Hello,

I am a tenth-grader physics student, and I have a mechanics-realted problem you may think is quite trivial, but I have been unsuccessfully trying to solve it for a whole day, so I would thank you if you answered.
Please consider the attached graph; the question relates to it:
Two trains, a red one and a yellow one, travel towards each other on a straight railway-track. When the distance between them is 800 m, the drivers notice the danger and hit the brakes simultaneously. The graph shows their velocity in respect to time from the moment of hitting the brakes, if each train had travelled alone.
(Here there are a couple of pretty eaay questions; I'm skipping right to the problem.)
Did the trains collide into each other? If they did – when did it happen? And if the didn't – what was the distance between them when they stopped?
The other attached image shows my feeble attempts to solve this question.
The last thing I should do about these trains is draw (on the same coordinate system) a schematic displacement-time graph for both trains from t=0 to the moment of stopping/collision.

Thanks for your help!

Attachments:
Image 1 (the calculations) - http://img468.imageshack.us/img468/2295/trains1dy3.jpg
Image 2 (the graph) - http://img157.imageshack.us/img157/200/trains2ab7.jpg

P.S: I just noticed a couple of small alignment problems in Image 1. I'm sorry about them, I think my OpenOffice needs a bit of configuration.

I don't quite understand how you set up the problem, and I don't understand why you tried to set it up that way in the first place, so I can't check your work and find the error without spending a lot more time than I'm willing to do. You're not using the standard symbol s for distance so your formulas are confusing.

You may be confused because the acceleration of one of the trains has a negative value, but don't be. The formula works either way.

You know the rate of acceleration for each train. Therefore you can easily calculate how long it will take each train to reach a complete stop.

You know the formula for calculating distance traveled as a function of acceleration and time. Since you now have both of those variables for both trains you can calculate the distance that each train will travel in the time required to stop. If the sum of the stopping distance of both trains < 800m they will not collide.

I am generally loth to assist high school students with their homework since that is their parents' job. But in your case my feeling is that you got lost trying to express the problem in mathematical notation. I tried to get you restarted on a sound basis and the rest is up to you.

Good luck.

Hello,

I am a tenth-grader physics student, and I have a mechanics-realted problem you may think is quite trivial, but I have been unsuccessfully trying to solve it for a whole day, so I would thank you if you answered.
Please consider the attached graph; the question relates to it:
Two trains, a red one and a yellow one, travel towards each other on a straight railway-track. When the distance between them is 800 m, the drivers notice the danger and hit the brakes simultaneously. The graph shows their velocity in respect to time from the moment of hitting the brakes, if each train had travelled alone.
(Here there are a couple of pretty eaay questions; I'm skipping right to the problem.)
Did the trains collide into each other? If they did – when did it happen? And if the didn't – what was the distance between them when they stopped?
The other attached image shows my feeble attempts to solve this question.
The last thing I should do about these trains is draw (on the same coordinate system) a schematic displacement-time graph for both trains from t=0 to the moment of stopping/collision.

Thanks for your help!

Attachments:
Image 1 (the calculations) - http://img468.imageshack.us/img468/2295/trains1dy3.jpg
Image 2 (the graph) - http://img157.imageshack.us/img157/200/trains2ab7.jpg

P.S: I just noticed a couple of small alignment problems in Image 1. I'm sorry about them, I think my OpenOffice needs a bit of configuration.
Hi Omri,
I'd attack this problem by first determining the stopping distance for each train. If the distances add up to less than 800, you know that they didn't collide and how much distance is left between them. If the distances add up to more than 800, you know that they did not have enough space to stop, so they did collide.

When you find that they did collide, that's when you proceed with your calculations to solve for t.

Some problems with your calculations:
The yellow train's acceleration should be -1.5
&Delta;x for the red train will be negative, so you need to solve for when the difference between the displacements is +/-800, not the sum.

I always found it useful to just take one of the objects as the rest frame, and adjust the other's velocity and acceleration to reflect what an observer on one (does not matter which one) train would see. its easier to think about if you only have one object with velocity/acceleration.

my physics teacher was always interested in how I would look at problems. I, for some reason, could do the problems better if I took my own route, rather than the traditional. so maybe you just need to think about it in a few different ways.

Thanks everyone,

I noticed I made a typing mistake in the yellow train's acceleration; I know it should be -1.5, as you can see from the dx expression I wrote for it.

You may have misunderstood my problem: I realise the trains collided, and I managed to calculate the displacements; but then I created and equation - dx[red]+dx[yellow]=800, which gave me the square root of a negative number.

And a short reply for the first replier - since the railway is a straight line and it is constant acceleration, the displacement equals to the distance, therefore s = delta x. Also you said that parents should help their sons and daughters with their homework; that is impossible in my case for two reasons: one, I attend a boarding school, and two, since I take physics as my main subject, and since the school specialises in natural and exact sciences (it also has arts and music departments, but they are irrelevant), the level will soon be to high for my parents to help me - they don't have high physics education.

Thanks again!

Have you tried something like this:

Xy=0 + 45t - 0.5*1.5*t^2
Xr=800 - 30t + 0.5*1*t^2

(I think I got those right. Better check.)

