I hate to be the party pooper, but since the moderator has obviously yet to notice this post, I'll just say that you shouldn't put homework on the forum, and this is obviously a homework problem. (Unless you're making playing some sort of dairy farm/railroad tycoon game) But I have no power or authority on the forums, so go right ahead

EDIT - I take it back, I won't be mean since you're only asking for help and not the answer. Sry, that time of the month (not really, but you get the idea).

'There's something out there....something stupid...'
- Daria

DirectInput8 with the Keyboard , DirectInput8 with the Mouse , Using DirectX Audio 8

[edited by - masonium on September 16, 2002 9:06:58 PM]

You're right I noticed this when I tried to graph the quad on my TI89. Although I believe the the two roots now are 22.5403s and 177.46s. So assuming as its already been stated that the problem started at t=0, if there is a collision at 22.5403s I would assume there would not be another at 177.46s

22.5403

200 + 15.0(22.5403) = 538.1045m

Also @ AP: Yes Im sorry I overlooked the fact that the quad was for the variable t,time.

I'd also like to thank everyone who posted and put my mind at ease.

*Oh, posted late... seems someone already found the mistake as well*

[edited by - Zorbfish on September 16, 2002 9:09:41 PM]

Yes, masonium I know this is a homework problem. But I adhered to the rules and didn''t just look for the answer. I provided some of my work and asked if anyone could give me any direction. They did, I thank them, and Im sorry if this was the wrong place to post this. I just assumed so many great mathemiticians (sp?) come to these forums that I could at least gets some insight into what I was doing wrong.

But this is a game development site, called gamedev.net, not helpmewithmyphysicshomework.net, and physics problems should not be posted unless they relate directly to game programming. Just my 2 cents.

I would say that the first half of this thread was conducted very well and I wouldn't see a need to close it. Zorbfish is correct in that he/she did not ask for the solution, showed that they had attempted to find the answer and asked for a hint. The AP handled this very nicely by leading Zorbfish toward the answer. My only suggestion to improve this would be to Zorbfish. Once you have attempted the problem using the hint, it would be worthwhile posting your new working. That way other posters can more easily pick up where you may have made a mistake and direct you toward that point, rather than posting the working themselves (Burning_Ice)!

y2kiah: I respect the fact that some people don't want to see homework problems posted to this forum... and I agree that we should discourage it. However, since we have not been able to find a way for cutting people fingers off when they type in homework problems, or when they post answers to such problems, we (the moderators) have had to resort to trying to educate forum members to handle homework questions in a constructive manner that encourages the student to learn and understand rather than provide them with direct answers. That way, when they tackle maths/physics problems related to games (and we hope you all do at some time), they'll have better research skills with which to tackle the problem.

Finally, Zorbfish. Might I suggest that when you have worked out the problem, you sit down and write out an algorithm for solving similar problems in the future! This will make it easier to remember for your exams/tests, rather than having to remember each particular problem you solved, hoping to find one that matches the new question!

Regards,

Timkin

[edited by - Timkin on September 19, 2002 2:13:47 AM]

Just a few small things I''d like to direct your attention to before this thread goes under

quote:

---

a). Will the cows nearby witness a collision?

---

Well, it really depends whether or not the cows are facing the track at the time the collision occurs. You know them cows, they could be facing any direction at any time, and that''s not even taking into account that they lower their heads to feed, or actually happen to walk around the pasture. If, however, we knew the cows'' direction and position relative as a function of t as well, then we would be able to determine there direction as the time of the collision. In addition, we would have to specify a FOV for these cows, as it isn''t 0°.

But there is something else. Let''s say the cows are close enough to actually hear the crash taking place or feel a shock wave. If we determine them to not be facing the collision location at the time it takes place, will they turn around to witness the crash before the event is complete - if they even turn around at all? If these cows are hard of hearing, will the shock wave produced by the collision be strong enough to distract the cows and subsequently induce them to turn around and "witness" this crash before it is over?

There are so many variables involved in answering the questions put forth that I''m surprised the question is even asked outside a Graduate-level class!

quote:

---

b). if so where will it take place?

---

This question completely depends on the solving of part a), which looks completely hopeless at this point...

quote:

---

c). on a single graph sketch a x t graph of both the trains positions.

---

Did your teacher say if you should also graph the location and direction of the cows as a function of time? If so, you might have to look around for the equations that describe the behavior of cows in discrete mathamatical terms.

In the end, it''s the cows'' fault.

Timkin and forum members,

I'd like to thank Timkin for his concise and (in my opinion) proper analysis of this thread. I concur with his analysis in full, and want to emphasize that it is up to every forum participant to be proactive in helping homework help-seekers to learn to solve problems for themselves, rather than merely providing answers to them.

I would also like to acknowledge that many of the forum members are doing the right thing, in my opinion, such as Zorbfish (who followed the forum guidelines for homework questions that are posted), and y2kiah and masonium (who are helping us moderators to educate the masses about the forum policies). The maturity of folks in this forum is actually quite amazingly great most of the time. I APPRECIATE THIS! .

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

[edited by - grhodes_at_work on September 18, 2002 11:31:46 PM]

Thank you. And I''d also like to make it known that I did in fact spend two days trying to solve this problem and that posting on this forum was just a shot in the dark.
For the whole first day I spent time working on the problem as I thought it should be solved. Unfortunately that''s when I hit the road block. Without any significant example provided in my text I asked around to all my peers and none of them had started the work. (they procrastinate to the last minute) I even had asked my instructor but apparently no help can be obtained until the problems are discussed AFTER the assignment has been collected. Also due to class schedule conflicts I couldn''t go to any of the physics help groups on campus.So with nowhere to turn to I started searching with google but could not find any such example of this type of problem.
Day two and no luck, I still didn''t have it done. I began searching for just math/physics help sites and either I''m looking in the wrong places, or there''s not many. So if anyone know''s a few good ones or even just forums on such subjects plz don''t hesistate to tell me. By the end of the day I had been thinking about posting here, despite knowing how its not looked on unfavorably.
So all in all, Im just glad I known how to approach such a problem now. And don''t worry about anymore posts, because this was just a last straw for me.

Ok, maybe I''m just totally missing something...but wouldn''t this question have been 100 times easier if you just used relative velocity to figure it out?

Vrel = 10m/s
A = -.1m/s^2

therefore to figure out how far the passenger train would travel before it slows down to the same velocity as the freight train, just use

d = 10(100) + .5(-.1)(100)^2

Now why did I use 100 for t? Well, if the acceleration is -.1m/s^2, then it would take 100 seconds for the RELATIVE velocity to reach 0m/s.

This equation gives 500 for the distance that the passenger train would travel. So, obviously the trains will collide.

Next, to figure out the time that they collide, just use the same equation, but leave t as the unknown:

200 = 10t + .5(-.1)t^2

Using the quadratic equation, this gives the two roots 177.4 and 22.5 (just like the ones stated above).

Now since the passenger train travelled 500m relative to the freight train after 100s, obviously 177.4s must be too much.

Ok, so maybe it''s not 100 times easier...but I just prefer relative velocity questions. :D

Massif - "It means mountain."