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Q.

An elevator is going up. The variation in the velocity of the elevator is as given in the graph. What is the height to which the elevator takes the passengers take velocity in m/s and time in sec

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a

3.6 m

b

28.8 m

c

36.0 m

d

72.0 m

answer is C.

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Detailed Solution

The velocity-time graph of the elevator is given, and we know that the area under the curve on the time axis gives the displacement.

Height to which the elevator takes the passengers = Area of the trapezium

⇒H = 1/2​ × Sum of the parallel sides × Distance between them

      =1/2 × (12 + 8) × 3.6 = 1/2 ​× 20 × 3.6 = 36m

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