Physics question

[GML3G0]

So here's the problem.

Train moving at 36 m/s hit's brakes and decellerates at a rate of 3 m/s^2 to avoid colliding with a train 100m in fron of it which is moving at a constant velocity of 11 m/s in the same dirrection. Will they collide?

I say yes, they will, but according to my teacher they won't...

Here's what I did:

You want the faster train to slow down to 11 m/s before it reaches the same distance as the other train.

So...

V(final) = V(initial) + at
11 = 36 + (-3)t
t= 8 and 1/3 s


Train 1

d = 36 ( 8 1/3) + 0.5 (-3) (8 1/3) ^2
d = 195.83333...

Train 2

d = 11 (8 1/3)
d = 91.666...

195.8333.... = 91.6666... + 100 ? No, so the first train overtook the first, so obviously they collided...

Teacher's reasoning was:

When you put the values in the distance formula, it takes 12s for the fast train to come to a halt, and then plug 12s into the distance formulas for both trains and the slower train went further. OK, but what about before that time????

That's why his method doesn't work...

Train 1: ||----------|1|--------|2|---|3|--|4|-|5|

Train 2: ||---------------|1|--|2|--|3|--|4|--|5|--|6|--|7|

So who is right?

 

[spidey07]

Train 1 is 100 meters away when they started.

no collision

-edit- I can't read. If your math is right they did collide.

 

[alexeikgb]

Your assumption that it's enough for the fast train to slow down to 11 m/s to avoid collision is wrong, they will collide if the fast train stops decelerating at 11m/s.
BUT he should continue to decelerate if he really want to aviod the collision thus 12 sec and teacher is right,

 

[GML3G0]

I added 100 to make up for the difference. Read it again.

 

[silverpig]

You're right. The trains will collide.

 

[Armitage]

At least from a quick look, you are right. It's irrelevant how long it takes the trailing train to stop - what matters is how long it takes for it to reduce its speed <= the speed of the first train.

 

[GML3G0]

Originally posted by: alexeikgb
Your assumption that it's enough for the fast train to slow down to 11 m/s to avoid collision is wrong, they will collide if the fast train stops decelerating at 11m/s.
BUT he should continue to decelerate if he really want to aviod the collision thus 12 sec and teacher is right,

I was just using that as a basis to find what time I should look be looking at... regardless

at the 8.3333 second mark, the train that is slowing down passed the other.

 

[silverpig]

Originally posted by: alexeikgb
Your assumption that it's enough for the fast train to slow down to 11 m/s to avoid collision is wrong, they will collide if the fast train stops decelerating at 11m/s.
BUT he should continue to decelerate if he really want to aviod the collision thus 12 sec and teacher is right,

How is a train going to hit a train in front of it when it's going slower?

That's like rear ending someone on the freeway when you're going 30 and they're doing 100. You can't do it.

 

[silverpig]

The only way that it matters how long it takes the train to stop is if BOTH trains are decelerating.

 

[mugs]

Originally posted by: alexeikgb
Your assumption that it's enough for the fast train to slow down to 11 m/s to avoid collision is wrong, they will collide if the fast train stops decelerating at 11m/s.
BUT he should continue to decelerate if he really want to aviod the collision thus 12 sec and teacher is right,

If they're both going the same speed and haven't collided yet, then they won't collide.

I haven't done any of the math, but your reasoning is sound and the teacher's is not (to OP)

 

[Armitage]

Originally posted by: alexeikgb
Your assumption that it's enough for the fast train to slow down to 11 m/s to avoid collision is wrong, they will collide if the fast train stops decelerating at 11m/s.
BUT he should continue to decelerate if he really want to aviod the collision thus 12 sec and teacher is right,

If they are both traveling 11 m/s how will they collide?

 

[Tiamat]

mine was wrong. totally screwed up lol

 

[BigJ]

The trains will be at the same spot when t = 20/3s, which means there is a collision.

To see if they collide, set the displacements equal to each other, adding 100 to d2 to make up for the 100m head start.

d1 = d2 + 100
36*t + -1.5*t^2 = 11*t + 100
-1.5*t^2 + 25*t - 100 = 0

Use the pythagorean theorem to find out t.

t = 20/3 and 30/3.

Since T exists, that means they will collide.

If you wanted to avoid colliding, you would have to decelerate greater than a rate of 3.125m/s^2.

 

[Armitage]

Here's a quick plot showing the issue: http://pics.bbzzdd.com/users/Armitage/train.png

Just plot the position of both trains:
P = P0 + V*t + 0.5*A*t^2

For Train 1:
P0 = 100
V = 11
A = 0

For train 2:
P0 = 0
V = 36
A = -3

You can see from the plot that from about6.3 seconds to about 10 seconds the position of train 2 > position of train 1 - this means they collided. At 12 seconds when train 2 has stopped, it is indeed further behind, but that's irrelevant.

 

[GML3G0]

yay, I'm right.

wow, you even graphed it. lol.

well, thanks.

