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. $\sqrt{\frac{3 v_{0}}{2 }}$
• B. $\left(\frac{v_{0}^{2}}{3}\right)^{1/3}$
• C. $\left(\frac{2 v_{0}^{2}}{3}\right)^{1/3}$
• D. $\left(\frac{3 v_{0}^{2}}{2}\right)^{1/3}$

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
$\displaystyle \int _{\text{v}_{0}}^{0} \text{vdv} = \displaystyle \int _{0}^{\text{x}} - \text{x}^{2} \text{dx}$
$\left(\frac{\left(\text{v}\right)^{2}}{2}\right)_{\left(\text{v}\right)_{0}}^{0} = - \left[\frac{\left(\text{x}\right)^{3}}{3}\right]_{0}^{\text{x}}$
$- \frac{\text{v}_{0}^{2}}{2} = - \frac{\left(\text{x}\right)^{3}}{3} \, ⇒ \, \text{x} = \left(\frac{3 \text{v}_{0}^{2}}{2}\right)^{\frac{1}{3}}$