A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t2 – 5t)cm/sec2. What will be the distance from the origin at the end of 4 seconds?

A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t2 – 5t)cm/sec2. What will be the distance from the origin at the end of 4 seconds? Correct Answer 30(2/3)

Let x cm be the distance of the moving particle from the origin and v cm/sec be its velocity at the end of t seconds. Then the acceleration of the particle at time t seconds = dv/dt and its velocity at that time = v = dx/dt. By question, dv/dt = 3t2 – 5t Or, dv = 3t2 dt – 5t dt Or, ∫dv = 3∫t2 dt – 5∫t dt Or, v = t3 – (5/2)t2 + c ……….(1) Given, v = 5, when t = 0; hence putting these values in equation (1) we get, c = 5 Thus v = t3 – (5/2)t2 + 5 Or, dx/dt = t3 – (5/2)t2 + 5 ………..(2) Or, dx = t3 dt – (5/2)t2 dt + 5 dt Integrating this we get, x = (1/4)t4 – (5/2)t3/3 + 5t + k ……….(3) By the problem, x = 0, when t = 0; hence, from (3) we get, k = 0. Thus, x = (1/4)t4 – (5/6)t3 + 5t ……….(4) Thus, the velocity of the particle at the end of 4 seconds, = t = 4 = (1/4)44 – (5/6)43 + 5(4) = 30(2/3) cm
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