Two cars start from rest at a red stop light. When the light turns green, both c

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 3.4 m/s2 for 4.1 seconds. It then continues at a constant speed for 6.1 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 140.47 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. What is the acceleration of the blue car once the brakes are applied? What is the total time the blue car is moving? What is the acceleration of the yellow car?

Explanation / Answer

blue car -
Initial Velocity of blue car = 0
Final Velocity , v = ?
Acceleration, a = 3.4 m/s^2
t = 4.1 s

v = u + a*t
v = 0 + 3.4 * 4.1 m/s
v = 13.94 m/s

Speed of blue car after 4.1 s, v = 13.94 m/s

Distance travelled by Blue car before applying brakes ,
S = u*t2 + 1/2 * at2^2 + v*t2
S = 0 + 1/2 * 3.4* 4.1^2 + 13.94 * 6.1 m
S =  113.6 m

Now,
Inital Speed = 13.94 m/s
Final Speed = 0
Distance travelled before blue car comes to stops = 140.47 - 113.6 = 26.87 m

v^2 = u^2 - 2*a*s
0 = 13.94^2 - 2*a*26.87
a = 0.976 m/s^2
Acceleration of the blue car once brakes are applied ,a = 3.616 m/s^2

(b)
Time taken by blue car to stop, V = u - a*t
0 = 13.94 - 3.616 * t
t = 3.86 s

Total time blue Car is moving, t = 4.1 s + 6.1 s + 3.86 s
Total time blue Car is moving, t = 14.06 s

(c)

Initial Velocity, u = 0
Distance, s = 140.47 m
time, t = 14.06 s

s = u*t + 1/2 * at^2
140.47 = 0 + 1/2 * a * 14.06^2
a = 1.42 m/s^2
Acceleration of the yellow car, a = 1.42 m/s^2

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