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# A particle moves along a straight line and its velocity depends on time as v=4t−t2.Then for first 5 sec

A
average speed is 135 m/s
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B
average speed is 53 m/s
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C
average speed is 3215 m/s
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D
average speed is 35 m/s
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Solution

## The correct option is A average speed is 135 m/sGiven, v(t)=4t−t2, we will check if the particle has changed its direction or not within the asked time interval, i.e. the time when velocity of the particle becomes zero. ⇒ 4t−t2=0⇒ t=0,4 sec Also, we know that xf−xi=∫t0v(t)dt=∫t0(4t−t2)dt ⇒ xf−xi=2t2−t33 Let particle starts from origin i.e xi=0 So, at t=0,xf=0, at t=4,xf=2×42−433=96−643=323 m at t=5,xf=2×52−533=150−1253=253 m Hence, average speed is 323+(323−253)5=135 m/s  Suggest Corrections  0      Similar questions
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