Question

A particle is accelerating at $a=2x$ and initially it was at $x=2\text{m}$ moving with $3\text{m/s}$. Then find out the velocity at position $x=5\text{m}$.

• A. $7\text{m/s}$
• B. $\sqrt{31}\text{m/s}$
• C. $\sqrt{28}\text{m/s}$
• D. $\sqrt{51}\text{m/s}$

Answer 1

The correct option is D $\sqrt{51}\text{m/s}$

Given:
Acceleration, $a=2x$

$\text{At}t=0;x=2\text{m};v=3\text{m/s}$

Since, we know that

$a=v\frac{dv}{dx}=2x$

$\Rightarrow vdv=2xdx$

Integrating both side, we get

${\int }_{v=3}^{v}vdv={\int }_{x=2}^{x}2xdx$

$\Rightarrow {\left(\frac{v^2}{2}\right)}_3^v={\left(x^2\right)}_2^x$

$\Rightarrow \frac{v^2-9}{2}=x^2-4\Rightarrow v^2=2x^2-8+9$

$\therefore v^2=2x^2+1$

So at $x=5\text{m}$

$v^2=(2\times 5^2)+1=51$

$\Rightarrow v=\sqrt{51}\text{m/s}$

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