SOLUTION: A jogger started a course and jogged at an average speed of 4 mph. One hour later, a cyclist started the same course and cycled at an average speed of 11 mph. How long after the jo
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: A jogger started a course and jogged at an average speed of 4 mph. One hour later, a cyclist started the same course and cycled at an average speed of 11 mph. How long after the jo
Log On
Question 118112: A jogger started a course and jogged at an average speed of 4 mph. One hour later, a cyclist started the same course and cycled at an average speed of 11 mph. How long after the jogger started did the cyclist overtake the jogger? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A jogger started a course and jogged at an average speed of 4 mph. One hour later, a cyclist started the same course and cycled at an average speed of 11 mph. How long after the jogger started did the cyclist overtake the jogger?
:
Let t = travel time of the jogger
then
(t-1) = travel time of the cyclist
:
When the cyclist overtakes the jogger they will have traveled the same distance
Write a distance equation; distance = speed * time
:
Cyclist dist = jogger distance
11(t-1) = 4t
:
11t - 11 = 4t
:
11t - 4t = + 11
:
7t = 11
:
t =
:
t = 1.57 hrs or 1 = 1 hr 34.2 min
:
:
Check solution; find the distance of each:
1.57 * 4 = 6.28 mi
.57 * 11 = 6.27 mi