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(D) Given that the displacement $x$ is proportional to the cube of time $t$,we can write: $x = K t^3$,where $K$ is a constant. To find the velocity $v

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$a \propto t$

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(D) Given that the displacement $x$ is proportional to the cube of time $t$,we can write: $x = K t^3$,where $K$ is a constant. To find the velocity $v$,we differentiate the displacement with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(K t^3) = 3 K t^2$. To find the acceleration $a$,we differentiate the velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(3 K t^2) = 6 K t$. Since $6$ and $K$ are constants,we have $a \propto t$. Therefore,the acceleration of the particle is directly proportional to time.

The displacement of a particle is proportional to the cube of time elapsed. How does the acceleration of the particle depend on time?

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