The coordinate of a particle in meters is given by x(t) = 16t − 3.0t3, where the time t is in seconds. The particle is momentarily at rest at t

t = -4/3 s, t = 4/3 s

Explanation:

I am ASSUMING that you want to know the time(s) the particle is at rest.

and that your position equation is x(t) = 16t − 3.0t³

IF those assumptions are true

THEN

velocity is the derivative of position

v(t) = 16 - 9t²

the particle will be at rest when velocity is zero

0 = 16 - 9t²

9t² = 16

t² = 16/9

t = ± 4/3

a=?

Vf=15.0m/s

Vi=6.0m/s

t=3.0s

a=Vf-Vi/t

a=15.0-6.0/3.0

a=9/3

a=3m/s^2

Explanation: