Chapter 3.4, Problem 24E

To determine

To calculate: The body’s speed each time when speed is zero.

Answer to Problem 24E

Body’s speed, $v(1)=0m/\sec ,v(2)=-1m/\sec .$

Explanation of Solution

Given information: In this question, the velocity of body at any time is given $v(t)=2t^3-9t^2+12t-5$ .we will find body’s speed when acceleration is zero.

Formula used: We know that, Instantaneous acceleration is $a(t)=\frac{dv}{dt}$.

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

  $\begin{array}{l}v(t)=2t^3-9t^2+12t-5\\ a(t)=\frac{dv}{dt}=6t^2-18t+12=0\\ t=\frac{18\pm \sqrt{{18}^2-4*6*12}}{2*6}\\ t=1\sec ,2\sec \\ v(1)=2-9+12-5=0m/\sec \\ v(2)=16-36+24-5=-1m/\sec .\end{array}$

Thus, the correct answer is, Body’s speed, $v(1)=0m/\sec ,v(2)=-1m/\sec .$