# Train kinematics question

Say a locomotive at a constant 18 mi/h comes onto a track (from some other track) that has a passenger train att 100 mi/h on it. At the instant the train is 0.42 miles away from the locomotive, it starts to decelerate. What is the resulting constant acceleration that is needed to just avoid a collision? I have 18t + 0.42 = 100t + 0.5a(t^2).

However, I can't solve this! I know there's something simple I haven't seen, so please help.

Thanks.

## Answers and Replies

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Andrew Mason
Homework Helper
OVB said:
Say a locomotive at a constant 18 mi/h comes onto a track (from some other track) that has a passenger train att 100 mi/h on it. At the instant the train is 0.42 miles away from the locomotive, it starts to decelerate. What is the resulting constant acceleration that is needed to just avoid a collision? I have 18t + 0.42 = 100t + 0.5a(t^2).

However, I can't solve this! I know there's something simple I haven't seen, so please help.
I assume the locomotive and train moving in the same direction. What is the relative speed and separation as a function of time? What is the condition for no collision?

AM

Last edited:
Andrew Mason said:
I assume the locomotive and train moving in the same direction. What is the relative speed and separation as a function of time? What is the condition for no collision?

AM
Well, they are in the same direction, and the point is to set their position functions equal so that they are a centimeter (or something that negligible) away from touching)

The only info my problem gives is that one is at 18 miles an hour, the other at 100 mi/hr, and the separation as 0.42 miles. This is getting to me, because evidently I would have to solve for time AND acceleration.

Do I need calculus to solve the problem by any chance?

By the way, the answer in my book is 3.26 ft/sec^2, but I don't know how to arrive at that...

Units! Units! Units!

ALWAYS use ft and seconds for units in kinematics. CONVERT everything before you even start the problem.

Andrew Mason
Homework Helper
OVB said:
Do I need calculus to solve the problem by any chance?

By the way, the answer in my book is 3.26 ft/sec^2, but I don't know how to arrive at that...
Draw a graph of the speed vs. time of each. The area under each graph between two times is the distance covered in that time interval. So, do an expression for those areas and apply the condition (ie where the area under the train graph is less than the area under the locomotive graph).

AM