The velocity u and displacement r of a body a | vedclass

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Question

$u^{2}=\mathrm{kr} \quad$ or $\quad u=\sqrt{\mathrm{kr}}$
$\frac{\mathrm{du}}{\mathrm{dt}}=\sqrt{\mathrm{k}} \frac{1}{2} \mathrm{r}^{-1 / 2} \frac{\ma

Vedclass · Accepted Answer

$\frac{k}{2}\,{r^o}$

Vedclass · Answer

$u^{2}=\mathrm{kr} \quad$ or $\quad u=\sqrt{\mathrm{kr}}$
$\frac{\mathrm{du}}{\mathrm{dt}}=\sqrt{\mathrm{k}} \frac{1}{2} \mathrm{r}^{-1 / 2} \frac{\mathrm{dr}}{\mathrm{dt}}=\sqrt{\mathrm{k}} \frac{1}{2} \mathrm{r}^{-1 / 2} \cdot \mathrm{u}$

$=\sqrt{\mathrm{k}} \frac{1}{2} \mathrm{r}^{-1 / 2} \cdot \sqrt{\mathrm{k} \mathrm{r}}^{1 / 2}=\frac{\mathrm{k}}{2} \mathrm{r}^{0}$

Velocity after $1\, sec$ $=0+\frac{\mathrm{k}}{2} \mathrm{r}^{0} 	imes 1=\frac{\mathrm{k}}{2} \mathrm{r}^{0}$

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 $1\, second$ ? (Given that the displacement is zero at $t = 0$)

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