Question

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

Answer 1

Let displacement be $s$

So, $s$ is prop to $t^3$

$s=k{t}^3$ Where k is the proportionality constant

On differentiation

$\frac{ds}{dt}=3k{t}^2$

$\Rightarrow v=3k{t}^2$

On again differentiation

$\frac{d^{2}s}{d{t}^2}=6kt$

$\Rightarrow a=6kt$

So,$a$ is proportional to $t$

Where $6k$ is the proportionality constant

Acceleration is directly proportional to time, it varies linearly with time and increases with respect to time.