Question

The distance traveled $s$ by an accelerated particle of mass $M$ is given by the following relation (in MKS units) $s=6t+3t^2$. The velocity of the particle after $2s$ in the corresponding unit.

• A. $6$
• B. $12$
• C. $18$
• D. $24$

Answer 1

The correct option is C

$18$

Step 1: Given data

Distance travelled by particle, $s=6t+3t^2$

Time $\left(t\right)=2sec$

Step 2: Compute the velocity

It is understood that the rate of change of distance with respect to time is known as velocity.

It means, $v=\frac{ds}{dt}$

So,

$\begin{array}{rcl}\frac{d}{dt}s& =& \frac{d}{dt}\left(6t+3t^2\right)\\ \frac{ds}{dt}& =& 6+6t\\ v& =& 6+6t\\ v_{t=2}& =& 6+6\times 2\\ v_{t=2}& =& 18m/sec\end{array}$

The velocity of the particle after $2s$ is, $\begin{array}{rcl}{v}_{t=2}& =& 18m/sec\end{array}$.

Hence, option (C) is the correct answer.