SOLUTION: At a high school cross country meet, Jared jogged 8 mph for the first part of the race, then increased his speed to 12 mph for the second part. If the race was 21 mi. long and Jare
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: At a high school cross country meet, Jared jogged 8 mph for the first part of the race, then increased his speed to 12 mph for the second part. If the race was 21 mi. long and Jare
Log On
Question 404707: At a high school cross country meet, Jared jogged 8 mph for the first part of the race, then increased his speed to 12 mph for the second part. If the race was 21 mi. long and Jared finished in 2hr., how far did he jog the faster pace? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! At a high school cross country meet, Jared jogged 8 mph for the first part of
the race, then increased his speed to 12 mph for the second part.
If the race was 21 mi. long and Jared finished in 2hr., how far did he jog the faster pace?
:
Let t = time jogged at the faster pace
then
(2-t) = time at the slower pace
:
Write a distance equation: dist = speed * time
:
slow dist + fast dist = 21 mi
8(2-t) + 12t = 21
16 - 8t + 12t = 21
4t = 21 - 16
4t = 5
t =
t = 1.25 hrs at 12 mph
:
Find the distance at this speed
1.25 * 12 = 15 mi at the faster speed
:
:
Check this solution by finding the dist at each speed (2-1.25 = .75 hr)
.75 * 8 = 6 mi
1.25* 12 = 15 mi
----------------
total = 21 mi