Thread: Question that has me stumped.

Carlos leaves Los Angeles on a cross-country car trip at 8:00 AM. He averages 50 Miles per hour.

Juanita plans to take exactly the same route, but does not leave 9:00 AM. He averages 60 miles per hour.

Develop a diagram or table to determine at what time Juanita will pass Carlos.

I've figured out that in between 1 and 2PM the cars will have both traveled 300 miles - but I am finding difficulty in determining when the vehicles will be physically passing.

Thanks for reading...

how about plotting a graph from the values you have, where they intersect would be where the cars pass, that is if you plot a distance-time graph

Worked like a charm!

Originally Posted by Sarlowbat

Carlos leaves Los Angeles on a cross-country car trip at 8:00 AM. He averages 50 Miles per hour.

Juanita plans to take exactly the same route, but does not leave 9:00 AM. He averages 60 miles per hour.

Develop a diagram or table to determine at what time Juanita will pass Carlos.

I've figured out that in between 1 and 2PM the cars will have both traveled 300 miles - but I am finding difficulty in determining when the vehicles will be physically passing.

Thanks for reading...

Sarlowbat,

If you need to use a table how about

Hour Carlos Juanita

1 50 0 9am
2 100 60 10
3 150 120 11
4 . .
5 . .
6 . .

Also, consider if you let x equal the number of hours it will take for Juanita
to catch Carlos you would have

50mph*x number of hours equals 60 mph*x number of hours minus 60 because Juanita stats 1 hour later.
50x=60x-60
60=10x
6=x Six hours or 2pm.

Originally Posted by Sarlowbat

Carlos leaves Los Angeles on a cross-country car trip at 8:00 AM. He averages 50 Miles per hour.

Juanita plans to take exactly the same route, but does not leave 9:00 AM. He averages 60 miles per hour.

You can do this one in your head, Sarlow:

When J leaves, she's 50 miles behind C (C travelled 50 miles from 8 to 9)

Since J travels at 10 mph faster than C (60 mph - 50 mph), then it will
take 50 / 10 = 5 hours to catch up.