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: 36

Answer

The detailed explanations are below.
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Work Step by Step

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