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(B) Given that the velocity $v$ is proportional to the cube root of displacement $x$: $v = k x^{1/3}$,where $k$ is a constant. We know that accelerati

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inversely proportional to its velocity

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(B) Given that the velocity $v$ is proportional to the cube root of displacement $x$: $v = k x^{1/3}$,where $k$ is a constant. We know that acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$. First,find $\frac{dv}{dx}$: $\frac{dv}{dx} = k \cdot \frac{1}{3} x^{-2/3} = \frac{k}{3 x^{2/3}}$. Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula: $a = (k x^{1/3}) \cdot \left( \frac{k}{3 x^{2/3}} \right) = \frac{k^2}{3 x^{1/3}}$. Since $v = k x^{1/3}$,we can write $x^{1/3} = \frac{v}{k}$. Substituting this into the expression for $a$: $a = \frac{k^2}{3 (v/k)} = \frac{k^3}{3v}$. Therefore,$a \propto \frac{1}{v}$,which means the acceleration is inversely proportional to its velocity.

If the velocity $v$ of a particle moving on a straight line is proportional to the cube root of its displacement $x$,then its acceleration $a$ is

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