Chapter 3.4, Problem 21E

To determine

To calculate: Find (a) the instantaneous velocity at any time t.

(b) the acceleration of the particle at any time t.

(c) the time when particle at rest.

(d) the time when particle change the direction.

Answer to Problem 21E

The correct answer is (a) $\frac{ds}{dt}=(t-2)(3t-10)$.

(b) $\frac{d^{2}s}{d{t}^2}=6t-16$.

(c) $t=2\sec ,\frac{10}{3}\sec$

(d) $t=2\sec ,\frac{10}{3}\sec$

Explanation of Solution

Given information: Position of particle at any time t is given $s={(t-2)}^2(t-4)$ .where s is measured in m, and t is in sec.

Formula used: We know that, Instantaneous velocity is $V(t)=\frac{ds}{dt}$ .Instantaneous acceleration is $a(t)=\frac{d^{2}s}{d^{2}t}$ . And for time where particle change the direction is $V(t)=\frac{ds}{dt}=0$.

Calculation:

1. For instantaneous velocity

  $\begin{array}{}$ s={(t-2)}^2(t-4)$\end{array}$

&emap; $\frac{ds}{dt}=2(t-2)(t-4)+{(t-2)}^2$dsdt=(t−2)(2t−8+t−2)dsdt=(t−2)(3t−10)

2. For instantaneous acceleration

  $\begin{array}{}$ s={(t-2)}^2(t-4)$\end{array}$

&emap; $\frac{ds}{dt}=2(t-2)(t-4)+{(t-2)}^2$dsdt=(t−2)(2t−8+t−2)dsdt=(t−2)(3t−10)d2sdt2=(t−2)*3+(3t−10)d2sdt2=6t−16

(c) Time when particle at rest,

  $\begin{array}{}$ s={(t-2)}^2(t-4)$\end{array}$

&emap; $\frac{ds}{dt}=2(t-2)(t-4)+{(t-2)}^2$dsdt=(t−2)(2t−8+t−2)dsdt=(t−2)(3t−10)=0t=2sec,103sec

(d) See the below graph carefully,

  

Red graph shows distance-time and blue graph shows velocity-time. From the blue graph we can say particle change the direction $t=2\sec ,\frac{10}{3}\sec$ , because here velocity is zero.