Question

The equation of motion of a particle moving along a straight line is $s=2t^3–9t^2+12t$, where the units of $s$and $t$ are cm and sec. The acceleration of the particle will be zero after

• A. $\frac{3}{2}sec$
• B. $\frac{2}{3}sec$
• C. $\frac{1}{2}sec$
• D. Never

Answer 1

The correct option is A

$\frac{3}{2}sec$

Step 1: Given data:

Equation of motion, $s=2t^3–9t^2+12t······\left(1\right)$

Step 2: Formula used:

The relationship between acceleration and displacement is-

$a=\frac{d^{2}s}{d{t}^2}$

Step 3: Calculation of time:

According to the formula, obtain the second derivative of equation $\left(1\right)$ to get the acceleration,

The first derivative of equation $\left(1\right)$ is,

$\begin{array}{rcl}\frac{ds}{dt}& =& \frac{d\left(2t^3-9t^2+12t\right)}{dt}\\ \frac{ds}{dt}& =& 6t^2-18t+12.......\left(2\right)\end{array}$

Again differentiate equation (2) with respect to time to obtain acceleration-

$\begin{array}{rcl}\frac{d^{2}s}{d{t}^2}& =& 12t-18\\ a& =& 12t-18\end{array}$

According to the problem, the particle's acceleration would be zero. So,

$\begin{array}{rcl}12t-18& =& 0\\ 12t& =& 18\\ t& =& \frac{3}{2}\end{array}$

Thus, the acceleration of the particle will be zero after $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.sec$.

Hence, option A is the correct answer.