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(A) Given that the displacement $S$ is proportional to the cube of time $t$,we can write $S = kt^3$,where $k$ is a constant.To find the velocity $v$,w

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Increasing with time

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(A) Given that the displacement $S$ is proportional to the cube of time $t$,we can write $S = kt^3$,where $k$ is a constant.To find the velocity $v$,we differentiate $S$ with respect to time $t$:$v = \frac{dS}{dt} = \frac{d}{dt}(kt^3) = 3kt^2$.To find the acceleration $a$,we differentiate the velocity $v$ with respect to time $t$:$a = \frac{dv}{dt} = \frac{d}{dt}(3kt^2) = 6kt$.Since $a = 6kt$,the acceleration $a$ is directly proportional to time $t$ $(a \propto t)$.Therefore,the magnitude of the acceleration increases with time.

The displacement of a body is given to be proportional to the cube of time elapsed. The magnitude of the acceleration of the body is

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