Problem below:

"A freight train is leaving the station 2 miles away from the an at-grade crossing. The train accelerates at 1.0 ft/s2, decelerates at 2.0 ft/s2, and has a top speed of 50 mph. A passenger car 2 miles away is traveling at a constant speed toward the at-grade crossing. What is the minimum constant speed the passenger car can maintain to arrive at the crossing before the warning signals are activated? The warning signals are activated when the freight train is 2000 feet away from the at-grade crossing."

I am trying to figure out how much time it takes for the freight train to reach the at-grade crossing, and use that time to determine the min velocity of the passenger car. But the inclusion of both acceleration and deceleration is throwing me off. Any help?

  • $\begingroup$ This doesn't appear to be a well-posed problem. If the intent is that the train should accelerate to its top speed and maintain it, then you should be able to determine how long it takes to get to the signal activation point. But it's not clear why we need to know the decel rate. $\endgroup$
    – rogerl
    Commented Feb 23, 2019 at 21:53
  • $\begingroup$ I agree with you in that the problem is poorly worded. I don't have the answer sheet, but the 4 choices for answers are as follows: a) 40 mph b) 45 mph c) 50 mph d) 55 mph $\endgroup$
    – civilbee
    Commented Feb 23, 2019 at 22:13

1 Answer 1


The train accelerates to its maximum speed at 50 seconds, at which point its position is $\frac{1}{2}t^2 = 1250$ feet. Its velocity at that time is $50\cdot \frac{5280}{3600} = \frac{220}{3}$ f/s. It needs to cover $2\cdot 5280 - 1250 - 2000 = 7310$ feet before the signal is triggered, and it takes $7310/(220/3) = \frac{2193}{22}$ seconds to do that.

The car is $10560$ feet away from the crossing, so if it is to get to the crossing when the signal is triggered, its minimum speed must be $10560/(2193/22)$ f/s, or about $72.23$ m/h.

So there must be some other assumption being made.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .