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 a particle is described as $s=2-3t+4t^3$

Step 2: Formula used-

The relation between the velocity and displacement is given as follows-

$v=\frac{ds}{dt}$

Acceleration will be calculated using the formula-

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

Step 3: Compute the time $t$ where velocity is zero

It is understood that velocity is the rate of change of displacement. It means $v=\frac{ds}{dt}$. So,

$\begin{array}{rcl}s& =& 2-3t+4t^3\\ \frac{ds}{dt}& =& -3+12t^2\\ v& =& -3+12t^2----\left(1\right)\end{array}$

So, the time $t$ where velocity is zero,

$\begin{array}{rcl}-3+12t^2& =& 0\\ 12t^2& =& 3\\ t^2& =& \frac{1}{4}\\ t& =& \frac{1}{2}\end{array}$

Step 4: Compute acceleration of the particle

We know that the change in velocity with respect to time represents acceleration. So, $a=\frac{dv}{dt}$

$\begin{array}{rcl}v& =& -3+12t^2\\ \frac{dv}{dt}& =& 24t\\ a_{t=\frac{1}{2}}& =& 24\times \frac{1}{2}\\ a_{t=\frac{1}{2}}& =& 12\\ & & \end{array}$

So, the acceleration of the particle at the point would be $12unit$.

Hence, option D is the correct answer.