Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.
Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.
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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 5 m/s2 for 4.3 seconds. It then continues at a constant speed for 9.7 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 286 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 far does the blue car travel before its brakes are applied to slow down?
2. What is the acceleration of the blue car once the brakes are applied?
3. What is the total time the blue car is moving?
4. What is the acceleration of the yellow car?
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The blue car
Calculate the velocity of the blue car after 4.3 seconds = 5 * 4.3 = 21.5 m/s
Calculate the distance of the blue car after 4.3 seconds : 1/2 * 5 * 4.3^2 = 46.225 m
Calculate the distance of the blue car for the next 9.7 seconds : 21.5 * 9.7 = 208.55 m
The blue car has traveled : 46.225 + 208.55 = 254.775 meters
The car stopped after 286 meter, so : 286 - 208.55 - 46.225 = 31.225 meters remaining, so:
31.225 = (0 + 21.5)/2 x time
31.225 = 10.75 * t
t = 2.9 s
The blue car total time : 4.3s + 9.7s + 2.9s = 16.9 seconds
The yellow car
It travels 286 meters in 16.9 seconds, so
286 = 1/2 * a * 16.9^2
286 = 142.805a
a = 2m/s2
Question 1 : 254.775 m
Question 2 : as the car moves constantly, the acceleration is 0
Question 3 : 16.9 seconds
Question 4 : 2m/s2
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I was looking forward about the car riding and this post supplies thhose fact very keenly. Thanks! Keep it up!!..... 
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