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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 distance travelled by 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 distance travelled by the particle during first 3 seconds after the start? Correct Answer 24 cm

Let, x be the distance travelled by the particle in time t seconds. Then, v = dx/dt = 3t2 – 4t + 5 Or ∫dx = ∫ (3t2 – 4t + 5)dt So, on integrating the above equation, we get, x = t3 – 2t2 + 5t + c where, c is a constant.  ……….(1) Therefore, the distance travelled by the particle at the end of 3 seconds, = t = 3 – t = 0 = (33 – 2*32 + 5*3 + c) – c = 24 cm.

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