# Thread: Rate Problem? Need some direction

1. ## Rate Problem? Need some direction

Hi. I'm working on what I think is a rate problem but I'm having trouble defining my variables to solve the problem. Will you please help? I've tried a few things but it's all been junk.

Question:
Leila left home at noon, traveling at 24mph. An hour later her brother Josh, driving at 36 mph, set out to overtake her. How long did it take him to meet her?

Thank you so much for your help! My group really appreciates it.

2. These sequences represent distance travelled after n hours where n=0,1,2,...

Leila:0, 24, 48, 72, ...
Josh:0, 0, 36, 72, ...

Looks like four hours is the answer..

Do you follow?

3. Beautiful work! The logic you provide is very comforting! How would I write them into a system of equations?

4. No matter how I look at it, I system of equations doesn't occur to me. :-(

5. Originally Posted by yvonnehr
Question:
Leila left home at noon, traveling at 24mph. An hour later her brother Josh, driving at 36 mph, set out to overtake her. How long did it take him to meet her?
let $t = 0$ be noon.

distance $L$ travels ... $D_L = 24t$

distance $J$ travels ... $D_J = 36(t-1)$

set the two expressions equal and solve for $t$

6. Originally Posted by skeeter
let $t = 0$ be noon.

distance $L$ travels ... $D_L = 24t$

distance $J$ travels ... $D_J = 36(t-1)$

set the two expressions equal and solve for $t$
I see! And D is the same for both because we are only concerned about the point in which they meet?! Right?

7. Yes. However, note that the question specifically asked "How long did it take him to meet her?"- that is, how long had he been traveling. Since t was taken to be 0 at noon, an hour before he left, the answer to the question is t- 1.

Because, his travel time is what is asked, I would have said "let t be the time Josh travels, in hours. Then Leila travels for t+ 1 hours. Josh will have traveled 36t miles and Leila 24(t+1). Since they "meet" when they have gone the same distance,
36t= 24(t+ 1)". Solving that for t, you should get 1 less than when you solve skeeter's equation.