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
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options
Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience
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