Question

If a particle moves such that the displacement is proportional to the square of the velocity acquired, then its acceleration is

• A. Proportional to $s^2$
• B. Proportional to $\frac{1}{s^2}$
• C. Proportional to $\frac{1}{s}$
• D. A constant

Answer 1

The correct option is D

A constant

Solution:

Step 1: Given information:

If a particle moves in a way that its displacement increases in direct proportion to its speed,

Step 1: Formula used and apply the condition:

Know that according to the given condition,

$s\alpha v^2$

Now,

$v=k\sqrt{s}$

Now differentiate the formula,

$$\begin{gathered} \frac{dv}{dt}=k\left(\frac{1}{2\sqrt{s}}\right)\times \left(\frac{ds}{dt}\right)Because,\left(\frac{ds}{dt}\right)=k\sqrt{s} \\ \frac{dv}{dt}=k\left(\frac{1}{2\sqrt{s}}\right)\times k\sqrt{s} \\ \frac{dv}{dt}=k^2/2 \end{gathered}$$

Therefore this is a constant.

Hence, the correct option is (D).