(1-e^{-1})=180-\frac{180}{e})\approx113.78ft/s$ $t=10$ $v(10)=180(1-e^{-0.2\times10})=180(1-e^{-2})=180(1-\frac{1}{e^2})=180-\frac{180}{e^2}\approx155.64ft/s$ (c) We can graph the following function using a graphing calculator. See the image above. (d) As we can clearly see from the graph the velocity $v(t)$ is approaching $180$ as $t$ gets bigger. So the terminal velocity of the diver is $180ft/s$
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