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.