A particle is moving in a straight line and its distance s cm from a fixed point in the line after t seconds is given by s = 12t – 15t^2 + 4t^3. What is the acceleration of the particle after 3 seconds?

(a) 41 cm/sec^2

(b) 42 cm/sec^2

(c) 43 cm/sec^2

(d) 44 cm/sec^2

I got this question in an internship interview.

The origin of the question is Calculus Application in portion Application of Calculus of Mathematics – Class 12

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The correct option is (b) 42 cm/sec^2

Easiest explanation: We have, s = 12t – 15t^2 + 4t^3 ……….(1)

Differentiating both side of (1) with respect to t we get,

(ds/dt) = 12 – 30t + 12t^2

And d^2s/dt^2 = -30 + 24t

So, acceleration of the particle after 3 seconds is,

[d^2s/dt^2]t = 3 = – 30 + 24(3)

= 42 cm/sec^2.