An automobile starting from rest speeds up to 40 ft per sec with a constant acceleration of 4 ft per sec^2, ru

An automobile starting from rest speeds up to 40 ft per sec with a constant acceleration of 4 ft per sec^2, runs at this speed for a time, and finally comes to rest with a deceleration of 5 ft per sec^2. If the total distance traveled is 1000 ft, find the total time required.

a = vf-vi/t
a = 40-0/t
t = 40-0/4
t = 10s

that's only the time it took to accelerate to 40ft/s
the distance that it traveled in that 10s can be derived by this equation:

vf^2 - vi^2 = 2ad

since vi = 0 we can neglect it, plugging in the a and the vf we find that d = 200ft

now to find the time and distance it took to decelerate
using the same method used above, except with different vf and vi now:

a=vf-vi/t
-5 = 0-40/t
t = -40/-5
t = 8s

and finding the distance:
vf^2-vi^2 = 2ad
0^2-40^2 = 2(-5)d
d = 160ft

adding up the distance it took to accelerate and decelerate and subtracting it from the total distance traveled we come to the conclusion that:

1000ft - 200ft - 160ft = 640ft traveling at a constant 40ft/s

v = d/t
t = d/v
t = 640/40 = 16s

adding up all the times (acceleration, deceleration, constant speed):

10s + 8s + 16s = 34s

the total time of travel is 34s