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A particle start from rest. Its acceleration (a) versus time (t) is as shown in the figure. The maximum speed of the particle will be

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a
110 m/s
b
55 m/s
c
550 m/s
d
660 m/s

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detailed solution

Correct option is B

The area under acceleration-time graph gives change in velocity. As acceleration is zero at the end of 11 sec, we havevmax= Area of ∆OAB         = 12×11×10 = 55 m/s


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