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Colliding trains

  • Thread starter lep11
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  • #1
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Homework Statement


Collision - The engineer of a passenger train traveling at 25 m/s sights a freight train whose caboose is 200 m ahead on the same track. The freight train is traveling at 15 m/s in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant acceleration of -0.1 m/s^2, while the freight train continues with constant speed. Take x=0 as the location of the front of the passenger train when the engineer applies the brakes. (a) Will the cows nearby witness a collision? (b) If so, where will it take place?


Homework Equations


v0passenger=25 m/s
x0freight=200m



The Attempt at a Solution


When the collision takes place, xpassenger=xfreight
v0passenger*t +½*a*t2=x0freight +vfreight*t

Solving this quadratic equation for t gives me t=22.54s AND t=177.459...s
I choose t=22.54s because "it happens earlier"
Now x0freight+vfreight*t= 200m+(15m/s)*22.54s≈540m
 
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Answers and Replies

  • #2
Here's a tip, use relative velocities. i.e. If you the passenger train is moving at 10 m/s, what is the speed of the freight train relative to the passenger train?
 
  • #3
Sorry, I may have been unclear, if the passenger train is moving at 10m/s what would the freight trains velocity by so that their relative velocities are still the same?
 
  • #4
haruspex
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Solving this quadratic equation for t gives me t=22.54s AND t=177.459...s
I choose t=24.54s because "it happens earlier"
Now x0freight+vfreight*t= 200m+(15m/s)*24.534s=760m
22.54 became 24.54 then 24.534?
Check your answer by calculating where each will be after 22.54 (or however many) seconds.
 
  • #5
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22.54 became 24.54 then 24.534?
Check your answer by calculating where each will be after 22.54 (or however many) seconds.
Corrected now.
 
  • #6
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But why did I get two solutions for t and which one to pick and why?
 
  • #7
haruspex
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Corrected now.
Looks right now.
There are two solutions because the equations don't 'know' that the trains cannot pass through each other, or that the passenger train's acceleration will cease when it comes to rest. So the second solution corresponds to the passenger train overtaking the freight train, going into reverse and meeting it a second time.
 
  • #8
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Would this be considered a good way to solve the problem?
 
  • #9
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The quadratic which i am getting is t2-200t-4000=0.

Solving it,I get t=218s , -18s .

lep11...If you have the answer key,please check the answer .

Or may be I am making some mistake...


Edit: The equation should be t2-200t+4000=0.
 
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  • #10
There is no need for time, although perfectly acceptable. If you look at relative velocities, we can say the freight train is stationary and making the same reduction to the passenger train gives a velocity of 10m/s, then you can use the formula Vf^2=Vo^2-a(x-xo). Where x is the position of the freight train and xo is the position of the passenger train.
 
  • #11
haruspex
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There is no need for time, although perfectly acceptable. If you look at relative velocities, we can say the freight train is stationary and making the same reduction to the passenger train gives a velocity of 10m/s, then you can use the formula Vf^2=Vo^2-a(x-xo). Where x is the position of the freight train and xo is the position of the passenger train.
You do not know the collision speed. The collision is necessarily represented as the two trains being in the same place at the same time. That makes it something of a challenge to solve the problem without reference to time.
 
  • #12
haruspex
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The quadratic which i am getting is t2-200t-4000=0..
I believe you have a sign wrong.
 
  • #13
Did you mean "not necessarily"?

Velocity it relative regardless, the given values are relative to the ground, if we instead work with relative to each other, and then take the equation I posted, which is derived from solving the position formula for t, there is no need to actually find a value for t. You just by pass that step and simply find whether the passenger has a velocity


Edit: I read your post again, I don't see why we care how fast the one train is going when/if it collides. The question only asks if they collide. So you simply have to decide whether it has stopped before it gets to a certain point.
 
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  • #14
haruspex
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Did you mean "not necessarily"?
No. The fact of collision is defined by their being in the same place at the same time.
Velocity it relative regardless, the given values are relative to the ground, if we instead work with relative to each other, and then take the equation I posted, which is derived from solving the position formula for t, there is no need to actually find a value for t. You just by pass that step and simply find whether the passenger has a velocity<=0 at the point in question.
You do not know the relative velocity when they collide (without first solving the problem).
 
  • #15
We don't need the velocity when they collide, to find out if they collide, but the calculations are one in the same. If you get a positive, non-zero velocity from the formula I posted, then that value IS the relative velocity of the passenger train.*when they collide *
 
  • #16
haruspex
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We don't need the velocity when they collide, to find out if they collide, but the calculations are one in the same. If you get a positive, non-zero velocity from the formula I posted, then that value IS the relative velocity of the passenger train.*when they collide *
Your equation included two velocities (presumably the initial and final relative velocities) and an unknown distance x. Since you do not yet know the collision velocity (i.e. the final relative velocity), you have two unknowns but only one equation.
Have you obtained an answer to the question using your equation, and checked that it matches the answer already obtained in this thread?
I see in one of your posts you mentioned the (relative) velocity becoming zero. That does not give you when they collide - it gives you the point at which they are moving at the same speed, which is of little relevance.
 
  • #17
If you define one train as stationary at x=0 and keep the same relative velocity for the other train you the, it becomes a simple does the train stop in x distance w/ a certain initial velocity and a certain acceleration. If the freight train is stationary, then the displacement is not unknown. It is 200m. With the collision point being x=0.
 
  • #18
haruspex
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If you define one train as stationary at x=0 and keep the same relative velocity for the other train you the, it becomes a simple does the train stop in x distance w/ a certain initial velocity and a certain acceleration. If the freight train is stationary, then the displacement is not unknown. It is 200m. With the collision point being x=0.
Ok, I see. You're only addressing part (a). I should have realised that. The problem with that approach is that it doesn't help much with part (b), does it?
 

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