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Question

The figure shows the v - t graph of a particle moving in straight line. Find the time (in s) when particle returns to the starting point.

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Solution

The particle returns to its starting point, when the net displacement of the particle is zero.

As the net displacement of the particle is given by area under vt graph, we can say:

Area under vt graph =0
Consider t as the time when area becomes zero, then
12×25×2012×(t25)×(v)=0
Here, v is the velocity of particle at time t.
Now we need to calculate v in terms of t.
For that, we can equate slopes of both adjacent triangles:
205=vt25
v=4(t25)

25012×(t25)×4(t25)=0
2502(t25)2=0
(t25)2=125
t25=±55
t=25±55

As t>25, we have to take t=25+55
t=25+5×2.2=36 s

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