Question

A particle moves along a straight line such that its displacement at any time $t$ is given by $S=t^3–6t^2+3t+4metres$. The velocity when the acceleration is zero is

• A. $3m{s}^{-1}$
• B. $-12m{s}^{-1}$
• C. $42m{s}^{-1}$
• D. $-9m{s}^{-1}$

Answer 1

The correct option is D

$-9m{s}^{-1}$

Step 1: Given data

The distance travelled by the particle, $S=t^3–6t^2+3t+4metres$

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.

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

Step 3: Find velocity

Velocity (v) is rate of change of the distance or displacement with respect to the time,

$$\begin{gathered} v=\frac{ds}{dt} \\ v=3t^2-12t+3 \\ v=3{\left(2\right)}^2-12\times 2+3 \\ v=-9m/s \end{gathered}$$

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