University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.5 - Derivatives of Trigonometric Functions - Exercises - Page 152: 55


- Velocity $=-\sqrt2(m/s)$. - Speed $=\sqrt2(m/s)$. - Acceleration $=\sqrt2(m/s^2)$ - Jerk $=\sqrt2(m/s^3)$

Work Step by Step

$$s=f(t)=2-2\sin t$$ a) Velocity $v$: $$v(t)=f'(t)=(2-2\sin t)'=0-2\cos t=-2\cos t$$ So, $$v(\pi/4)=-2\cos(\pi/4)=-2\times\frac{\sqrt2}{2}=-\sqrt2(m/s)$$ b) Speed: The speed at time $t=\pi/4$ $=|v(\pi/4)|=\sqrt2(m/s)$. c) Acceleration $a$: $$a(t)=v'(t)=(-2\cos t)'=-2(-\sin t)=2\sin t$$ So, $$a(\pi/4)=2\sin(\pi/4)=2\times\frac{\sqrt2}{2}=\sqrt2(m/s^2)$$ d) Jerk $j$: $$j(t)=a'(t)=(2\sin t)'=2\cos t$$ Therefore, $$j(\pi/4)=2\cos(\pi/4)=2\times\frac{\sqrt2}{2}=\sqrt2(m/s^3)$$
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