The kinetic energy of a body moving along a s | vedclass

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(D) From the graph,the kinetic energy $(K.E.)$ is directly proportional to time $(t)$:$K.E. = \frac{1}{2} m v^{2} = k t$ (where $k$ is a constant)$v^{

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inversely proportional to velocity.

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(D) From the graph,the kinetic energy $(K.E.)$ is directly proportional to time $(t)$:$K.E. = \frac{1}{2} m v^{2} = k t$ (where $k$ is a constant)$v^{2} \propto t \Rightarrow v \propto \sqrt{t}$Let $v = \alpha \sqrt{t}$,where $\alpha$ is a constant.The acceleration $a$ is the rate of change of velocity:$a = \frac{d v}{d t} = \frac{d}{d t} (\alpha \sqrt{t}) = \frac{\alpha}{2 \sqrt{t}}$The force $F$ acting on the body is given by Newton's second law:$F = m a = m \left( \frac{\alpha}{2 \sqrt{t}} \right)$Since $v = \alpha \sqrt{t}$,we have $\sqrt{t} = \frac{v}{\alpha}$.Substituting this into the expression for force:$F = m \left( \frac{\alpha}{2 (v / \alpha)} \right) = \frac{m \alpha^{2}}{2 v}$Thus,$F \propto \frac{1}{v}$.Therefore,the force is inversely proportional to the velocity.

The kinetic energy of a body moving along a straight line varies with time as shown in the figure. The force acting on the body is:

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