If the position of an object is given by x = 2.00t^3, where x is measured in meters and t in seconds, find (a) the average velocity and (b) the average acceleration between t = 1.0 s and t = 2.0 s. Then find (c) the instantaneous velocity v and (d) the instantaneous acceleration a at t = 1.0 s. Next find (e)v and (f)a at t = 2.0 s.

Answer 1

x = 2t^3
v = 6t
a = 6

To find an average, just divide the change by the interval.

(a)
x(1) = 2
x(2) = 16
avg v = ∆x/∆t = (16-2)m/(2-1)s = 14m/s

I think you can handle the rest, no?

Answer 2

No