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?

Solution

1. Given the acceleration $$\begin{array}{rl}a& =v\frac{dv}{dx}\\ & =-k{x}^2\end{array}$$, we set up the integral ${\int }_0^{v}vdv={\int }_{x_0}^0-k{x}^{2}dx$.
2. Evaluating the integrals gives $\frac{1}{2}{v}^2={\left[-\frac{k}{3}{x}^3\right]}_{x_0}^0$.
3. Simplifying the expression results in $\frac{1}{2}{v}^2=\frac{k}{3}{x}_0^3$.
4. Solving for $v$, we find $$\begin{array}{rl}v& =\sqrt{\frac{2k}{3}{x}_0^3}\\ & =\sqrt{\frac{2k}{3}}{x}_0^{3/2}\end{array}$$.
5. Hence the answer is (A).