Two decelerating trains on a collision course

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SUMMARY

The discussion centers on the collision dynamics of two trains, a red train traveling at 20 m/s and a green train at 40 m/s, both decelerating at 1 m/s² from a starting distance of 950 meters. It is established that the trains will collide, with the red train coming to a complete stop and the green train reaching a velocity of 10 m/s at the moment of collision. The solution involves calculating the distance each train travels before stopping and determining their velocities at the time of impact through relative motion equations.

PREREQUISITES
  • Understanding of kinematic equations, specifically v = v₀ + at and x - x₀ = v₀t + (1/2)at².
  • Knowledge of concepts related to relative motion in physics.
  • Familiarity with the principles of uniform acceleration and deceleration.
  • Ability to solve quadratic equations for motion analysis.
NEXT STEPS
  • Study the application of kinematic equations in collision scenarios.
  • Learn about the principles of relative motion in detail.
  • Explore advanced topics in dynamics, including systems with variable acceleration.
  • Investigate graphical methods for analyzing motion and collision problems.
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and dynamics, as well as educators seeking to enhance their understanding of collision problems in motion analysis.

marksyncm
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Homework Statement


A red and a green train are headed towards each other on a collision course.

Red train velocity = 20m/s
Green train velocity = 40m/s

When the trains are a distance of 950 meters apart, they begin to decelerate at a steady pace of 1m/s^2.

1) Will the trains collide?
2) If so, what will be their respective speeds at the time of the collision?

Homework Equations



##v=v_o+at## and ##x-x_0 = v_0t + \frac{at^2}{2}##

The Attempt at a Solution


[/B]
Actually, I was able to obtain a solution (the trains collide; the red train has a velocity of 0m/s and the green train a velocity of 10m/s), but I am not sure if there is an easier way to do it. Here are the steps I took (without equations):

# Check if the trains collided:

1) Calculate total distance traveled by each train before it comes to a halt.
2) Is combined distance traveled by both trains equal to or greater than 950 meters? Yes - then the trains collided.

# Calculate the velocity of each train at collision time (this is where I'm most uncertain)

First, I decided to determine if the slower of the two trains (the red one) came to a complete stop prior to the collision. To do this:

1) Calculate the time ##t## it takes the red train to come to a stop.
2) Check the distance ##d_1## traveled by the red train in the above time.
3) Check the distance ##d_2## traveled by the green train in time ##t##.
4) Is ##d_1+d_2 < 950##? If so, the trains collided when the red train had a velocity of 0 m/s and the green train had whatever velocity it has when it traverses a distance of ##950-d_1##.
5) If ##d_1+d_2 > 950## , then both trains had a velocity ##>0## at the time of collision. These velocities can be calculated by determining the time of impact using the relativity of motion - in this case, ##60t - t^2 = 950## - then checking the velocity of each train at that time.

I'm wondering if there's a simpler approach that works regardless of whether both or only one of the two trains were moving at the time of impact? It seems that when I try the "relativity of motion" approach for cases where one of the trains came to a stop before the collision, I get a solution that's a complex number, and the real part of that number is actually a correct solution. It seems intuitive to me that the relativity of motion approach should not work in these cases (because the combined deceleration of the system changes after one of the trains comes to a stop), but I can't shake this feeling that there's a simpler approach to these problems.
 
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I can't see a problem with your approach. It's a good point that the motion is potentially in two phases: both trains decelerate for a time, then only one decelerates.
 
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Thank you.
 

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