University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.4 - The Derivative as a Rate of Change - Exercises - Page 144: 1


a) The body's displacement is $-2m$. The average velocity is $-1m/s$. b) At $t=0$: speed $=3m/s$, acceleration $=2m/s^2$ At $t=2$: speed $=1m/s$, acceleration $=2m/s^2$ c) The body changes direction at $t=3/2s$.

Work Step by Step

$$s(t)=t^2-3t+2$$ a) The body's displacement, $\Delta s$, over the interval $[0,2]$ is $$\Delta s=s(2)-s(0)=(2^2-3\times2+2)-(0^2-3\times0+2)$$ $$\Delta s=0-2=-2 m$$ The average velocity, $v_a$, of the body over the interval $[0,2]$ is $$v_a=\frac{\Delta s}{2-0}=\frac{-2}{2}=-1m/s$$ b) We first look for the velocity of the body at time $t$, $v_t$, and the acceleration of the body at time $t$, $a_t$: $$v(t)=\frac{ds}{dt}=\frac{d}{dt}(t^2-3t+2)=2t-3$$ $$a(t)=\frac{dv}{dt}=\frac{d}{dt}(2t-3)=2$$ We now examine the speed and acceleration at 2 endpoints of $[0,2]$: - At $t=0$: $$v(0)=2\times0-3=-3m/s$$ So the speed of the body is $|v(0)|=3m/s$ $$a(0)=2m/s^2$$ - At $t=2$: $$v(2)=2\times2-3=1m/s$$ So the speed of the body is $|v(2)|=1m/s$ $$a(2)=2m/s^2$$ c) We know that if - $v(t)\gt0$: the body moves forward. - $v(t)\lt0$: the body moves backward. So the body will change direction when $v(t)$ changes sign. To find where $v(t)$ changes sign, we first locate the point where $v(t)=0$ then examine the sign before and after that point to determine whether $v(t)$ changes sign there. $$v(t)=0$$ $$2t-3=0$$ $$t=\frac{3}{2}s$$ On $[0,3/2)$ $v(t)\lt0$ and on $(3/2,2]$, $v(t)\gt0$. Therefore, there is a change of sign in $v(t)$ at $t=3/2s$, which also means the body has changed direction.
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