A particle is moving in a straight line. The variation of position $$x$$ as a function of time $$t$$ is given as $$x = (t^3 - 6t^2 + 20t + 15)$$ m. The velocity of the body when its acceleration becomes zero is:

We need to find the velocity of a particle when its acceleration is zero, given $$x = t^3 - 6t^2 + 20t + 15$$.

Find the velocity.

$$v = \frac{dx}{dt} = 3t^2 - 12t + 20$$

Find the acceleration.

$$a = \frac{dv}{dt} = 6t - 12$$

Find when acceleration is zero.

$$6t - 12 = 0 \implies t = 2$$ s

Find velocity at $$t = 2$$.

$$v(2) = 3(4) - 12(2) + 20 = 12 - 24 + 20 = 8$$ m/s

The correct answer is Option (2): 8 m/s.