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(A) Let $v$ be the velocity and $s$ be the distance covered by the particle. Given that $v \propto \sqrt{s}$,which implies $v = k\sqrt{s}$ for some co

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Constant

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(A) Let $v$ be the velocity and $s$ be the distance covered by the particle. Given that $v \propto \sqrt{s}$,which implies $v = k\sqrt{s}$ for some constant $k$. The acceleration $a$ is given by $a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \frac{dv}{ds}$. Since $v = k\sqrt{s}$,we have $\frac{dv}{ds} = k \cdot \frac{1}{2\sqrt{s}} = \frac{k}{2\sqrt{s}}$. Substituting these into the expression for acceleration: $a = (k\sqrt{s}) \cdot \left( \frac{k}{2\sqrt{s}} \right) = \frac{k^2}{2}$. Since $k$ is a constant,$a = \frac{k^2}{2}$ is also a constant. Therefore,the acceleration of the particle is constant.

If the velocity of a moving particle is directly proportional to the square root of the distance it has covered,what is its acceleration?

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