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 147: 34


The detailed explanations are below.

Work Step by Step

$s=t^2-3t+2$, $0\le t\le5$ 1) Find $v$ and $a$: $$v=\frac{ds}{dt}=\frac{d}{dt}(t^2-3t+2)=2t-3$$ $$a=\frac{dv}{dt}=\frac{d}{dt}(2t-3)=2$$ The graphs of $s, v$ and $a$ are shown below. 2) We examine each element of the graph: The curve $s$: - The curve $s$ starts from $2$ at $t=0$, decreases constantly before reaching its minimum value at $t=1.5$. So from $t=0$ to $t=1.5$, the object moves backwards. - From $t=1.5$ to $t=5$, the curve increases constantly before reaching its maximum value, which is $12$ at $t=5$. So during these times, the object moves forward. That being said, at $t=1.5$, the object has changed its direction. Overall, the object is furthest from the origin at $t=5$, when its position reaches the value $12$. The line $v$: - The line $v$ increases constantly on $[0,5]$. From $t=0$ to $t=1.5$, $v$ is negative and increases constantly before reaching $0$ at $t=1.5$. The object was moving backwards at the highest speed at the starting time. Then as it continued to move backwards $(v\lt0)$, it went more and more slowly before stopping at $t=1.5$, which is also when it changed direction. - From $t=1.5$ to $t=5$, $v$ is positive and increases constantly. After the object stopped moving backwards at $t=1.5$, it started to move forward $(v\gt0)$ and its velocity rose constantly and reached its maximum at $t=5$. The line $a$: The line $a$ is a horizontal line $a=2$ throughout the whole interval. Since $a$ is positive, we know that the movement trend of velocity is increase on the whole interval, and since $a$ does not change its value at all, this increase trend is constant.
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