A particle is moving along a straight line and its position is given by the relation x=( t3-6t2-15t+40)mm. Find:- (a). The time at which velocity is zero. (b). Position & displacement of the particle at that point. (c). Acceleration for the particle at that line.

To find the time at which velocity is zero, we need to first find the expression for velocity by differentiating the given position equation with respect to time.

(a) Velocity (v) can be found by differentiating the position equation x = t^3 - 6t^2 - 15t + 40 with respect to time (t):

v = dx/dt = d/dt(t^3 - 6t^2 - 15t + 40)

Differentiating term by term, we get:

v = 3t^2 - 12t - 15

To find the time at which velocity is zero, we need to solve for the value of t when v = 0:

0 = 3t^2 - 12t - 15

We can solve this quadratic equation to find the values of t when velocity is zero.

(b) To find the position and displacement of the particle at that point, substitute the value of t when velocity is zero into the given position equation x = t^3 - 6t^2 - 15t + 40.

(c) Acceleration (a) can be found by differentiating the velocity equation v = 3t^2 - 12t - 15 with respect to time (t):

a = dv/dt = d/dt(3t^2 - 12t - 15)

Differentiating term by term, we get:

a = 6t - 12

To find the acceleration for the particle at the specific time when velocity is zero, substitute the value of t into the acceleration equation.