The position of a particle is given by the function x=(5t3−8t2+12)m, where t is in s. at what time does the particle reach its minimum velocity?

The particle reach its minimum velocity at time 1.06 sec.

The function is given as

x=5t^3-8t^2+12

Differentiating the above equation with respect to time, to obtain the velocity

dx/dt=v=15t^2-16t

For maximum and minimum values, put dx/dt=0

15t^2-16t=0

On solving the equation, t=0, 1.06

Therefore at time t=1.06 sec, the particle has the minimum value of velocity.

what are the statements? x