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 v−t graph, we can say:
Area under v−t graph =0
Consider t as the time when area becomes zero, then
12×25×20−12×(t−25)×(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=vt−25
⇒v=4(t−25)
∴250−12×(t−25)×4(t−25)=0
250−2(t−25)2=0
(t−25)2=125
t−25=±5√5
t=25±5√5
As t>25, we have to take t=25+5√5
∴t=25+5×2.2=36 s