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A particle moves in a straight line and its velocity v at time t seconds is given byv = 3t2 – 4t + 5 cm/second. What will be the acceleration of the particle during first 3 seconds after the start?

A particle moves in a straight line and its velocity v at time t seconds is given byv = 3t2 – 4t + 5 cm/second. What will be the acceleration of the particle during first 3 seconds after the start? Correct Answer 14cm/sec2

n:Let f be the acceleration of the particle in time t seconds. Then, f = dv/dt = d(3t2 – 4t + 5)/dt = 6t – 4   ……….(1) Therefore, the acceleration of the particle at the end of 3 seconds, = t = 3 = (6*3 – 4) cm/sec2 = 14cm/sec2

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