Train Collision: Will They Meet? Find Out!

[Dave25]

Homework Statement

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?

Homework Equations

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

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.

 

[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.

 

[Dave25]

Ok. Thanks a lot.

 

[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?

 

[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.

 

[RoyalCat]

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.

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 (b²-4ac), 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).

 

[Dave25]

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

 

[LowlyPion]

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).

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.)

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

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.

 

[RoyalCat]

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.

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. :)