Chapter 3.4, Problem 23E

To determine

To find: The body’s acceleration each time when velocity is zero.

Answer to Problem 23E

Body’s acceleration, $a(1)=-6m/{\sec }^2,a(3)=6m/{\sec }^2$

Explanation of Solution

Given information: In this question, the position of body at any time is given $s(t)=t^3-6t^2+9t$ .we will find body acceleration when velocity is zero.

Formula used: We know that, Instantaneous velocity is $V(t)=\frac{ds}{dt}$ .Instantaneous acceleration is $a(t)=\frac{d^{2}s}{d^{2}t}$.

Calculation: For finding acceleration we will find first instantaneous velocity and equate to zero, then find time where velocity is zero. And then we will find acceleration for individual value of $t$.

  $\begin{array}{l}s(t)=t^3-6t^2+9t\\ v(t)=\frac{ds}{dt}=3t^2-12t+9=0\\ t=\frac{12\pm \sqrt{{12}^2-4*3*9}}{2*3}\\ t=1\sec ,3\sec \\ a(t)=\frac{d^{2}s}{d{t}^2}=6t-12\\ a(1)=-6m/{\sec }^2,a(3)=6m/{\sec }^2\end{array}$

Thus, the correct answer is, body’s acceleration, $a(1)=-6m/{\sec }^2,a(3)=6m/{\sec }^2$