A component is uniformly accelerated in a production plant from 3m/s to 8 m/s in 4s and at a constant velocity of 8 m/s for the following 2s .
Determine the displacement for a time of 6s.
Using Calculus:
For a constant acceleration of a m/s^2
v = at + c , where v is in m/s and t is in seconds
when t = 0 sec, v = 3 m/s
3 = 0 + c
so v = at + c
when t = 4 sec, v = 8 m/s
8 = 4a + 3
a = 5/4
so v = (5/4)t + 3
s = (5/8)t^2 + 3t + k
at t = 0 , s = 0 , so k = 0
s = (5/8)^2 + 3t
so after 4 seconds, the displacement was
s = (5/8)(16) + 12 = 22 m
it then went for another 2 sec at a constant 8 m/s or 16 m
so .....