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 fast is the blue car going 2.8 seconds after it starts?
How fast is the blue car going 2.8 seconds after it starts?
1 answer:
First, create an illustration of the motion of the two cars as shown in the attached picture. The essential equations used is
For constant acceleration:
a = v,final - v,initial /t
The solutions is as follows:
a = v,final - v,initial /t
3.8 = (v - 0)/2.8 s
v = 10.64 m/s
After 2.8 seconds, the speed of the blue car is 10.64 m/s.
For constant acceleration:
a = v,final - v,initial /t
The solutions is as follows:
a = v,final - v,initial /t
3.8 = (v - 0)/2.8 s
v = 10.64 m/s
After 2.8 seconds, the speed of the blue car is 10.64 m/s.
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