The variation in velocity of a particle moving along a straight line is shown in the figure. If the particle is moving along x-axis with initial position x=10m, find the final position of particle.
A
35m
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B
25m
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C
45m
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D
55m
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Solution
The correct option is A35m For finding displacement, area below the time axis is considered with −ve sign, due to velocity in opposite direction.
Initial position, xi=10m
Displacement (S)=area under(v−t)graph S=Area of trapezium−Area of triangle ⇒S=[12×(4+10)×5]−[12×4×5] ∴S=25m
In terms of position x, displacement is given as: S=xf−xi ∴xf=xi+25=10+25=35m