The Car and the Train (Posted on 2008-01-30)

	Submitted by Brian Smith
Rating: 3.0000 (2 votes)
Solution:	(Hide)
Charlie and Kenny both provide solutions in the comments. My solution is below. Let L be the length of the train. Let V be the speed of the train. Let A be the rate of acceleration of the car. Let t1 be the amount of time elapsed from the start to then the car matches speed with the train. Let t2 be the amount of time elapsed from the start to when the car reaches the speed limit. Let D be the distance traveled. Let S be the speed limit. When the back of the train reaches the car, the following formulas can be made: [1] V = t1 * A [2] (1/2) * A * t1^2 + L = V * t1 When the car catches up to the front of the train, the following formulas can be made: [3] D = V * t2 [4] D = (1/2) * A * t2^2 [5] S = t2 * A Equating equations [3] and [4] yields: [6] A * t2 = 2 * V Substituting equation [6] into equation [5] yields [7] S = 2 * V Comparing equation [6] with equation [1] yields: [8] 2*t1 = t2. Substituing equation [1] into equation [2] yields: [9] L = (1/2) * A * t1^2 Substituting equation [8] into equation [4] yields: [10] D = 2 * A * t2^2 Substituting equation [10] into equation [9] yields [11] D = 4 * L Equations [7] and [11] give the answers to the problem: The speed limit is twice the speed of the train and the distance traveled equals four times the length of the train.

	Subject	Author	Date
	Puzzle Thoughts	K Sengupta	2023-04-24 00:13:24
	Alternate solution for first part	Kenny M	2008-02-01 00:13:30
	solution	Charlie	2008-01-30 12:46:29