SOLUTION: Joe leave his home to run to school at a Constant rate of 6 mph. His brother John was not ready on time and left their house 20 minutes after Joe are in a constant rate of 8 mph.

Question 1037612: Joe leave his home to run to school at a Constant rate of 6 mph. His brother John was not ready on time and left their house 20 minutes after Joe are in a constant rate of 8 mph. How long will it take for John to catch up to Joe? How far from home Will they be when John catches Joe
Found 2 solutions by ankor@dixie-net.com, josgarithmetic:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Joe leave his home to run to school at a Constant rate of 6 mph.
His brother John was not ready on time and left their house 20 minutes after Joe are in a constant rate of 8 mph.
How long will it take for John to catch up to Joe?
:
They are using mph, change 20 min to hr
let t = running time of John, when he catches Joe
we know Joe ran 20 min longer, therefore
(t+ = running time of Joe when he is caught
:
When John catches Joe the will have traveled the same distance
Write a distance equation dist = speed * time
8t = 6(t+)
distribute the 6
8t = 6t + 2
8t - 6t = 2
2t = 2
t = 1 hr for John to catch Joe
:
How far from home Will they be when John catches Joe
Dist = speed * time
dist = 8 * 1
dist = 8 mi to school
:
Confirm this using Joe's time
6 * 1 = 8 mi also
:
Sorry, I hit the submit key before I was done.
Answer by josgarithmetic(39613) (Show Source): You can put this solution on YOUR website!
Travel Rates Rule, RT=D relating speed, time, distance.

RATE TIME DISTANCE JOE 6 t+1/3 d JOHN 8 t d

The situation described gives you two equations in two unknown variables.

You could equate the two equal distances: and solve for t, first.

----------- 1 hour for John to catchup.
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