A body of mass m starts from rest and acquire | vedclass

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(D) The body starts from rest,so initial velocity $u = 0$. Acceleration $a = \frac{v - u}{t_1} = \frac{v}{t_1}$. Distance covered in time $t$ is given

Vedclass · Accepted Answer

$\frac{1}{2}m\frac{v^2}{t_1^2}t^2$

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(D) The body starts from rest,so initial velocity $u = 0$. Acceleration $a = \frac{v - u}{t_1} = \frac{v}{t_1}$. Distance covered in time $t$ is given by $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}at^2 = \frac{1}{2}at^2$. Work done $W = F \cdot s = (ma) \cdot (\frac{1}{2}at^2) = \frac{1}{2}ma^2t^2$. Substituting $a = \frac{v}{t_1}$,we get $W = \frac{1}{2}m \left( \frac{v}{t_1} \right)^2 t^2 = \frac{1}{2}m \frac{v^2}{t_1^2} t^2$. Thus,the work done is proportional to $\frac{1}{2}m \frac{v^2}{t_1^2} t^2$.

$A$ body of mass $m$ starts from rest and acquires a velocity $v$ in time $t_1$. The work done in time $t$ is proportional to:

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