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 velocity of the particle?

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 velocity of the particle? Correct Answer 29 cm/sec

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 – 5tdt 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) Thus, the velocity of the particle at the end of 4 seconds, = t = 4 = (43 – (5/2)42 + 5 ) cm/sec = 29 cm/sec