SOLUTION: I know that this motion problem involves a system of two equations and I could solve it if I could only figure out how to set it up correctly. I have been using the Distance = Ra

Question 5092: I know that this motion problem involves a system of two equations and I could solve it if I could only figure out how to set it up correctly. I have been using the Distance = Rate x Time formula to try and set it up, but I just can't figure it out and it's driving me nuts! Here's the problem:
"An express train and a local train leave Gray's Lake at 3PM and head for Chicago 50 miles away. The express travels twice as fast as the local and arrives 1 hour ahead of the local. Find the speed of each train."
Here's what I've tried to do to set this up. I've tried to set up a table so I can see what my equations should be. I'll try to approximate below.
Using the Rate x Time = Distance formula. Let s=The speed the train is traveling and let t=the time that it takes to get to Chicago.
Rate Time Distance
Local s x t =50
Express 2(s)x t-1 =50
The above is as far as I get, but I don't think it's right and thus won't get me to the speed that each train is traveling at.
Help!
Karen Henry

Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
I think your mistake is not putting parentheses around the time (t-1).

As you said,
let s = speed of the local
2s = speed of the express

t= time of the local
t-1 = time of the express

Distances are equal at 50 miles, so (rate times time) = (rate times time)
s*(t) = 2s *(t-1)

Since s the speed of the train does not equal zero, you can divide both sides by s:

t= 2(t-1)
t=2t -2
-t = -2
t=2 hours for the local
t-1 = 1 hour for the express

After you do that, then remember that D=RT, so
Rate for the local = =25 mph
Rate for the express = = 50 mph.

R^2 at SCC

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