Question

A particle is projected with velocity $v_0$ along axis. The deceleration is proportional to the square of the distance from the origin, i.e., $a=\alpha x^2,$ the distance from the particle stop is :

• A. $\frac{\sqrt{3}{v}_0}{2\alpha }$
• B. ${\left(\frac{\sqrt{3}{v}_0}{2\alpha }\right)}^{1/3}$
• C. $\frac{\sqrt{2}{v}_0}{3\alpha }$
• D. ${\left(\frac{3{v_0}^2}{2\alpha }\right)}^{1/3}$

Answer 1

The correct option is D ${\left(\frac{3{v_0}^2}{2\alpha }\right)}^{1/3}$

Given, initial velocity $=v_0$

Final velocity $=0$

Deceleration , $a=-\alpha x^2$ ...(i)

Let the distance travelled by the particle be s.

Now, we know that

$a=\frac{dv}{dt}=\frac{dv}{dt}\times \frac{dt}{dx}=\frac{vdv}{dx}$

or $a=v\frac{dv}{dx}$

From Eqs. (i) and (ii), we get

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

or $vdx=-\alpha x^{2}dx$

On integrating with limit $v_0\to 0$ and $0\to s$

${\int }_{v_0}^{0}vdv={\int }_0^s-\alpha x^{2}dx$

or ${\left(\frac{v^2}{2}\right)}^{0}n=-\alpha {\left(\frac{x^3}{3}\right)}^s$

$\frac{v_0^2}{2}=\frac{\alpha (s{)}^3}{3}$

$\frac{v_0^2}{2}=\frac{\alpha s^3}{3}$

$\frac{3v_0^2}{2\alpha }=s^3$

$s={\left(\frac{3v_0^2}{2\alpha }\right)}^{1/3}$