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(A) Given the displacement $s = \sqrt{t} = t^{1/2}$.The velocity $v$ is the derivative of displacement with respect to time:$v = \frac{ds}{dt} = \frac

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$(Velocity)^3$

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(A) Given the displacement $s = \sqrt{t} = t^{1/2}$.The velocity $v$ is the derivative of displacement with respect to time:$v = \frac{ds}{dt} = \frac{1}{2} t^{-1/2} = \frac{1}{2\sqrt{t}}$.From this,we can express $\sqrt{t}$ in terms of $v$:$\sqrt{t} = \frac{1}{2v}$,which implies $t = \frac{1}{4v^2}$.The acceleration $a$ is the derivative of velocity with respect to time:$a = \frac{dv}{dt} = \frac{d}{dt} (\frac{1}{2} t^{-1/2}) = \frac{1}{2} 	imes (-\frac{1}{2}) t^{-3/2} = -\frac{1}{4} t^{-3/2}$.Substitute $t = \frac{1}{4v^2}$ into the expression for $a$:$a = -\frac{1}{4} (\frac{1}{4v^2})^{-3/2} = -\frac{1}{4} (4v^2)^{3/2} = -\frac{1}{4} (8v^3) = -2v^3$.Therefore,the magnitude of acceleration is proportional to the cube of the velocity,i.e.,$a \propto v^3$.

$A$ particle moves in a straight line such that $s = \sqrt{t}$,then its acceleration is proportional to

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