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: 35

Answer

The detailed explanations are below.
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Work Step by Step

$s=t^3-6t^2+7t$, $0\le t\le4$ 1) Find $v$ and $a$: $$v=\frac{ds}{dt}=\frac{d}{dt}(t^3-6t^2+7t)=3t^2-12t+7$$ $$a=\frac{dv}{dt}=\frac{d}{dt}(3t^2-12t+7)=6t-12$$ 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 $0$ at $t=0$, goes up before reaching its maximum value at $t=0.709$. Then it goes down before being back to $s=0$ at $t=1.586$. This means that from $t=0$ to $t=0.709$, the object moved forwards, but then it stopped at $t=0.709$, changed direction and moved backwards continuously and was back to its original position at $t=1.586$. - From $t=1.586$ to $t=3.291$, the curve continued to decreases before reaching its minimum value at $t=3.291$. Then it went up again until $t=4$. This shows that the object continued to move backwards when it was back at origin at $t=1.586$ up until $t=3.291$. Then it stopped and changed direction one more time, moved forward until $t=4$. Overall, the object is furthest from the origin at $t=3.291$. The line $v$: - The line $v$ started at $7$ at $t=0$. From $t=0$ to $t=0.709$, $v$ is positive but it went down continuously and reached $0$ at $t=0.709$. Then it continued to go down (now $v$ is negative) until it reached the bottom at $t=2$. The object was moving forward at the highest speed at the starting time. Then as it continued to move forward $(v\gt0)$, it went more and more slowly before stopping at $t=0.709$, which is also when it changed direction the first time. Then from $t=0.709$ to $t=2$, the object moved backwards $(v\lt0)$ at an increasing velocity, reaching its maximum at $t=2$. - From $t=2$ to $t=3.291$, $v$ is negative but it went up continuously and reached $0$ at $t=3.291$. Then it continued to go up (now $v$ is positive) until it reached its maximum again at $t=4$. After reaching its highest speed of moving backwards, the object still moved backward $(v\lt0)$ but its speed got slower and slower and became $0$ and the object stopped at $t=3.291$, which is where the object changed direction the second time. Then from $t=3.291$ to $t=4$, the object moved forward $(v\gt0)$ at an increasing velocity, reaching its maximum speed again at $t=4$. The line $a$: The line $a$ goes up throughout the whole interval. From $t=0$ to $t=2$, $a$ is negative but goes up continuously. This shows that the velocity of the object is going down but more and more slowly as $t$ increases. From $t=2$ to $t=4$, $a$ is positive and goes up continuously. This shows that the velocity of the object is going up more and more quickly as $t$ increases.
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