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

• A. $\frac{3}{2}sec$
• B. $\frac{2}{3}sec$
• C. $\frac{1}{2}sec$
• D. $1sec$

Answer 1

The correct option is A

$\frac{3}{2}sec$

Step 1: Given data:

Equation of motion, $s=2t^3–9t^2+12t$

Let at $\text{'}t\text{'}sec$ the acceleration of the particle will be zero.

Step 2: Formula used:

The relation between acceleration and displacement is given by-

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

Where $a$ is acceleration, $s$ is a displacement.

Step 3: Calculation of time

The given equation of motion is-

$s=2t^3–9t^2+12t\dots \dots \dots \dots \dots \left(1\right)$

Using the formula, the acceleration can be calculated as

Differentiate the equation of motion in equation (1) with respect to time.

$\begin{array}{rcl}\frac{ds}{dt}& =& 6t^2–18t+12\dots \dots \dots \dots \left(2\right)\end{array}$

Again differentiating the equation (2) with respect to time.

$\begin{array}{rcl}\frac{d^{2}s}{d{t}^2}& =& 12t–18\dots \dots \dots \left(3\right)\end{array}$

We know that from the relation between acceleration and displacement that

$a=\frac{d^{2}s}{d{t}^2}\dots \dots \dots \dots \dots \left(4\right)$

Equating the equation (3) and equation (4)

$\begin{array}{rcl}a& =& 12t–18\end{array}$

But it is given that the acceleration of the particle will be zero at time $\text{'}t\text{'}sec$

Substituting the known values that the acceleration $a=0$ for finding time.

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

Hence, option A is the correct answer.