Train A is at Station T, and Train B is at station U. The distance between Stations T and U is 800 km. Train A started to move from Station T to Station U at 1 pm with initial time t=0, and train B started to move from Station U towards station T at 2:45 pm, with an initial acceleration of 11m/(s^2). As the two trains meet, which one will be the nearest to station T?
the equation of speed of train A is v= (t^2) + 6t +9
the equation of speed of train B is v= 3(t^2) + 11t +10
The first one with the simplest way of solving this question wins the 10 points.
2006-07-21 09:51:41 · 20 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I meant by my question when they are at the same point.
2006-07-21 10:10:04 · update #1
This is not a homework, this is a trick question that i wanted to publish.
2006-07-21 10:11:09 · update #2
Neither one, when the trains meet they will both be at the same distance from station T.
2006-07-21 09:57:29 · answer #1 · answered by GOL10 2 · 4⤊ 1⤋
To bandaid or whatever: I've never met a more cocky rooster who's wrong..There are always 2 one-way tracks between 2 stations and not a one two-way track. Also the accn of B may enable it to catch up a greater distance than what A chugged thru b4 2.45
To Vivek: Ure right mostly but when u differentiate speed of A u get 2t+6 as the acceleration, and the way uve dun ur calculation of distance moved by A from 1.00 to 2.45, u assume acceleration is zero but it's only so when t= -3 which cannot be while if A's speed has to be constant accn shud always be zero. Therefore both the speeds of A and B in terms of t are at the point of passing. Besides after integrating uve 4gotten bout dealing wi8 the constant.
Correct me if I'm wrong.
Afterthought: At time 0s, distance from T is 0 so constant's taken care of..sorry!
2006-07-22 04:56:12 · answer #2 · answered by life_boat 2 · 0⤊ 0⤋
OK This though is a lengthy looking problem is very easy.
You need to use the formula for distance D= integration of v over dt,
So v= v(t) given integrate this with the given time interval to get distance.
Now the train A has already run for 105 mins before train B started. So the distance travel led by train A in 105 minutes must be calculated 1st. For this integrate the velocity equation. I guess your velocity unit is in m/s. So integrate this and that will be the distance traveled by train A.
Now suppose that comes 200 Kms.
So at 2:45 pm distance between two trains is 600 kms. Now you can use the relative speed concept. The combined speed here will be some of two given speed and integrate it over any given t. The distance here will be 600kms with units conversion applied.
Now you need to solve for t. This t will be cubic in equation.
Once you get t you can integrate individually and find the distance.
This will give you answer.
2006-07-21 19:28:43 · answer #3 · answered by Vivek 4 · 0⤊ 0⤋
Okay...if the trains are both at the same point when they meet (obviously), then they are the same distance from station T, regardless of what speeds they were going, or how far apart the stations are.
2006-07-21 17:32:35 · answer #4 · answered by mathgirl 3 · 0⤊ 0⤋
Train A, since the caboose of A will be closer than the engine car of B, which are the two points of each train that is closest to Station T.
2006-07-21 17:14:56 · answer #5 · answered by Wied 3 · 0⤊ 0⤋
Duh! When they meet they will be equal distances from T! I can't believe the number of people that didn't see that... perhaps all your numbers and notation scared them. :)
Now this is assuming you think of the trains as points like in most math problems (e.g. consider only the front of the train). In reality, when the front of the two trains meet, the rear of train A is closer to station T since it is heading away from T and train B is heading toward T...
By the way, care to figure out what their velocities will be when they meet? At the rate they are accelerating that's going to be one spectacular crash. I hope they are on separate tracks. :)
2006-07-21 16:57:36 · answer #6 · answered by Puzzling 7 · 0⤊ 0⤋
Train A will always be the closest to Station T when they meet.
Just imagine the trains on the same track at a stand still nose-to-nose. The train that started at Station T will still be closer to Station T.
Here's a "sorta" illustration
Station U.........trainB-trainA.........Station T
2006-07-21 17:04:12 · answer #7 · answered by weigh_with_words 1 · 0⤊ 0⤋
train A of course since train A is already at station T at the start when they collide train a will be closest the speed really doesnt matter
2006-07-21 16:57:52 · answer #8 · answered by Anonymous · 0⤊ 0⤋
When they meet they are at the same point, hence they are at the same distance to each station.
2006-07-21 17:10:08 · answer #9 · answered by LUIS 6 · 0⤊ 0⤋
isnt there enough complication in life already, without making up new ones? Jeez! My brain hurts just looking at that!
2006-07-21 16:56:03 · answer #10 · answered by Big hands Big feet 7 · 0⤊ 0⤋