Distance rate time

[brian12]

I've been staring at this problem for a couple of hours now. It seems to me that a piece of information was left out of the book. But I guess I could be wrong.

Allan can drive his car over a route in 5 hours, and Carla can drive her car over the same route in 4 hours. How long would it take to meet if they started at opposite ends at the same time?

Can this problem be solved without a distance or rate?

[FWT]

Allan can drive his car over a route in 5 hours, and Carla can drive her car over the same route in 4 hours. How long would it take to meet if they started at opposite ends at the same time?

Can this problem be solved without a distance or rate?

Yeh: use what they show in this "distance" lesson:

Allan:
distance: d
time: 5
rate: d/5

Carla:
distance: d
time: 4
rate: ??

Finish the Carla talbe above.

They come at each other so you add their rates. What's the total rate? How long to go d at that rate? etc.

[brian12]

I appreciate the help. I see I was making the mistake of setting the distance equal to 2d last night. That's what I get for trying to study when I'm tired. Quick question, that page you linked to, I found it a day or two ago and noticed it said "you cannot add rates". I solved an earlier problem in the same book by adding rates, but I figured it was just a coincidence I found the correct answer. Is there some kind of special circumstance where you are allowed to add rates?

[FWT]

that page you linked to...said "you cannot add rates"....Is there some kind of special circumstance where you are allowed to add rates?

You can't add rates for two different parts, like you can add distances and times. Like "1st leg is 100 mi in 2 hrs & 2nd leg is 300 mi in 5 hrs", but you can't add "20 mph on one street & 40 mph on another street" to get "60 mph for the two streets". That's what they mean when they said "you can't add rates".

But if you're doing combined speeds, like where they're going toward each other so the "net" speed is the combined/added speed, then it's okay. But when it's like that, you can't count both distances or times so "2d" was wrong. It's kind of a context thing I guess.