Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises: 49

Answer

(a) The velocity as function of $t$ is $v(t)=3t^2-3$ The acceleration as function of $t$ is $a(t)=6t$ (b) The acceleration after 2s is $12m/s^2$. (c) When the velocity is 0, the acceleration is $6m/s^2$.

Work Step by Step

The equation of motion $$s=t^3-3t$$ (a) The first derivative of function $s$ would be the velocity as function of $t$. So, the velocity as function of $t$ is $$v(t)=s'=\frac{d}{dt}(t^3)-3\frac{d}{dt}(t)$$$$v(t)=3t^2-3$$ (v is in $m/s$) Similarly, the second derivative of function $s$ would be the acceleration as function of $t$. So, the acceleration as function of $t$ is $$a(t)=s''=3\frac{d}{dt}(t^2)-\frac{d}{dt}(3)$$$$a(t)=3\times2t-0=6t$$ (a is in $m/s^2$) (b) To find the acceleration after $2s$, we only need to apply $t=2$ to the acceleration function: $$a(2)=6\times2=12(m/s^2)$$ That means the acceleration after 2s is $12m/s^2$ (c) First, we need to find after how many seconds, the velocity of the particle would be $0$. In other words, what is the value of $t$ so that $v(t)=0$. $$v(t)=0$$$$3t^2-3=0$$$$t^2=1$$ That means $t=1$ or $t=-1$. However, since $t$ refers to time here, $t\geq0$. Therefore, $t=1$. Or after 1 second, the velocity of the particle is 0. Now, we would apply $t=1$ to the function of $a(t)$ to find the acceleration when the velocity is 0: $$a(1)=6\times1=6(m/s^2)$$ Therefore, when the velocity of 0, the acceleration is $6m/s^2$
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