If a particle moving along a line follows the law $s=\sqrt{1+t}$, then the accelertion is proportional to

cube of the velocity

We have,

$s=\sqrt{1+t}$

$\Rightarrow \frac{d s}{d t}=\frac{1}{2 \sqrt{1+t}}$

$\Rightarrow \frac{d s}{d t}=\frac{1}{2 s}$

$\Rightarrow \frac{d^2 s}{d t^2}=-\frac{1}{2 s^2} \frac{d s}{d t}=-\frac{1}{2 s^2} \times \frac{1}{2 s}=-\frac{1}{4 s^3}$

$\Rightarrow \frac{d^2 s}{d t^2}=-2 \times\left(\frac{1}{2 s}\right)^3=-2\left(\frac{d s}{d t}\right)^3$