The if the trains collide they will do so when Xy=Xr. Use that to combine the two equations above into a single quadratic and solve with the quadratic equation. You will of course have to verify which of the two is a real solution, or indeed if either is.

displacement is the surface area under the velocity graph. heh, this what integrals are used to calculate, areas under graphs, heh.
however our graph is a simple straight line so we dont really need integrals :)

anyways, the trains do collide as the total area(rectangular triangle between axises and the velocity graph) for the two graphs is greater than 800. 45 * 30 / 2 + 30 * 30 / 2 > 800

so, total displacement d = 800;
time t is the same for both at point of collision.

from the graph
a_1 = 45/30 = 1.5
a_2 = 30/30 = 1
v_1 = 45
v_2 = 30

s_1 = v_1 * t - a_1 * t^2 / 2
s_2 = v_2 * t - a_2 * t^2 / 2
d = s_1 + s_2

800 = 45 t - 0.75 t^2 + 30t - 0.5 t^2
800 = 75 t - 1.25 t^2
0 = - 1.25 t^2 + 75t - 800

t_1 = (-75 + sqrt(5625 - 4000)) / -2.5
t_1 = 13.875

t_2 > 30 - t_2 is the solution if they continue deccelerating and start going backwards in case they miss each other they will go past each other again going backwards.

t_1 is indeed the solution

what you did wrong is play with +/- a lot when defining your task and got confused. there is no need for that. it is simple, when both trains absolute displacement sum up to 800 they will collide.

exactly like kevin said, you need to add the inital displacement of 800 into one of the deltas if you really want to be picky with directions.

like he said
Xy=0 + 45t - 0.5*1.5*t^2
Xr=800 - 30t + 0.5*1*t^2

and then do Xy = Xr.

obviously the two displacements from just velocities 45t and 30t need to add up to 75t, you have 15t in your equasion

vx220: Thanks, but there is one thing I misunderstand - you said that:
s = v_0-a*t^2 / 2
Isn't it supposed to be
s = v_0+a*t^2 / 2
?
And also, you said I need the absolute displacement - why is that? Am I right when I assume you do it to get the distance (which cannot be negative)?

Thanks again!

Thanks everyone,

I noticed I made a typing mistake in the yellow train's acceleration; I know it should be -1.5, as you can see from the dx expression I wrote for it.

You may have misunderstood my problem: I realise the trains collided, and I managed to calculate the displacements; but then I created and equation - dx[red]+dx[yellow]=800, which gave me the square root of a negative number.
Hi Omri,
the expression for the red train's displacement is negative, so your equation should be:
dx[yellow] + (-dx[red]) = 800


This is really the same thing that kevin and vx220 said, but closer to your original approach.
You could also take cato's approach, and express the problem in terms of how the distance from the front of the red train to the yellow train changes with time.

In that case:
V0 = 75
a = -2.5
s = 800
t = ?


No matter which approach you take, you'll end up with the same quadratic in t, with two real roots.

Thanks eveyone, I just managed to reach the final solution using your advice.
The way that finally worked out for me is the one offered by Pete, although I am going to try using the other ways of attacking the problem.

Thanks a lot! :)

vx220: Thanks, but there is one thing I misunderstand - you said that:
s = v_0-a*t^2 / 2
Isn't it supposed to be
s = v_0+a*t^2 / 2
?
And also, you said I need the absolute displacement - why is that? Am I right when I assume you do it to get the distance (which cannot be negative)?

Thanks again!

if the starting speed is positive then acceleration is negative - deceleration.
we could say s = vt + at^2/2 if we set 'a' to be -1.5.
or like i do, s = vt - at^/2 but my both of my 'a' are set to be positive numbers 1.5 and 1. that's why i subract them.
doesnt really make any difference how you put it, you just have to know what you're doing.

it is easier without confusing yourself with +/- signs. you KNOW the trains are going towards each other, there's no need to express direction in equations. you simply calculate how much distance each of the trains will cover in any direction - absolute distance. and total the two distances to 800..

if, for example, you need to make a computer program displaying the movement of the two trains you would have to do it using kevins way - you have to calculate correct x for each of the trains in a cartesian coordinate system for display on computer screen.

dunno, such stuff is kinda inbuilt in my way of thinking and comes natural so its perhaps tough to explain for me. heh - i tried. i did a lot of programming 2d game engines as a kid - it was fun for me, heh. movement, inertia, gravity, acceleration, turning, friction, collision all mostly in 2d... anyway, all the stuff you need to make a spaceship fly around or a car drive around etc. did some stuff in 3d, but back then i used qbasic which was simply too slow for 3d. 386 and 486 pcs werent really capable of good 3d anyway, lol.

The reason I do it the way I do is to avoid -/+ confusion as well. lol. If you set up a coord axis with a definate origin and direction it makes the sign of the velocities and accelerations obvious for me. But whatever works. ;)

One other thing for Omri, when you set Xr = Xy and solve for a single quadratic of the form 0 = a + b*t + c*t^2 , what you are effectively doing is subtracting one side of Xr = Xy from both sides. For example:

Xr = Xy
0 = Xy - Xr

but Xy - Xr is just the separation S .

Also, I would mention that most of the changes in signs of the various methods in this thread really boil down to choices of origin and direction of coord axis.

buh, you simply have to know what you're doing intuitively. you have to know what you are supposed to get and how it is supposed to look. and then simply solve the quadratic equasion. you can play with signs all day long and get the same results.

i only give the "scientific" advantage to kevins way as it is a proper calculation in a coordinate system which could be used to calculate and display both x positions at any point in time in a coordinate system - a computer screen for example. and the resulting coordinates could be used to calculate interactions with other moving entities in this coordinate system, providing you have the same type formulae for their movement. blah...

given there's only two trains, you can do whatever you want with signs and ways to put it