A particle is moving in a straight line is at a distance x from a fixed point in straight line at time t seconds, where x = 2t³ – 12t + 11. What is average acceleration of particle at end of 3 seconds?

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