Question

a=x^2 find velocity when x is 2 m given at x=0 body is at rest

Answer 1

Step 1: Given that:

Acceleration in the body;

$a=x^2$

At $x=0$ , the velocity of the body(u) = 0

Step 2: Calculation of the velocity of the body:

In differential form, acceleration ($a$ ) is given as; $a=\frac{dv}{dt}$

Thus,

$\frac{dv}{dt}=x^2$ ...........(1)

Now, the differential form of velocity is $v=\frac{dx}{dt}$

Therefore, from equation (1), we have;

$\frac{dv}{dx}\times \frac{dx}{dt}=x^2$

$\frac{dv}{dx}\times v=x^2$

$vdv=x^{2}dx$

Now, integrating the above for position x=0 to x=2m and for velocity v=0 to v, we have;

${\int }_0^{v}vdv={\int }_0^{2m}{x}^{2}dx$

${\left[\frac{v^2}{2}\right]}_0^v={\left[\frac{x^3}{3}\right]}_0^2$

$[\frac{v^2}{2}-0]=[\frac{2^3}{3}-0]$

$\frac{v^2}{2}=\frac{8}{3}$

$v^2=\frac{16}{3}$

$v=\frac{4}{\sqrt{3}}$

$v=\frac{4\sqrt{3}}{3}m{s}^{-1}$

Thus,

The velocity of the body will be $\frac{4\sqrt{3}}{3}m{s}^{-1}$ .