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 abrate of 3.7m/s^2 for 4.5 seconds. it then continues at a constant speed for 9 seconds, before applying brakes such that the car's speed decreases uniformly coming to rest 238.64 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

1. how fast is the blue car going 2.2 seconds after it starts?

2. how fast is the blue car going 11.3 seconds after it starts?

3. how far does the blue car travel before its brakes are appliedbto slow down

4. what is the acceleration of the blue car once the brakes are applied

5. what is the total time the blue car is moving?

6. what is the acceleration of the yellow car?

First, create an illustration of the motion of the two cars as shown in the attached picture. The essential equations that are used for the 6 questions are:

For constant acceleration:
a = v,final - v,initial /t
2ad = v,final² - v,initial²
d = v,initial*t + 1/2*at²

For constant velocity:
d = constant velocity*time

The solutions for each are as follows:

1. a = v,final - v,initial /t
    3.7 = (v - 0)/2.2 s
    v = 8.14 m/s
    After 2.2 seconds, the speed of the blue car is 8.14 m/s.

2. After 11.3 seconds, the blue car is now in the second segment of the motion at constant velocity. Thus,

   a = v,final - v,initial /t
  3.7 = (v₁ - 0)/4.5 s
  v₁ = 16.65 m/s

3. Total distance = d1 + d2 + d3
    d1 = d = v,initial*t + 1/2*at²
    d2 = constant velocity*time
    
   Total distance =  0*(4.5) + 1/2*(3.7)(4.5)² + (16.65)(9) + d3= 238.64
   d3 = 51.3275 m

4. 
    2a3d = v,final² - v,initial²
    2a3(51.3275) = 0² - 16.65²
    a3 = -2.7 m/s²

5. Total time = t1 + t2 + t3
     
    a3 = (v3 - v2)/t3
    -2.7 = (0 - 16.65)/t3
    t3 = 6.167 s

    Total time = 4.5 + 9 + 6.167 = 19.67 s

6. Since the yellow car just caught up to the blue car in time, the total time would also be 19.67 s.

d = v,initial*t + 1/2*at²
238.64 = 0*(19.67) + 1/2*a*(19.67)²
a = 1.234 m/s²