Question

A particle is projected with velocity $v_0$ along $x$ -axis. The deceleration on the particle is proportional to the square of the distance from the origin $i.e.a=-x^2$ . The distance at which particle stops is -

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

Answer 1

Answer: (D)

Given initial velocity = v₀
final velocity = 0
$a=-{\text{x}}^2$
now we know that $a=\frac{\text{dv}}{\text{dt}}$ and $\text{v}=\frac{\text{dx}}{\text{dt}}$
thus $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{dv}}{\text{dt}}\cdot \frac{\text{dx}}{\text{dx}}=\frac{\text{vdv}}{\text{dx}}$
$\frac{\text{vdv}}{\text{dx}}=-{\text{x}}^2$
$\text{vdv}=-{\text{x}}^2\text{dx}$
integrating we get
${\int }_{{\text{v}}_0}^0\text{vdv}={\int }_0^{\text{x}}-{\text{x}}^2\text{dx}$
${\left(\frac{{\left(\text{v}\right)}^2}{2}\right)}_{{\left(\text{v}\right)}_0}^0=-{\left[\frac{{\left(\text{x}\right)}^3}{3}\right]}_0^{\text{x}}$
$-\frac{{\text{v}}_0^2}{2}=-\frac{{\left(\text{x}\right)}^3}{3}\Rightarrow \text{x}={\left(\frac{3{\text{v}}_0^2}{2}\right)}^{\frac{1}{3}}$