Right choice is (c) 24cm/sec^2
Explanation: We have, x = t^3 + 6t^2 – 15t + 18
Let, ‘v’ be the velocity of the particle and ‘a’ be the acceleration of the particle at the end of t seconds. Then,
v = dx/dt = d/dt(t^3 + 6t^2 – 15t + 18)
So, v = 3t^2 + 12t – 15
Therefore, a = dv/dt = d/dt(3t^2 + 12t – 15)
So, a = 6t + 12
Thus, acceleration of the particle at the end of 2 seconds is,
[dv/dt]t = 2 = 6(2) + 12 = 24cm/sec^2.