A particle is moving in a straight line is at a distance x from a fixed point in the straight line at time t seconds, where x = 2t^3 – 12t + 11. What is the acceleration of the particle at the end of 2 seconds?

(a) 22 cm/sec^2

(b) 24 cm/sec^2

(c) 26 cm/sec^2

(d) 28 cm/sec^2

This question was addressed to me in an international level competition.

Enquiry is from Calculus Application in portion Application of Calculus of Mathematics – Class 12

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Right option is (a) 22 cm/sec^2

Explanation: We have, x = 2t^3 – 12t + 11 ……….(1)

Let v and f be the velocity and acceleration respectively of the particle at time t seconds.

Then, v = dx/dt = d(2t^3 – 12t + 11)/dt

= 6t^2 – 12 ……….(2)

And f = dv/dt = d(6t^2 – 12)/dt

= 12t ……….(3)

Putting the value of t = 2 in (3),

Therefore, the displacement of the particle at the end of 2 seconds,

12t = 12(2)

= 24 cm/sec^2