JEE MAIN - Physics (2022 - 25th July Morning Shift - No. 3)
A person moved from A to B on a circular path as shown in figure. If the distance travelled by him is $$60 \mathrm{~m}$$, then the magnitude of displacement would be :
(Given $$\left.\cos 135^{\circ}=-0.7\right)$$
_25th_July_Morning_Shift_en_3_1.png)
42 m
47 m
19 m
40 m
Explanation
Distance travelled = 60 m
$$\Rightarrow$$ Angle covered = 135$$^\circ$$
Displacement $$ = 2R\sin \left( {{{135^\circ } \over 2}} \right)$$
$$ = 2\left( {{{60} \over {135}} \times {{180} \over \pi }} \right){\left[ {{{1 - \cos (135^\circ )} \over 2}} \right]^{1/2}}$$
$$ = 2\left( {{{80} \over \pi }} \right){(0.85)^{1/2}}$$
$$ \approx 47$$ m
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