A particle moves along the x-axis such that its velocity $v$ varies with position $x$ as $v=\alpha \sqrt{x}$, where $\alpha$ is a positive constant. Find the acceleration of the particle.

Solution

1. The acceleration $a$ is given by $a=v\frac{dv}{dx}$.
2. Given $v=\alpha x^{1/2}$, we find $\frac{dv}{dx}=\frac{\alpha }{2}{x}^{-1/2}$.
3. Substituting these into the acceleration formula gives $a=(\alpha x^{1/2})(\frac{\alpha }{2}{x}^{-1/2})$.
4. Simplifying the expression yields $a=\frac{{\alpha }^2}{2}$.
5. Hence the answer is (A).