JEE Main · PhysicsHard

$A particle moves along the x-axis such that its acceleration is given by $a = -k x^2$, where $k$ is a positive constant. If the particle starts from rest at $x = x_0$, what is its velocity when it reaches the origin?$

Show correct answer & step-by-step solution

Correct answer: A — $\sqrt{\frac{2k}{3}} x_0^{3/2}$

Solution

Use the relation

a = v \frac{d v}{d x}

to integrate the acceleration function. Integrating

\int_{0}^{v} v^{'} d v^{'} = \int_{x_{0}}^{0} - k x^{2} d x

yields

\frac{v ^{2}}{2} = \frac{k x _{0}^{3}}{3}

, hence

v = \frac{2 k}{3} x_{0}^{3/2}