JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 19)
A person starts his journey from centre 'O' of the park and comes back to the same position following path OPQO as shown in the figure. The radius of path taken by the person is 200 m and he takes 3 min 58 sec to complete his journey. The average speed of the person is _____________ ms$$-$$1. (take $$\pi$$ = 3.14)
_30th_June_Morning_Shift_en_19_1.png)
Хариулт
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Тайлбар
$$
\begin{aligned}
& V_{a v}=\frac{S_{\text {total }}}{T_{\text {total }}}=\frac{200+200+\frac{2 \pi \times 200}{4}}{238} \\\\
&=\frac{400+100 \pi}{238}=\frac{714}{238}=3 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$