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