# Train collision prevention problem

#### opus

Gold Member
1. The problem statement, all variables and given/known data
When a train traveling at 161 km/h rounds a bend, the engineer sees that a locomotive has entered onto the track from the side and is a distance D=676m ahead. The locomotive is moving at 29 km/h. The engineer of the high speed train immediately applies the brakes.
What must the magnitude of the resulting constant deceleration be if a collision is to be just avoided?

2. Relevant equations
All constant acceleration equations of motion.

3. The attempt at a solution
My approach currently is to take $v=v_0 + at$ for the train, solve for t, and plug that expression in for t into the position function for the train. I do this because I don't know anything about time, and it isn't mentioned in the problem.

So I get the position function for the train is equal to $x_2=x_0 + \frac{-v_1^2 + v_2^2}{2a}$, solving for a, I get $a=\frac{-v_1^2 + v_2^2}{2(x_2-x_1)}$.

Now here is my next step, and I'm unsure on it's validity. It has given me the incorrect solution, but I'm trying to figure out what part is wrong with this.
Since we want to know the deceleration magnitude that the train must have to not collide with the locomotive, we can say that when they are at the same position, their speeds must be equal. For this reason, I have taken the $v_2$ in my equation, and substituted in the velocity of the locomotive for $v_2$ I then plug in the value of the initial speed for the train, and 676 meters for $x_2-x_1$ So it looks like:

$a_{train}=\frac{-(44.72~m/s)^2+(8.05 m/s)^2}{2(676~m)}$
This gives me an incorrect solution, but not a drastically unreasonable one. I get 1.43 m/s and the correct solution is 0.994 m/s.
So where in my reasoning, have I made an error?
Thank you.

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#### TSny

Homework Helper
Gold Member
and 676 meters for $x_2-x_1$
The locomotive moves while the train is catching up to it.

#### neilparker62

Homework Helper
Perhaps easier to use relative velocity - then the locomotive can be considered at rest whilst the train catches up to it. And the final (relative) velocity must be 0 in $${v_f}^2={v_i}^2+2aΔx$$

#### RPinPA

Homework Helper
EDIT: I'm completely wrong with what's going on here, and it took me three paragraphs of rambling to get to where OP started. See my next post for a corrected response.
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It's not clear what the locomotive is doing. Is it braking at the same rate of deceleration? Is it continuing to move at constant velocity? The problem doesn't say, but if it doesn't make any attempt to stop then the train is going to hit it no matter what he does.

So I get the position function for the train is equal to $x_2 = x0 + \frac{-v_1^2 + v_2^2}{2a}$
You skipped a few steps here so it's not clear which assumption you made.

OK, I just re-read the problem one more time and have a new interpretation: The locomotive is moving FORWARD, in the same direction as the high-speed train. So it doesn't need to decelerate, in fact it doesn't want to. And the high-speed train doesn't need to come to a full stop, it only needs to be slower than the locomotive. Because of those factors, less deceleration is needed than what you calculated (if you assumed a different state of things, as I did).

I get 1.43 m/s and the correct solution is 0.994 m/s.
No you didn't, and no that is not the correct solution. Those units are velocities.

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#### RPinPA

Homework Helper
And now I've reread your equation and I think you got the analysis right, where I got it wrong. So let me look more deeply at your solution.

My approach currently is to take $v=v_0 + at$ for the train, solve for t, and plug that expression in for t into the position function for the train. I do this because I don't know anything about time, and it isn't mentioned in the problem.
Which is why the preferred kinematic equation here is $v_f^2 = v_i^2 + 2ad$, which it looks like you derived and used (mostly) correctly.

So the issue here is what @TSny said at #2, which is that the distance is more than 676 m since the locomotive continues to move. Continuing from your same starting method, use the same value of $t$ to get an expression for how far the locomotive moves, and add that to the 676 m.

#### opus

Gold Member
The locomotive moves while the train is catching up to it.
In deriving the equation I wanted to use, I made effort to continually keep this in consideration. Now, at the equation that I ultimately ended up with, I got 2(676m) in the denominator, so my assumption (perhaps an incorrect one), was that this new distance 2*676m is the distance that was covered by the train by the time it reached the appropriate acceleration. Is this faulty reasoning?

Perhaps easier to use relative velocity - then the locomotive can be considered at rest whilst the train catches up to it. And the final (relative) velocity must be 0 in $${v_f}^2={v_i}^2+2aΔx$$
Ok and here is a key aspect that I am not understanding.
As far as I know, the train is accelerating and the locomotive has no acceleration. So is it not the case that the train is in a "non-inertial" reference frame and the locomotive is in an "inertial" reference frame, so we can't just combine their velocities?

And now I've reread your equation and I think you got the analysis right, where I got it wrong. So let me look more deeply at your solution.

Which is why the preferred kinematic equation here is $v_f^2 = v_i^2 + 2ad$, which it looks like you derived and used (mostly) correctly.

So the issue here is what @TSny said at #2, which is that the distance is more than 676 m since the locomotive continues to move. Continuing from your same starting method, use the same value of $t$ to get an expression for how far the locomotive moves, and add that to the 676 m.
Let me give this a try.

#### TSny

Homework Helper
Gold Member
In deriving the equation I wanted to use, I made effort to continually keep this in consideration. Now, at the equation that I ultimately ended up with, I got 2(676m) in the denominator, so my assumption (perhaps an incorrect one), was that this new distance 2*676m is the distance that was covered by the train by the time it reached the appropriate acceleration. Is this faulty reasoning?
The factor of 2 here is just part of the formula $a=\frac{-v_1^2 + v_2^2}{2(x_2-x_1)}$.

The 2 does not mean that the train travels a distance of $2(x_2-x_1)$. The distance the train travels is $(x_2-x_1)$, and by the time the train reaches the locomotive, this distance is equal to 676 m plus the distance that the locomotive has moved during the time it takes the train to catch up.
As far as I know, the train is accelerating and the locomotive has no acceleration. So is it not the case that the train is in a "non-inertial" reference frame and the locomotive is in an "inertial" reference frame, so we can't just combine their velocities?
@neilparker62 was suggesting going to the frame of reference of the locomotive, which would be an inertial frame. The acceleration of the train in this new frame is the same as the acceleration of the train in the original frame. The velocities will change in going to the new frame.

#### opus

Gold Member
Excellent thank you guys. So this minimum needed is 0.994 m/s^2 in magnitude!

#### neilparker62

Homework Helper
@neilparker62 was suggesting going to the frame of reference of the locomotive, which would be an inertial frame. The acceleration of the train in this new frame is the same as the acceleration of the train in the original frame. The velocities will change in going to the new frame.
To be quite honest this would be an implied suggestion since I have no 'schooling' in the use of frames of reference other than by instinct. Happily instinct was correct in this instance - thanks for explaining why!

"Train collision prevention problem"

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