A particle starts from 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 distance from origin at end of 4 seconds?

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,
= [x]_{t = 4} = (1/4)4⁴ ? (5/6)4³ + 5(4) [putting t = 4 in (4)]
= 30(2/3) cm