A body of mass m starts moving from rest alon | vedclass

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(A) Given the velocity $v = a\sqrt{s}$.Since $v = \frac{ds}{dt}$,we have $\frac{ds}{dt} = a\sqrt{s}$.Separating variables,we get $\int s^{-1/2} ds = \

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$\frac{1}{8} m a^4 t^2$

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(A) Given the velocity $v = a\sqrt{s}$.Since $v = \frac{ds}{dt}$,we have $\frac{ds}{dt} = a\sqrt{s}$.Separating variables,we get $\int s^{-1/2} ds = \int a dt$.Integrating both sides,$2\sqrt{s} = at + C$. Since the body starts from rest at $t=0$,$s=0$,so $C=0$.Thus,$\sqrt{s} = \frac{at}{2}$,which implies $s = \frac{a^2 t^2}{4}$.Differentiating $s$ with respect to $t$,we get velocity $v = \frac{ds}{dt} = \frac{a^2 t}{2}$.Differentiating $v$ with respect to $t$,we get acceleration $a_{acc} = \frac{dv}{dt} = \frac{a^2}{2}$.According to the work-energy theorem,the total work done is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2} m v^2$.Substituting $v = \frac{a^2 t}{2}$,we get $W = \frac{1}{2} m \left( \frac{a^2 t}{2} \right)^2 = \frac{1}{2} m \left( \frac{a^4 t^2}{4} \right) = \frac{1}{8} m a^4 t^2$.

$A$ body of mass $m$ starts moving from rest along the $x-$axis such that its velocity varies as $v = a\sqrt{s}$,where $a$ is a constant and $s$ is the distance covered by the body. The total work done by all the forces acting on the body in the first $t$ seconds after the start of the motion is:

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