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A runner starts from O and goes to O following path OQRO in lhr. What is net displacement and average speed?
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Verified Answer
The correct answer is:
$0,3.57 \mathrm{~km} / \mathrm{hr}$
$\because$ Runner starts from O and goes to O .
So, net displacement is zero.
$\begin{aligned} \because \text { Average speed } & =\frac{\text { Total distance }}{\text { Total time }} \\ & =\frac{\mathrm{OQ}+\mathrm{QR}+\mathrm{RO}}{\text { total time }}\end{aligned}$
$\begin{aligned} & \frac{1 \mathrm{~km}+(2 \pi \mathrm{r})\left(\frac{1}{4}\right)+1 \mathrm{~km}}{1 \mathrm{hr}} \\ = & 2+\frac{\pi}{2}=3.57 \mathrm{~km} / \mathrm{hr}\end{aligned}$
So, net displacement is zero.
$\begin{aligned} \because \text { Average speed } & =\frac{\text { Total distance }}{\text { Total time }} \\ & =\frac{\mathrm{OQ}+\mathrm{QR}+\mathrm{RO}}{\text { total time }}\end{aligned}$
$\begin{aligned} & \frac{1 \mathrm{~km}+(2 \pi \mathrm{r})\left(\frac{1}{4}\right)+1 \mathrm{~km}}{1 \mathrm{hr}} \\ = & 2+\frac{\pi}{2}=3.57 \mathrm{~km} / \mathrm{hr}\end{aligned}$
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