SOLUTION: So I asked my teacher to help me with this but he said he couldn't because he wanted us to figure this out on our own and I kind of forgot how to do this. Here's the problem:

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Question 812711: So I asked my teacher to help me with this but he said he couldn't because he wanted us to figure this out on our own and I kind of forgot how to do this.
Here's the problem:
two cars are traveling in the same direction. The first car is going 45 mi/h and the second car is going 60 mi/h. The first car left 2 hours before the second car. Explain how you could solve an equation to find out how long it would take for the second car to catch up to the first car?

Found 2 solutions by josmiceli, josgarithmetic:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
First you have to find the 1st car's head start
+d%5B1%5D+=+45%2A2+
+d%5B1%5D+=+90+ mi
----------------
Imagine you start a stopwatch when the 2nd car leaves
Let +t+ = the time in hours on the stopwatch when
2nd car catches the 1st car
Let +d+ = the distance in miles that the 2nd car
travels in time +t+
----------------------
1st car's equation:
+d+-+90+=+45t+
2nd car's equation:
+d+=+60t+
-----------
This is 2 equations and 2 unknowns
so it's solvable
---------------
By substitution:
+60t+-+90+=+45t+
You can finish

Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Make a data table: speed, time, distance.
Since the variables as unknown seem to be time, look carefully about time to assign variables and expression, and build from there.

Slow car begins first and goes for time x+2 hours.
Fast car begins later than the first, and goes for x hours.

Now put what we have into the data table and analyze.

WHICH_________________speed___________time_____________distance
SLOW__________________45______________x+2______________d
FAST__________________60______________x________________d

Remember, Rate*Time=Distance, and you want to know for what x value will the distance traveled for SLOW and FAST car to be the same.