Question

A particle moves along a straight line according to the law $s=16–2t+3t^3$, where $s$ is the distance of the particle from a fixed point at the end of $tsec$. The acceleration of the particle at the end of $2s$ is

• A. $36m/s^2$
• B. $34m/s^2$
• C. $36m$
• D. None of these

Answer 1

The correct option is A

$36m/s^2$

Step1: Given data

Distance, $s=16–2t+3t^3$

Time, $t=2s$

Step 2: Find Acceleration, $a$

Acceleration is the rate of change of the velocity, $v$ with respect to the time and the velocity is the rate of the change of the distance or displacement with respect to the time.

$$\begin{gathered} s=16-2t+3t^3 \\ v=\frac{ds}{dt} \\ v=-2+9t^2 \\ a=\frac{dv}{dt} \\ a=\frac{d}{dt}\left(-2+9t^2\right) \\ a=18t \end{gathered}$$

At,

$$\begin{gathered} t=2s, \\ a=18\times 2 \\ a=36m{s}^{-2} \end{gathered}$$

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