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Question

A graph of acceleration versus time of a particle starting from rest at t=0 is as shown in figure. The speed of the particle at t=14 s is


A
2 ms1
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B
34 ms1
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C
20 ms1
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D
42 ms1
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Solution

The correct option is B 34 ms1
We know that area under the at graph gives change in velocity

Total area under the graph from t=0 to t=14 s is given by
A= Area under the graph from t=0 to t=12 s [Trapezium] - Area under the graph from t=12 s to t=14 s [triangle]
A=12×4×(12+6)12×2×2
A=362=34 m/s
Now, A= Change in velocity
A=v140
v14=34 m/s

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