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(C) Given $K \propto t$,we can write $K = \lambda t$,where $\lambda$ is a constant.Since $K = \frac{1}{2}mv^2$,we have $\frac{1}{2}mv^2 = \lambda t$,w

Vedclass · Accepted Answer

$(ii), (iv)$

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(C) Given $K \propto t$,we can write $K = \lambda t$,where $\lambda$ is a constant.Since $K = \frac{1}{2}mv^2$,we have $\frac{1}{2}mv^2 = \lambda t$,which implies $v = \sqrt{\frac{2\lambda}{m}} t^{1/2}$.The acceleration $a$ is given by $a = \frac{dv}{dt} = \sqrt{\frac{2\lambda}{m}} \cdot \frac{1}{2} t^{-1/2} = \sqrt{\frac{\lambda}{2m}} t^{-1/2}$.The force $F = ma = m \cdot \sqrt{\frac{\lambda}{2m}} t^{-1/2} = \sqrt{\frac{\lambda m}{2}} \cdot \frac{1}{\sqrt{t}}$.Thus,$F \propto \frac{1}{\sqrt{t}}$,which matches statement $(ii)$.Now,expressing $F$ in terms of speed $v$: Since $v \propto t^{1/2}$,we have $t^{1/2} \propto v$. Substituting this into the force equation,$F \propto \frac{1}{v}$.Thus,$F$ is inversely proportional to the speed of the body,which matches statement $(iv)$.Therefore,statements $(ii)$ and $(iv)$ are correct.

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