Kinematics involving two objects

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Homework Help Overview

The problem involves kinematics with two trains, focusing on the conditions necessary to avoid a collision. The original poster presents a scenario where a passenger train must decelerate to prevent crashing into a freight train ahead, with specific parameters such as speeds and reaction times provided.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the formulation of equations representing the positions of both trains over time, questioning the correctness of the equations and exploring alternative methods of analysis, such as using a different frame of reference.

Discussion Status

The discussion is ongoing, with participants providing different perspectives on how to approach the problem. Some have offered insights into solving the equations, while others express uncertainty about the initial setup and seek clarification.

Contextual Notes

There are constraints related to the reaction times of the engineer and the need for the passenger train to decelerate at a minimum rate to avoid collision. The original poster expresses confusion about the next steps in their solution process.

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Homework Statement


A passenger train is traveling at 27 m/s when the engineer sees a freight train 345 m ahead of his train traveling in the same direction on the same track. The freight train is moving at a speed of 5.9 m/s.
(a) If the reaction time of the engineer is 0.36 s, what is the minimum (constant) rate at which the passenger train must lose speed if a collision is to be avoided? (m/s/s)
(b) If the engineer's reaction time is 0.73 s and the train loses speed at the minimum rate described in Part (a), at what rate is the passenger train approaching the freight train when the two collide? (m/s)
(c) For both reaction times, how far will the passenger train have traveled in the time between the sighting of the freight train and the collision? (km)

Homework Equations


Xf= Xi + vt + 0.5at^2

The Attempt at a Solution



(a) Xf1 = 27(t-0.36) + 0.5a(t-0.36)^2
Xf2 = 345 + 5.9t

Dont know what to do now. I am really confused
 
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So you have two equations. One gives Xf1 (the final position of train 1) and one gives Xf2 (the final position of train2), both as a function of the unknown time to intercept and the unknown acceleration. That's two equations in two unknowns.

Without knowing how it's all going to work itself out, how about solving those two equations for t and a. See if any useful insights fall out.
 
Are the equations even right
 
The equations look right to me.

Personally, I would have approached the problem differently. I'm more comfortable using physical intuition rather than algebra. Transform to a frame of reference in which the front train is stationary see what acceleration is needed to attain a particular velocity over a particular separation distance. That becomes essentially a work = force times distance situation, easy to solve for velocity in terms of force, or for force in terms of velocity.
 

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