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 velocity of the particle?

(a) 27cm/sec

(b) 28 cm/sec

(c) 29 cm/sec

(d) 30 cm/sec

The question was posed to me in exam.

This question is from Linear First Order Differential Equations topic in portion Differential Equations of Mathematics – Class 12

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Correct choice is (c) 29 cm/sec

To elaborate: 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 – 5tdt

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)

Thus, the velocity of the particle at the end of 4 seconds,

= [v]t = 4 = (4^3 – (5/2)4^2 + 5 ) cm/sec [putting t = 4 in (2)]

= 29 cm/sec