How Far From the First Station Do Two Oppositely Accelerating Trains Meet?

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SUMMARY

Two trains, starting from rest and 5.0 km apart, accelerate towards each other with different rates: one at 2.5 m/s² and the other at 4.0 m/s². The correct approach to determine their meeting point involves using the equation d = (1/2)at², which accounts for the initial conditions of both trains. The trains will meet approximately 97.5 meters from the first station, but this calculation must consider the relative motion of both trains. The final solution requires equating their positions over time to find the exact meeting point.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = (1/2)at²
  • Familiarity with relative motion concepts
  • Basic knowledge of acceleration and displacement
  • Ability to solve quadratic equations
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  • Study the derivation and application of kinematic equations in physics
  • Learn about relative motion and how it applies to two objects moving towards each other
  • Explore examples of uniformly accelerated motion in one dimension
  • Practice solving problems involving multiple objects with different accelerations
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OmniNewton
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Homework Statement



Two trains start from rest from two stations 5.0 km apart and travel, on two parallel railroad tracks, in opposite directions. One train accelerates uniformly at 2.5 m/s^2 and the other has a constant acceleration of 4.0 m/s^2.
How far from the first station will they meet?


Homework Equations



1. d= at^2
2. v^2 = u^2+ 2ad
3. t= v-u/ a

The Attempt at a Solution


Train 1 relative to Train 2
Initial velocity = 0 m/s
Acceleration = 6.5 m/s^2
Displacement = 5000m

v^2 = u^2+ 2ad
v = Square root of u^2 + 2ad
v= Square root of 2ad
v= Square root of 2(6.5 m/s^2) (5000m)
v= Square root of 65000 m^2/s^2
v= + or - 255 m/s

t= v-u/ a
t= 255ms^-1/6.5m/s^-2
t= 39 seconds

Train 1

d= at^2 because initial velocity = 0
d= (2.5 ms^-1) (39 seconds)
d= 97.5m

Therefore they will meet 97.5m from the first station.

However this answer does not make sense to me because if I solve for distance with the other train it does not add to 5000m or 5.0 km. I do not know what I did wrong.

Thanks In Advanced
 
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Welcome to PF OmniNewton!

OmniNewton said:

Homework Statement



Two trains start from rest from two stations 5.0 km apart and travel, on two parallel railroad tracks, in opposite directions. One train accelerates uniformly at 2.5 m/s^2 and the other has a constant acceleration of 4.0 m/s^2.
How far from the first station will they meet?


Homework Equations



1. d= at^2
2. v^2 = u^2+ 2ad
3. t= v-u/ a

This bit in red is wrong. The actual equation is:
d = d0 + v0t + (1/2)at2

Where d0 and v0 are the initial position and speed respectively. If these are both zero, it's still d = (1/2)at2. You are missing the factor of 1/2.

OmniNewton said:

The Attempt at a Solution


Train 1 relative to Train 2
Initial velocity = 0 m/s
Acceleration = 6.5 m/s^2
Displacement = 5000m

v^2 = u^2+ 2ad
v = Square root of u^2 + 2ad
v= Square root of 2ad
v= Square root of 2(6.5 m/s^2) (5000m)
v= Square root of 65000 m^2/s^2
v= + or - 255 m/s

t= v-u/ a
t= 255ms^-1/6.5m/s^-2
t= 39 seconds

There is a fundamental conceptual error here. You are assuming that the first train travels 5 km. It doesn't. The situation is that the two trains start out 5 km apart and head towards each other along parallel tracks. Since they are approaching each other, they will meet somewhere in the middle of the two stations, and you have to figure out where and when this occurs.

In any case, you do not need this equation v2 = u2 + 2ad. The two trains will "meet" when their positions are the same. So, write down an expression for the position vs. time of train 1. Now write down an expression for the position vs. time of train 2 (hint: it does not have the same starting position, and the sign of its acceleration is opposite). Equate these two expressions, and solve for t. This gives you the time at which they meet. From this, you can get the position of each one.
 
It actually all worked out once I used that equation d= at^2/2

5000m worked because train 1 relative to train 2 equates to 5000m

train 1's displacement = x

train 2's displacement = 5000m - x

so ultimately

train 1's displacement relative to train 2 has to be 5000m.

Correct me if I am wrong however.

But thank you very much for the help you really helped me and it is very much appreciated :D

Thank You Very Much For The Welcome.
 
OmniNewton said:
It actually all worked out once I used that equation d= at^2/2

5000m worked because train 1 relative to train 2 equates to 5000m

train 1's displacement = x

train 2's displacement = 5000m - x

so ultimately

train 1's displacement relative to train 2 has to be 5000m.

Correct me if I am wrong however.

But thank you very much for the help you really helped me and it is very much appreciated :D

Thank You Very Much For The Welcome.

If it worked out, then I guess your math must have been right, but what you are *saying* is totally wrong. First of all, train 1's *position* is x1. Its displacement is its change in position. Train 2's position is x2 relative to its station (or 5000 - x2 relative to the first station). Their separation (the distance between them) is NOT 5000 m at any time other than t = 0. E.g. at t = 0, x1 = 0 and x2 = 5000, so their separation is x2 - x1 = 5000 m. But suppose at some later time, the first train has moved 100 m along the track, and the second has moved 400 m along the track (totally made up numbers). So, train1's position (and displacement) is 100 m. Train 2's position (measured relative to the first station) is 5000 - 400 = 4600 m. Train 2's displacement is 400 m. The separation between the trains ("relative displacement") is 4600 m - 100 m = 4500 m. So, you can see that their relative distance is NOT 5000 m after t = 0, since they are moving *towards* each other.
 

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