Motion in 1 dimension easy question?

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A car accelerates from rest at +1.5 m/s² for 5 seconds, reaching a speed of 7.5 m/s before braking at -2.0 m/s² for 3 seconds. During the acceleration phase, the car travels 18.75 meters, while during braking, it covers an additional 13.5 meters. The total distance traveled is approximately 32.25 meters. The equation used to calculate the distance during uniform acceleration is x = t (v_i + v_f) / 2. The discussion provides a clear breakdown of the motion phases and calculations involved.
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a car starts from rest travels for 5.0s with a uniform acceleration of + 1.5m/s^2. The driver then applies the brakes, causing uniform acceleration of -2.0m/s^2. If the brakes are applies for 3.0s (a) how fast is the car going atthe end of the braking period and b how far has it gone

I have already done part a and i have the answers for both parts i just don't know the steps in approaching part b
the answer to b is 32.2m
please help! thank you :D
 
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When it's speeding up, it starts at 0 m/s and after the speedup is at 7.5 m/s. That's an average of 3.75 m/s for 5 seconds, so the distance traveled during this part is 18.75 m.

When it begins to brake, it's going at 7.5 m/s, and then after the braking is at 1.5 m/s. That's an average of 4.5 m/s during this period of 3 seconds, so during the braking it moves 13.5 meters.

The sum is 32.25 meters.
 
but what equation did you use?
 
x = t (v_i + v_f) / 2
 
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