Two cars start from rest at a red stop light. when the light turns green, both cars accelerate forward. the blue car accelerates uniformly at a rate of 5.8 m/s2 for 3.3 seconds. it then continues at a constant speed for 9.8 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 255.79 meters from where it started. the yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop.
3)how far does the blue car travel before its brakes are applied to slow down?

Well, the car gets to 5.8*3.3=19.14 m/s, getting to 5.8*3.3*3.3=63.162 meters from the start. For 9.8 secons it goes 187.572 meters, a total of 250.734. So, in 5.056 meters, it stopped uniformly. The speed dropped from 19.14 to 0 in 5.056 meters. There was a formula which didn't use time, but I've forgotten it. It would have given us the deceleration and then one could find the time through a 2nd degree equation. The timing for the first two parts is already known.
Either way, we've calculate more than we need to know for this question. You can cut the answer from that 250.734 distance.

Iwas going to ask that quesytion  sorry dude i dont  know : (    : /

The answer would be a) -80of