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Question

(C) Given the relation $u^2 = kr$,we have $u = \sqrt{k} \cdot r^{1/2}$.Since $u = \frac{dr}{dt}$,we can write $\frac{dr}{dt} = \sqrt{k} \cdot r^{1/2}$

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$\frac{k}{2}$

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(C) Given the relation $u^2 = kr$,we have $u = \sqrt{k} \cdot r^{1/2}$.Since $u = \frac{dr}{dt}$,we can write $\frac{dr}{dt} = \sqrt{k} \cdot r^{1/2}$.Separating the variables,we get $r^{-1/2} \, dr = \sqrt{k} \, dt$.Integrating both sides with the initial condition $r = 0$ at $t = 0$: $\int_{0}^{r} r^{-1/2} \, dr = \int_{0}^{t} \sqrt{k} \, dt$.This gives $[2r^{1/2}]_{0}^{r} = \sqrt{k} \cdot t$,so $2\sqrt{r} = \sqrt{k} \cdot t$,which implies $\sqrt{r} = \frac{\sqrt{k}}{2} t$.Substituting $\sqrt{r}$ back into the velocity equation $u = \sqrt{k} \cdot \sqrt{r}$:$u = \sqrt{k} \cdot (\frac{\sqrt{k}}{2} t) = \frac{k}{2} t$.At $t = 1 \, s$,the velocity $u = \frac{k}{2} (1) = \frac{k}{2}$.

The velocity $u$ and displacement $r$ of a body are related as $u^2 = kr$,where $k$ is a constant. What will be the velocity after $1 \, s$? (Given that the displacement is zero at $t = 0$)

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