Question

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 3.8 m/s2 for 4.6 seconds. It then continues at a constant speed for 9.2 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 257.71 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.
How far does the blue car travel before its breaks are applied to  slow down?

Answer 1

First, create an illustration of the motion of the two cars as shown in the attached picture. The essential equations used are:

For constant acceleration:
a = v,final - v,initial /t
d = v,initial*t + 1/2*at²

For constant velocity:
d = constant velocity*time

The solutions is as follows:

   a = v,final - v,initial /t
  3.8 = (v₁ - 0)/4.6 s
  v₁ = 17.48 m/s

    Total distance = d1 + d2 + d3
    d1 = d = v,initial*t + 1/2*at²
    d2 = constant velocity*time
    
   Total distance =  0*(4.6) + 1/2*(3.8)(4.6)² + (17.48)(9.2) + d3= 257.71
   d3 = 56.69 m