Question

A car starts from rest and accelerates at 5m/s^2 for 4s, then maintains that velocity for 3 seconds and then decelerates to a rate of 2m/s^2 for 3 seconds. How far does the car travel?

I would have to add the total distance throughout the trip. And to find the distance I would use:
d= d_0 + v_0t + 1/2at^2

=0+ 0(4) +1/2(5)(4^2) = 40m
=40+ 0(3) +1/2(5)(3^2) = 62.5m
=62.5 + 0(3) +1/2(2)(3^2)= 71.5m

thus 40 + 62.5 + 71.5 = 174m

Answer 1

On the second period, acceleration is zero (it maintains velocity as a constant). D during the second period is velocity*time=20*3=60

Now during the final period,
d=vi*t+1/2 at^2=20*3-1/2 2*9=42m check that.

The total distance is the sum of distances...
40+60+42.

Now a word on your work. YOu got the second and third wrong, (see above), however, you use of the formula bothers me...
df=do + vi*t + 1/2 a t^2. do is the the distance already traveled if you use this formula. Then, the prior distances are already added, you do not add them again as you did.