Question

In a particular straight line motion, the distance ($x$) covered is proportional to the square root of the time taken. Then the acceleration of the particle is proportional to

• A. $\sqrt{v}$
• B. $v$
• C. $v^2$
• D. $v^3$

Answer 1

The correct option is D $v^3$

Distance covered $x\propto \sqrt{t}$

$⟹x=k\sqrt{t}$

Velocity of particle $v=\frac{dx}{dt}=\frac{k}{2\sqrt{t}}$

Acceleration of particle $a=\frac{dv}{dt}=\frac{-k{t}^{-3/2}}{4}$

$⟹a=\frac{-k}{4}\times \frac{2^3{v}^3}{k^3}$

$⟹a\propto v^3$