Question

The position of an object as a function of time is given as x = At3 + Bt2 + Ct + D. The constants are A = 2.3m/s3, B = 1.8m/s2, C = −4.3m/s, and
D = 2 m.(a) What is the velocity of the object at t = 12s?
(b) What is the acceleration of the object at t = 0.5s?

Answer 1

well I did the one you posted later
This time you take the derivative

x = A t^3 + B t^2 + C t + D

v = 3 A t^2 + 2 B t + C

a = 6 A t + 2 B

Answer 2

v = dx/dt = 3A t^2 + 2B t + C

solve for t=12

Answer 3

v = dx/dt = 3A t^2 + 2B t + C

solve for t=12


a = dv/dt = 6A t + 2B

solve for t=0.5