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 average acceleration of the particle at the end of 3 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 average acceleration of the particle at the end of 3 seconds? Correct Answer 30 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 acceleration of the particle at the end of 2 seconds, 12t = 12(2) = 24 cm/sec2 Now putting the value of t = 2 in (2), We get the displacement of the particle at the end of 2 seconds, 6t2 – 12 = 6(2)2 – 12 = 12 cm/sec  ……….(4) And putting the value of t = 3 in (2), We get the displacement of the particle at the end of 3 seconds, 6t2 – 12 = 6(3)2 – 12 = 42 cm/sec  ……….(5) Thus, change in velocity is, (5) – (4), =42 – 12 = 30cm/sec. Thus, the average acceleration of the particle at the end of 3 seconds is, = (change of velocity)/time = (30 cm/sec)/1 sec = 30 cm/sec2