The position of a particle is given by the function x=(4t3−6t2+12)m, where t is in s.
a.) at what time does the particle reach its minimum velocity
b.) what is (vx)min
c.) at what time is the acceleration zero

Answer

given,

x = 4 t³ - 6 t² + 12

velocity,

For minimum velocity calculation we have differentiate it and put it equal to zero.

.........(1)

putting it equal to zero

24 t - 12 =0

t = 0.5 s

At t = 0.5 s velocity will be minimum.

b) minimum velocity

    v = 12t² -12 t

    v = 12 x 0.5² -12 x 0.5

    v = -3 m/s

c) derivative of velocity w.r.t. time is acceleration

from equation 1

     a = 24 t - 12

time at which acceleration will be zero

     0 = 24 t - 12

     t = 0.5 s

At t = 0.5 s acceleration will be zero.

c) up

hope this