Question

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

• A. square of the velocity
• B. cube of the displacement
• C. cube of the velocity
• D. square of the displacement

Answer 1

The correct option is C cube of the velocity

$s=\sqrt{1+t}$

$\Rightarrow v=\frac{ds}{dt}=\frac{1}{2\sqrt{1+t}}$

$\Rightarrow a=\frac{dv}{dt}=-\frac{1}{4(1+t{)}^{3/2}}$

$\Rightarrow a=-\frac{1}{4(\sqrt{1+t}{)}^3}$

$\Rightarrow a=-2v^3$

$\Rightarrow a\propto v^3$