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^2This question was addressed to me in an international level competition.Enquiry is from Calculus Application in portion Application of Calculus of Mathematics – Class 12

A particle is moving in a straight line is at a distance x from a fixe

1.	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^2This question was addressed to me in an international level competition.Enquiry is from Calculus Application in portion Application of Calculus of Mathematics – Class 12
Answer» 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

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