Question

A particle moves along a straight line so that, distance $s$ in time $tsec$ is $s=t+6t^2–t^3$. After what time is the acceleration zero?

• A. $2sec$
• B. $3sec$
• C. $4sec$
• D. $6sec$

Answer 1

The correct option is A

$2sec$

Step 1: Given data

The distance travelled by the particle, $s=t+6t^2–t^3$

Acceleration,$\frac{d^{2}s}{d{t}^2}=0$

Step 2: Find the time taken by the particle before coming to rest

Acceleration is the rate of the change of the velocity with respect to the time. Acceleration is equal to the second derivative of the distance.

Velocity is the rate of change of the displacement with respect to the time.

$$\begin{gathered} \frac{ds}{dt}=1+12t–3t^2 \\ \frac{d^{2}s}{d{t}^2}=12–6t \\ \frac{d^{2}s}{d{t}^2}=0 \\ 12–6t=0 \\ t=2seconds \end{gathered}$$

Hence, option $\left(A\right)$ is correct.