University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 160: 99

Answer

$$v(6)=\frac{2}{5}(m/sec)$$ $$a(6)=-\frac{4}{125}(m/sec^2)$$

Work Step by Step

$$s=\sqrt{1+4t}=(1+4t)^{1/2}$$ a) Find the formula for the particle's velocity and acceleration: - Velocity: $$v=s'=\frac{1}{2}(1+4t)^{-1/2}(1+4t)'=\frac{1}{2\sqrt{1+4t}}\times4=\frac{2}{\sqrt{1+4t}}$$ - Acceleration: $$a=v'=\frac{(2)'\sqrt{1+4t}-2(\sqrt{1+4t})'}{1+4t}=\frac{0\sqrt{1+4t}-2s'}{1+4t}$$ $$a=\frac{-2\times\frac{2}{\sqrt{1+4t}}}{1+4t}=-\frac{4}{(1+4t)\sqrt{1+4t}}$$ b) For $t=6$ sec: $$v(6)=\frac{2}{\sqrt{1+4\times6}}=\frac{2}{\sqrt{25}}=\frac{2}{5}(m/sec)$$ $$a(6)=-\frac{4}{(1+4\times6)\sqrt{1+4\times6}}=-\frac{4}{25\sqrt{25}}=-\frac{4}{125}(m/sec^2)$$
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