Why is the following impossible? 2 trains

[feihong47]

1. Why is the following situation impossible? A freight train is lumbering along at a constant speed of 16.0 m/s. Behind the freight train on the same track is a passenger train traveling in the same direction at 40.0 m/s. When the front of the passenger train is 58.5 m from the back of the freight train, the engineer on the passenger train recognizes the danger and hits the brakes of his train, causing the train to move with acceleration -3.00 m/s2. Because of the engineer’s action, the trains do not collide.

I first calculated the distance it takes passenger train to stop:

0 = (40)²-2(3)$\Delta$x
Solving, $\Delta$x = 266.7 m to stop

Next I calculated the time it took to stop:

0 = 40 - 3(t), solving, t = 13.3 s

Next I calculated the distance of the freight train in those 13.3 seconds,

distance = 13.3 * 16 m/s = 213.3 m

Since the separation is 58.8 m before the passenger train braked, and the freight train goes 213.3 before the passenger train stops, I added 58.8 + 213.3 = 271.8 m

271.8 > 266.7, thus it appears the passenger train WILL stop just in time, but the problem obviously doesn't think so... what did I　do wrong?

 

[BruceW]

You did the problem correctly. The problem says 'the trains do not collide' - it agrees with you :)

 

[feihong47]

You did the problem correctly. The problem says 'the trains do not collide' - it agrees with you :)

The problem states that it is impossible that they do NOT collide, ie, they do collide.

 

[phyzguy]

You did the problem correctly. The problem says 'the trains do not collide' - it agrees with you :)

This is not correct. They will collide. Look at it in the frame of the freight train. The passenger train is 58.5 m away and closing at 24 m/s. It takes 8s (24/3.0) before the relative velocities of the trains drops to zero, and in this time, the front of the passenger train will have moved 96 m (24*8 - 3/2 *8^2) forward relative to the back of the freight train. Since 96>58.5, they will collide. To see the error in what the OP did, imagine that they are on two parallel tracks so that they don't actually collide. What happens is that the front of the passenger train will pass the back of the freight train. As the passenger train continues to decelerate below 16 m/s, the freight train will pull ahead, and when the passenger train comes to a stop, the back of the freight train will again be ahead of the front of the passenger train. This is what the OP calculated. But there was a period of time when the two trains overlapped, which means that they will collide if they are on the same track.

 

[feihong47]

This is not correct. They will collide. Look at it in the frame of the freight train. The passenger train is 58.5 m away and closing at 24 m/s. It takes 8s (24/3.0) before the relative velocities of the trains drops to zero, and in this time, the front of the passenger train will have moved 96 m (24*8 - 3/2 *8^2) forward relative to the back of the freight train. Since 96>58.5, they will collide. To see the error in what the OP did, imagine that they are on two parallel tracks so that they don't actually collide. What happens is that the front of the passenger train will pass the back of the freight train. As the passenger train continues to decelerate below 16 m/s, the freight train will pull ahead, and when the passenger train comes to a stop, the back of the freight train will again be ahead of the front of the passenger train. This is what the OP calculated. But there was a period of time when the two trains overlapped, which means that they will collide if they are on the same track.

thanks! makes sense.. so in those problems I HAVE to use relative velocities right?

 

[BruceW]

I see what me and the OP'er did wrong now.

faihong - you can calculate what happens from any inertial reference frame (eg you can use the original reference frame given, if you want to). But what we missed was that although the trains don't overlap when the passenger train stops, the trains would have overlapped before that time. This is more obvious from the inertial frame of the freight train, which is why phyzguy has explained the problem using that reference frame.

 

[phyzguy]

I see what me and the OP'er did wrong now.

faihong - you can calculate what happens from any inertial reference frame (eg you can use the original reference frame given, if you want to). But what we missed was that although the trains don't overlap when the passenger train stops, the trains would have overlapped before that time. This is more obvious from the inertial frame of the freight train, which is why phyzguy has explained the problem using that reference frame.

Agreed. You don't have to use relative velocities - it's just easier that way.

 

[newly]

Relative Velocities with Acc?

thanks! makes sense.. so in those problems I HAVE to use relative velocities right?

Hey Guys, but you can't use Galilean Transformation when the relative velocity between the systems is not constant... (a = -3 m/s2) am I right?
My 2 trains do not collide at all.. :-(
I did:
X (train pass.) = x (train freight)
v_i2*t + 1/2 a*t² = x₁ + v_i1*t
[STRIKE]and they do not meet.[/STRIKE]
[ THEY DO MEET, if you solve this equation... I had an error in my substitution ]
am i right?

 

[Chestermiller]

Suppose that, at time t = 0, the passenger train is at x = 0, and the freight train is at x = 58.5. At time t, the passenger train will be at x = 40 t -1.5 t², while the freight train will be at 58.5 + 16 t. Find the time t at which these two distances are equal. If there is no real solution, the trains will not collide. If there is a real solution, they will collide at the shorter of the two solutions.

 

[phyzguy]

Hey Guys, but you can't use Galilean Transformation when the relative velocity between the systems is not constant... (a = -3 m/s2) am I right?
My 2 trains do not collide at all.. :-(
I did:
X (train pass.) = x (train freight)
vi2 t + 1/2 a t ^ 2 = x1 + vi1 t
and they do not meet.

am i right?

No. You are making the same mistake as the OP did. Re-read through the thread to see your error.

Yes, you can use a Galilean transformation, because you are transforming out the constant velocity of the freight train. Anyway, there is no need to use any transformation. Try plotting the positions of the front of the passenger train and the back of the freight train in the rest frame. You will see that they overlap for a time.

You do know this thread is over two years old, don't you?

 

[newly]

No. You are making the same mistake as the OP did. Re-read through the thread to see your error.

Yes, you can use a Galilean transformation, because you are transforming out the constant velocity of the freight train. Anyway, there is no need to use any transformation. Try plotting the positions of the front of the passenger train and the back of the freight train in the rest frame. You will see that they overlap for a time.

You do know this thread is over two years old, don't you?

Thank you guys! The trains will collide.
(I've substituted 9.8 instead of 3 for a) :-) how silly, I was in free fall with my chair maybe:-)

Thanks phyzguy - I am not allowed to use the Galilean transformation yet... (but great tip, I will remember them now!)