Question

A point initially at rest moves along $X-axis$. Its acceleration varies with time as $a=(6t+5)m{s}^{-2}$. If it starts from origin, the distance covered in $2s$ is

• A. $20m$
• B. $18m$
• C. $16m$
• D. $25m$

Answer 1

The correct option is B

$18m$

Step 1: Given data

Acceleration of the point, $a=(6t+5)m{s}^{-2}$

Initial velocity of particle, $u=0m{s}^{-1}$

Time taken by the particle, $t=2s$

Step 2: Find the velocity of the particle

As, acceleration, $a=\frac{dv}{dt}$ [$v$ is the velocity]

Therefore, $\frac{dv}{dt}=6t+5,$

$dv=(6t+5)dt$

On integrating it,

$$\begin{gathered} {\int }_0^{v}dv={\int }_0^t(6t+5)dt \\ v=\frac{6t^2}{2}+5t+c \\ v=3t^2+5t+c \end{gathered}$$

At $t=0$, $v$ will be also $0$. So,

$$\begin{gathered} 0=0+0+c \\ c=0 \end{gathered}$$

Therefore,

Now, $$\begin{gathered} v=3t^2+5t \\ \frac{ds}{dt}=3t^2+5t \\ ds=(3t^2+5t)dt \\ \int ds={\int }_0^2(3t^2+5t)dt \\ s={\overline{)t^3+\left(\frac{5}{2}\right)t^2}}_0^2 \\ s=8+10 \\ =18m \end{gathered}$$ [ As$v=\frac{ds}{dt}$, $s$ is distance]

Hence, option ($B$) is correct.