The kinetic energy of a particle moving along a straight line is proportional to the time ‘t’ of its travel. Then its acceleration is proportional to

Given,

$\frac{1}{2}m{v}^2\propto t$

$\frac{1}{2}m{v}^2=kt$

$v=\sqrt{\frac{2k}{m}}\times t^{\frac{1}{2}}\_\_\_\_\_\_\left(1\right)$

The acceleration,

differentiating the equation 1,

$\frac{dv}{dt}=\sqrt{\frac{2k}{m}}\times \frac{1}{2}{\left(t\right)}^{\frac{-1}{2}}$

$\frac{dv}{dt}=\sqrt{\frac{2k}{m}}\times \frac{1}{2\sqrt{t}}$

$a=\frac{1}{\sqrt{t}}\sqrt{\frac{k}{2m}}$

The relationship between acceleration and velocity is straightforward.

As a result, the acceleration is inversely proportional to the time $\sqrt{t}.$

Hence the correct answer is $\frac{1}{\sqrt{t}}.$
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