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

(a) 30(4/3)

(b) 30(2/3)

(c) 30

(d) Unpredictable

This question was posed to me during an online exam.

My query is from Linear First Order Differential Equations topic in division Differential Equations of Mathematics – Class 12

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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