Question

The motion of a particle is described as $s=2-3t+4t^3$. What is the acceleration of the particle at the point where its velocity is zero?

• A. $0$
• B. $4unit$
• C. $8unit$
• D. $12unit$

Answer 1

The correct option is D

$12unit$

Step 1: Given data

The motion of particles, $s=2-3t+4t^3$

Step 2: Formula used

The relation between the velocity and displacement is given by:

$velocity\left(v\right)=\frac{ds}{dt}$

And the relation between velocity and acceleration is given by:

$\alpha =\frac{dv}{dt}$

Step 3: Compute the time where velocity is zero.

It is understood that the rate of change of motion with respect to time is known as velocity.

$s=2-3t+4t^3$……………….(1)

So, differentiate the equation $\left(1\right)$ with respect to time to obtain the velocity.

$\begin{array}{rcl}v& =& \frac{ds}{dt}=-3+12t^2\\ v& =& -3+12t^2······\left(2\right)\end{array}$

Substitute $v=0$ to obtain the time at which velocity becomes zero.

$\begin{array}{rcl}-3+12t^2& =& 0\\ 12t^2& =& 3\\ t^2& =& \frac{3}{12}\\ t& =& \pm \frac{1}{2}\\ t& =& \frac{1}{2}\end{array}$

Time can not be negative thus taking the positive value of time.

Step 4: Compute the acceleration at the point where velocity becomes zero

To obtain the acceleration, differentiate the equation $\left(2\right)$ with respect to time again,

$\begin{array}{rcl}\frac{dv}{dt}& =& 0+24t\\ \alpha & =& 24t·····\left(3\right)\end{array}$

Substitute $t=\frac{1}{2}$ in equation $\left(3\right)$ to obtain the acceleration at the point where velocity is zero.

$$\begin{gathered} \alpha =24\times \frac{1}{2} \\ \alpha =12 \end{gathered}$$

Therefore, the acceleration of the particle at the point where its velocity is zero will be $12unit$.

Hence, option D is the correct answer.