The correct choice is (b) 30 cm/sec^2
The explanation is: 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 acceleration of the particle at the end of 2 seconds,
12t = 12(2)
= 24 cm/sec^2
Now putting the value of t = 2 in (2),
We get the displacement of the particle at the end of 2 seconds,
6t^2 – 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,
6t^2 – 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/sec^2