Right choice is (b) 30(2/3)
Easiest explanation: 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 = 3t^2 – 5t
Or, dv = 3t^2 dt – 5t dt
Or, ∫dv = 3∫t^2 dt – 5∫t dt
Or, v = t^3 – (5/2)t^2 + c ……….(1)
Given, v = 5, when t = 0; hence putting these values in equation (1) we get, c = 5
Thus v = t^3 – (5/2)t^2 + 5
Or, dx/dt = t^3 – (5/2)t^2 + 5 ………..(2)
Or, dx = t^3 dt – (5/2)t^2 dt + 5 dt
Integrating this we get,
x = (1/4)t^4 – (5/2)t^3/3 + 5t + k ……….(3)
By the problem, x = 0, when t = 0; hence, from (3) we get, k = 0.
Thus, x = (1/4)t^4 – (5/6)t^3 + 5t ……….(4)
Thus, the velocity of the particle at the end of 4 seconds,
= [x]t = 4 = (1/4)4^4 – (5/6)4^3 + 5(4) [putting t = 4 in (4)]
= 30(2/3) cm