JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 2)
The velocity time graph of a body moving in a straight line is shown in the figure.
_24th_January_Evening_Shift_en_2_1.png)
The ratio of displacement to distance travelled by the body in time 0 to 10s is :
1 : 1
1 : 3
1 : 4
1 : 2
Explanation
From $v$-t graph
Displacement $=$ Area under curve considering sign also.
Distance $=$ Area under curve considering only magnitude
Displacement $= \Sigma$area $=8$$ \times $$2-4$$ \times $$2+4$$ \times $$4-2$$ \times $$4=16 \mathrm{~m}$
Distance $=\Sigma \mid$ area $\mid=48 \mathrm{~m}$
Displacement : distance $=1: 3$
Displacement $=$ Area under curve considering sign also.
Distance $=$ Area under curve considering only magnitude
Displacement $= \Sigma$area $=8$$ \times $$2-4$$ \times $$2+4$$ \times $$4-2$$ \times $$4=16 \mathrm{~m}$
Distance $=\Sigma \mid$ area $\mid=48 \mathrm{~m}$
Displacement : distance $=1: 3$
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