Intro Physics problem - 2 oncoming trains, help!

  • Thread starter matxjos
  • Start date
  • #1
16
0
Hey I'm in high school and I need help with this seemingly easy problem.
I have 3 constant acceleration equations but I don't know how to use them in conjunction with this problem.

A train leaves from Chicago toward Atl going 20 m/s. At the same time another train leaves from Atl going to Chicago at 25 m/s. The cities are 900km apart

When do the trains meet?
How long before they meet?
 

Answers and Replies

  • #2
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,721
2,979
What are the equations that you have? What do they describe? What kind of strategy do you think is appropriate? Give us your thoughts and we will help you.
 
  • #3
16
0
ok i have
x = x(starting distance) + v(starting vel.)t +.5at^2

v^2 = v(starting)^2 +2a(x-x[starting])

and v=v(Starting) +at

i found the amount of time is 300 for the 20m/s train and 268.3281573 for the 25 m/s train.
but how to found the time they meet?
 
  • #4
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,721
2,979
Just a moment. What did you use for the acceleration a? Also, when you say you "found the time", how did you find that time? What is that time?
 
  • #5
16
0
ah shoot you made me realize i used velocity for the acceleration dang it.
and yeah, i have no idea what time is supposed to be in. i'm guessing seconds.
and also one has to change the 900 km to 900000 metres i'm pretty sure.
i sort of have no clue on this problem...
 
  • #6
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,721
2,979
So what is the value of the acceleration and what happens when you plug that value in your equations?
 
  • #7
4,239
1
I have 3 constant acceleration equations but I don't know how to use them in conjunction with this problem.

A train leaves from Chicago toward Atl going 20 m/s. At the same time another train leaves from Atl going to Chicago at 25 m/s. The cities are 900km apart
Yes, you do have constant acceleration in this problem, but it is zero. In this problem, the acceleration terms drops out. The velocities are constant.

This equation, x = x(starting distance) + v(starting vel.)t +.5at^2 becomes

x = x0 + vt
 
  • #8
16
0
ummm so i used the v^2 = v(starting)^2 +2a(x-x[starting])

i got .222 = acceleration for the 20 m/s train
and .34722222 for the 25 m/s train...

now what?
 
  • #9
16
0
phrak how did you know the acceleration terms dropped out?

i used your equation though and i got
t=45 - 20 m/s tain
and
t=36 - 25 m/s train

but i left the distance 900 in km.
 
  • #10
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,721
2,979
Going back to post#6, what is the acceleration (phrak already told you) and what happens to your equations when you put that value in?
 
  • #11
16
0
oh ok i see now that the acceleration is 0...but how does this help me figure out when the trains meet and how long before the do?
 
  • #12
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,721
2,979
It helps you if you underdtand what you've got. The equation that you get for zero acceleration

x = x0 + vt

gives you the position of a train at any time t, where x0 is the position with respect to a reference point when the train begins to move and v is its velocity. Note that this reference cannot be one point for one of the trains and another point for the other train. It must be the same for both trains. So let's say you choose Chicago as the reference point. Can you rewrite the above equation twice to get two equations expressing the position (from Chicago) of each train at any time t? This is step 1. Complete this and we will move to step 2.
 

Related Threads on Intro Physics problem - 2 oncoming trains, help!

Replies
10
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
7K
Top