JEE MAIN - Physics (2020 - 5th September Evening Slot - No. 3)
The velocity (v) and time (t) graph of a body in
a straight line motion is shown in the figure.
The point S is at 4.333 seconds. The total
distance covered by the body in 6 s is :
_5th_September_Evening_Slot_en_3_1.png)
12 m
11 m
$${{49} \over 4}$$ m
$${{37} \over 3}$$ m
Explanation
4.333 sec = $${{13} \over 3}$$ sec
Distance = area under the v-t graph
= Area of Parallelogram + Area of triangle
= $${1 \over 2}\left( 4 \right)\left( {{{13} \over 3} + 1} \right)$$ + $${1 \over 2}\left( {6 - {{13} \over 3}} \right) \times 2$$
= $${{37} \over 3}$$ m
Distance = area under the v-t graph
= Area of Parallelogram + Area of triangle
= $${1 \over 2}\left( 4 \right)\left( {{{13} \over 3} + 1} \right)$$ + $${1 \over 2}\left( {6 - {{13} \over 3}} \right) \times 2$$
= $${{37} \over 3}$$ m
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