Chapter 3.4, Problem 20E

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 motion of particle and find value of t such that particle will

Change the direction.

Answer to Problem 20E

The correct answer is (a) $v(t)=\frac{ds}{dt}=-3t^2+14t-14$.

(b) $a(t)=\frac{d^{2}s}{d{t}^2}=-6t+14$

(c) At $t=1.451,3.215$ particle at rest.

(d) At $t=1.451,3.215$ particles change the direction.

Explanation of Solution

Given information: In this question, a particle moves along a line and its position at any time t is given, $s(t)=-t^3+7t^2-14t+8$ .using this we will find solution of given questions.

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:

(a) For instantaneous velocity,

  $\begin{array}{l}s(t)=-t^3+7t^2-14t+8\\ v(t)=\frac{ds}{dt}=-3t^2+14t-14\end{array}$

(b) For instantaneous acceleration,

  $\begin{array}{l}s(t)=-t^3+7t^2-14t+8\\ v(t)=\frac{ds}{dt}=-3t^2+14t-14\\ a(t)=\frac{d^{2}s}{d{t}^2}=-6t+14\end{array}$

(c) For finding the time, where particle at rest, means we will equate $V(t)=\frac{ds}{dt}=0$.

  $\begin{array}{l}s(t)=-t^3+7t^2-14t+8\\ v(t)=\frac{ds}{dt}=-3t^2+14t-14=0\\ t=\frac{-14\pm \sqrt{{14}^2-4*-3*-14}}{2*-3}\\ t=1.451,3.215\end{array}$

Here at $t=1.451\sec$ and at $t=3.215\sec$ particle at rest.

(d) For finding the time, where velocity of particle changes the direction means $V(t)=\frac{ds}{dt}=0$.

  $\begin{array}{l}s(t)=-t^3+7t^2-14t+8\\ v(t)=\frac{ds}{dt}=-3t^2+14t-14=0\\ t=\frac{-14\pm \sqrt{{14}^2-4*-3*-14}}{2*-3}\\ t=1.451,3.215\end{array}$

Here at $t=1.451\sec$ and at $t=3.215\sec$ particle change the direction. We can understand through graph also.

  

Velocity v/s time

Thus, the correct answer is (a) $v(t)=\frac{ds}{dt}=-3t^2+14t-14$

&emap; $\frac{ds}{dt}=-3t^2+14t-14$.

(b) $a(t)=\frac{d^{2}s}{d{t}^2}=-6t+14$

(c) At $t=1.451,3.215$ particle at rest.

(d) At $t=1.451,3.215$ particles change the direction