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Question

A velocity - time graph is shown above in figure (i) and (ii) find the acceleration and displacement.
101941_4268829ea7544952a868b253c9e064d6.png

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Solution

Figure (i) represents a velocity-time graph when the body starts from rest, and its velocity increases at a uniform rate.
The slope of the graph AC. i.e., ABBC gives the acceleration of body.
Acceleration=change in velocityTotal time
=(160)ms15s=3.2ms2
The area of triangle ABC, gives the displacement.
Thus, displacement in 5 s
12×AB×BC=12×16ms1×5s
=40m
Figure (ii) represents velocity - time graph, where the body is initially not at rest.
The slope of graph AC, i.e., (ABBC) gives the acceleration.
Acceleration=ABBC(255)ms14s=204ms2
=5ms2
The distance covered by the body in specified direction is area of trapezium ECAD.
Displacement=12(CE+AD)×ED
=12(5+25)ms1×4s=60m

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