 

[BigJ]

Originally posted by: Armitage
Here's a quick plot showing the issue: http://pics.bbzzdd.com/users/Armitage/train.png

Just plot the position of both trains:
P = P0 + V*t + 0.5*A*t^2

For Train 1:
P0 = 100
V = 11
A = 0

For train 2:
P0 = 0
V = 36
A = -3

You can see from the plot that from about6.3 seconds to about 10 seconds the position of train 2 > position of train 1 - this means they collided. At 12 seconds when train 2 has stopped, it is indeed further behind, but that's irrelevant.

The exact numbers are in the post above yours

 

[Armitage]

Originally posted by: BigJ

Originally posted by: Armitage
Here's a quick plot showing the issue: http://pics.bbzzdd.com/users/Armitage/train.png

Just plot the position of both trains:
P = P0 + V*t + 0.5*A*t^2

For Train 1:
P0 = 100
V = 11
A = 0

For train 2:
P0 = 0
V = 36
A = -3

You can see from the plot that from about6.3 seconds to about 10 seconds the position of train 2 > position of train 1 - this means they collided. At 12 seconds when train 2 has stopped, it is indeed further behind, but that's irrelevant.

The exact numbers are in the post above yours

Yep - except you actually solved a quadratic equation to find those roots - not a right triangle

 

[Fenixgoon]

woops, misread the question

 

[TuxDave]

Yeah, my high school physics teacher gave the same example and I proved her wrong after class. She corrected herself the next session.

 

[DrPizza]

This may help...
Change your frame of reference to the train that is moving ahead at 11m/s
From that frame of reference, the other train is approaching you at 25m/s. But, that train is slowing down at a rate of 3m/s each second.
That train is 100 meters away. Will it catch up?
It's going to slow down for 8 1/3 seconds before it is going at the same speed you are.
So, after 8 1/3 seconds, its speed, relative to you, is 0m/s
Average velocity during those 8 1/3 seconds is 12.5 m/s
Distance traveled during those 8 1/3 seconds is, unfortunately, slightly more than 100 meters.
You are correct.

Now, if the initial velocity of the train (relative to the track) was 35 m/s, then relative to the constant velocity train, it would be moving 24m/s. It would take 8 seconds to stop (relative to the 11m/s train), and travel 8*12=96 meters during that time, thus, not colliding.

<---- physics teacher

Additionally, if you have a graphing calculator, simply graph the position function of both trains.
Train A = 100 + 11t
Train B = 36t + 1/2 (-3)*t^2

You will see a straight line for the first train, and a parabola for the second train.
Clearly, there are two points of intersection.
One is at 6 2/3 seconds
Check... in 6 2/3 seconds, train A will be 100 + 11(6 2/3) = 173 1/3 meters down the track
In 6 2/3 seconds, Train B's velocity will have decreased by 6 2/3 * 3 or 20 m/s
So, it's velocity at that time is 16m/s
The average velocity during those 6 2/3 seconds is (36+16)/2 = 26m/s
And finally, 6 2/3 seconds * 26 m/s = 173 1/3 meters.
(And, I'm committing what I'd consider to be a grievous error in my class - the use of fractions instead of decimals to the correct number of significant digits)

Incidentally, if they were running on parallel tracks to avoid the collision, but train B slowed down anyway,
it would overtake train A at the 6 2/3 second mark, 173 1/3 meters down the track.
Train A would regain the lead at the 10 second mark, 210 meters down the track.

Incidentally, the crash would occur with the difference in speeds equal to 5 m/s, or about 11mph. That would do pretty significant damage.

Edit: yeah, I see the times were already answered, but mine is more elegant

2nd edit: Nice graph, Armitage... what software app do you use? It's certainly a lot nicer than staring at a graphing calculator.

 

[mugs]

Now the question is, can you convince your teacher that she's wrong without showing her this thread (which might piss her off)?

 

[spidey07]

Originally posted by: mugs
Now the question is, can you convince your teacher that she's wrong without showing her this thread (which might piss her off)?

I recommend a powerpoint presentation in class, complete with projector.

That will go over GREAT!

 

[GML3G0]

Originally posted by: mugs
Now the question is, can you convince your teacher that she's wrong without showing her this thread (which might piss her off)?

him*

I'll just show him Armitage's graph. lol. Thanks, Armitage.

I tried showing to him my reasoning and why he's wrong, but he was kind of stubborn. I eventually got him to see what I was talking about, and he was, "0o0o0o that's interesting," but I'm not sure if he's fully convinced.

Oh, and by the way, I wasn't trying to find the time in which they collided. I know it's not 8.333, I'm just saying if the faster train's position at that point was less, then they wouldn't have collided. I was just using that as a reference point. I know how to figure out the time of impact, I just wasn't looking for that. Anyways, thanks everyone.

 

[bernse]

Surprised your teacher didn't say they were just on parallel tracks and it was a trick question.

 

[Armitage]

Originally posted by: DrPizza
<snip>

Edit: yeah, I see the times were already answered, but mine is more elegant

Arguable - I personally like taking the difference of the position equations and finding the roots as BigJ did
But it is nice to see a physics teacher that gets it - very sad that the OP's teacher fubared such a basic problem.

2nd edit: Nice graph, Armitage... what software app do you use? It's certainly a lot nicer than staring at a graphing calculator.

I was lazy - that's just a screen capture from OpenOffice spreadsheet app. I really wish you could just directly save a graph to an image format.