A particle moves in a straight line so that $s=\sqrt{t}$, then its acceleration is proportional to
Answer & explanation
Correct answer: option 1
We have,
$s=\sqrt{t} \Rightarrow \frac{d s}{d t}=\frac{1}{2 \sqrt{t}}$ and $\frac{d^2 s}{d t^2}=-\frac{1}{4 t^{3 / 2}}$
$\Rightarrow \frac{d^2 s}{d t^2}=-\frac{1}{4}\left(\frac{2 d s}{d t}\right)^3=-2\left(\frac{d s}{d t}\right)^3$
Hence, Acceleration ∝ (Velocity)3.