# Train crash

1. Jun 20, 2009

### Dave25

1. The problem statement, all variables and given/known data

Two trains are heading towards each other on the same track. They are initially 2100 feet apart.
The first train has an initial velocity of 88 ft/s due East, the second train has an initial velocity of 107 ft/s due West. Both trains begin decelerating at a rate of a = - 3 ft/s2. Will the trains collide? If so, where?

2. Relevant equations

v^2 (final) = v^2 (initial) + 2 ax (I think this is the correct equation)

3. The attempt at a solution

I keep setting train 1 = to train 2 but both accelerations cancel out and give me 0. So my quadratic formula is 0t^2 + 19t + 2100. Am I going about this the wrong way? I need help please.

2. Jun 20, 2009

### LowlyPion

Write two equations of the form:

x = Xo + Vo*t + 1/2*a*t2

Being careful to note the signs of the velocity and acceleration with your selection of which direction is positive x.

3. Jun 20, 2009

### Dave25

Ok. Thanks a lot.

4. Jun 21, 2009

### mjoyce3

i am working on a similar problem, and am still very confused. for this one, what do you do after you solve for t?

5. Jun 21, 2009

### ideasrule

Assuming you solved for t correctly, you see how far the trains traveled during time t relative to each other. If that distance is greater than 2100 ft, the trains are screwed.

Last edited: Jun 21, 2009
6. Jun 21, 2009

### RoyalCat

Note that you can have an even simpler check as to whether they collide.
Assume that they do.
The quadratic equation for the east-bound train is x1(t).
And the quadratic equation for the west-bound train is x2(t).

Solve for t.
x1(t) = x2(t)

You've got a couple of options. The first is, that the quadratic equation has no solutions at all, meaning that they don't collide. The other, is that you get only negative values of t (Though my intuition says this is not the case) as solutions, which also means that they do not collide.

For the former case, though, you can see if a quadratic equation is solvable over the reals by looking at its discriminant (bÂ²-4ac), if it's negative, then the two trains surely never collide.

The other two cases are 1 positive solution (The moment of impact), 2 positive solutions (Won't happen here, though) and 1 positive/1 negative solution (Again, my intuition says that can't be the case here either).

7. Jun 21, 2009

### Dave25

Thanks everyone. I think I got the answer for this one.

8. Jun 21, 2009

### LowlyPion

Actually for the original problem there are 2 positive answers. The later one is tossed, because the deceleration will be non-linear due to the intervening collision. (Their phantoms, in the absence of an actual collision, or say if they were on parallel tracks, continue to slow until reversed and pass each other going the opposite direction at the later time.)
Plugging the time back in either equation yields the position x of impact, The idea being that they are in collision if they are at the same place.

9. Jun 21, 2009

### RoyalCat

Ah yes, correct, correct, in my head they were accelerating towards each-other when I was considering the case of 2 positive solutions. Thanks for pointing out my mistake. :)