Question

If a particle moving along a line follows the law $s=\sqrt{1+t}$ then the acceleration 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

We have, $s=\sqrt{1+t}$
$\Rightarrow \frac{ds}{dt}=\frac{1}{2\sqrt{1+t}}\Rightarrow \frac{ds}{dt}=\frac{1}{2s}\Rightarrow \frac{d^{2}s}{d{t}^2}=-\frac{1}{2s^2}\frac{ds}{dt}=-\frac{1}{2s^2}\times \frac{1}{2s}=-\frac{1}{4s^3}\Rightarrow \frac{d^{2}s}{d{t}^2}=-2\times {\left(\frac{1}{2s}\right)}^3=-{\left(\frac{ds}{dt}\right)}^3.$
Hence (c) is the correct answer.