When Will the Two Trains Collide?

AI Thread Summary
The discussion focuses on a physics problem involving a passenger train and a freight train on the same track. The passenger train, traveling at 25.0 m/s, applies brakes causing a deceleration of 0.100 m/s², while the freight train maintains a constant speed of 15.0 m/s, with its caboose 200 m ahead. Two potential collision times are calculated: approximately 177.46 seconds and 22.54 seconds. The consensus is that the 22.54 seconds is the more plausible solution, as it represents the first and likely only collision. The conversation emphasizes the importance of physical realism in determining the correct time for the collision.
Toranc3
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Homework Statement


The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. The freight train is traveling at 15.0 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 0.100 m/s^2, while the freight train continues with constant speed. Take x=0 at the location of the front of the passenger train when the engineer applies the brakes.

Where will the collision take place?




Homework Equations



y=yo+vo*t+1/2*a*t^(2)

-b+- Sqrt(b^(2)-4ac)/2a

The Attempt at a Solution



yp-passanger train and yf=freight train

Yp=yo+vo*t+1/2*a*t^(2)
yp=25m/s*t - 0.05m/s^(2)*t^(2)

Yf=yo + vo*t +1/2*a*t^(2)
yf=200m +15m/s*t

yp=yf

-0.05m/s^(2)*t^(2) + 10m/s*t -200m

I get t= 177.4596 seconds and t=22.5403 seconds

How do I know which time is correct?
 
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Toranc3 said:

Homework Statement


The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. The freight train is traveling at 15.0 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 0.100 m/s^2, while the freight train continues with constant speed. Take x=0 at the location of the front of the passenger train when the engineer applies the brakes.

Where will the collision take place?



Homework Equations



y=yo+vo*t+1/2*a*t^(2)

-b+- Sqrt(b^(2)-4ac)/2a

The Attempt at a Solution



yp-passanger train and yf=freight train

Yp=yo+vo*t+1/2*a*t^(2)
yp=25m/s*t - 0.05m/s^(2)*t^(2)

Yf=yo + vo*t +1/2*a*t^(2)
yf=200m +15m/s*t

yp=yf

-0.05m/s^(2)*t^(2) + 10m/s*t -200m

I get t= 177.4596 seconds and t=22.5403 seconds

How do I know which time is correct?

I haven't checked your numbers. But sure you get two solutions. The later one assumes the trailing train survives the first collision and passes through the leading train and then collides again as the trailing train catches up with it again. Which do you think is the physically plausible one?
 
Dick said:
I haven't checked your numbers. But sure you get two solutions. The later one assumes the trailing train survives the first collision and passes through the leading train and then collides again as the trailing train catches up with it again. Which do you think is the physically plausible one?

The 22.5403 seconds one!
 
Toranc3 said:
The 22.5403 seconds one!

I'd agree with that. The first collision is usually the last.
 
Last edited:
Dick said:
I'd agree with that. The first collision is usually the last.

Thank you!
 
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