1Used thatgetting 38.__ but answer given is 36.2 sec!!
Fig shows the v-t graph of particle moving in straight line.Find the time when particle returns to the starting point
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UP 0 DOWN 0 0 3
3 Answers
106use the fact that when the particle returns to the starting point, the (algebraic) area under the graph is zero.. i.e. part above x-axis is positive and part below x-axis is negative,
1Area above the x-axis= 0.5(25*20)= 250 sq units.
Now as the slope of the Graph after 20 s is -4,
from trigonometry,
we get the velocity when the displacement is 0= -4t
Area of the graph under the x-axis=0.5(t*4t)=2t2
For displacement to be 0,
Area above the x-axis= area below the x-axis
=> 2t2=250.
thus, t=5√5=11.25 approx
Therefore time after which particle returns to starting point=t+25= 36.25