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 = 2t3 – 12t + 11. What is the acceleration of the particle at the end of 2 seconds?

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 = 2t3 – 12t + 11. What is the acceleration of the particle at the end of 2 seconds? Correct Answer 22 cm/sec2

We have, x = 2t3 – 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(2t3 – 12t + 11)/dt = 6t2 – 12  ……….(2) And f = dv/dt = d(6t2 – 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/sec2
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