Question

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

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

Answer 1

The correct option is E

Constant

Step 1: Given information:

When a particle moves, its displacement increases proportionately to its gained velocity squared,

Step 2: Formula used and apply the condition:

Know that if displacement $\alpha {\left(velocity\right)}^2$,

It can be written as,

$s\alpha v^2$

Therefore,

$s=k{v}^2$

Now differentiate the function,

$$\begin{gathered} \frac{ds}{dt}=2kv\left(\frac{dv}{dt}\right) \\ \left(\frac{dv}{dt}\right)=\frac{1}{2k}=cons\tan t \end{gathered}$$

Therefore the acceleration is constant.

Hence, the correct option is (E).