SOLUTION: A jogger leaves a point on a fitness trail running at a rate of 7 miles per hour. Ten minutes later, a second jogger leaves from the same location running at 8 miles per hour. a)Ho
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: A jogger leaves a point on a fitness trail running at a rate of 7 miles per hour. Ten minutes later, a second jogger leaves from the same location running at 8 miles per hour. a)Ho
Log On
Question 821560: A jogger leaves a point on a fitness trail running at a rate of 7 miles per hour. Ten minutes later, a second jogger leaves from the same location running at 8 miles per hour. a)How long will it take the second jogger to overtake the first? b)How far will each have run at that point? Found 2 solutions by TimothyLamb, stanbon:Answer by TimothyLamb(4379) (Show Source):
You can put this solution on YOUR website! ---
s = d / t
d = s * t
---
at the overtake:
---
first jogger:
D = 7t
second jogger:
D = 8(t - 10/60)
---
7t = 8(t - 10/60)
7t = 8t - (8/6)
t = 4/3 = 1+1/3 hours
---
D = 7t
D = 7(4/3)
D = 9+1/3 miles
---
answer:
the overtake occurs:
1+1/3 hours = 1 hour and 20 minutes after the joggers leave the fitness trail
and both joggers will have run 9+1/3 miles at the overtake
---
Solve and graph linear equations:
https://sooeet.com/math/linear-equation-solver.php
---
Solve quadratic equations, quadratic formula:
https://sooeet.com/math/quadratic-formula-solver.php
---
Solve systems of linear equations up to 6-equations 6-variables:
https://sooeet.com/math/system-of-linear-equations-solver.php
You can put this solution on YOUR website! A jogger leaves a point on a fitness trail running at a rate of 7 miles per hour. Ten minutes later, a second jogger leaves from the same location running at 8 miles per hour.
a)How long will it take the second jogger to overtake the first?
1st jogger DATA:
rate = 7 mph ; time = x minutes ; distance = r*t = (7/60)x miles
-----
2nd jogger DATA:
rate = 8 mph ; time = x-10 min ; distance = r*t = (8/60(x-10)miles
--------------------------------------------------
b)How far will each have run at that point?
Solve for "x"::
distance = distance
(7/60)x=(8/60)(x-10)
------
7x = 8(x-10)
7x = 8x - 80
x = 80 minutes
---
Distance = (7/60)80 = (4/3)7 = 28/3 = 9 1/3 miles
----------------------------------------------------
Cheers,
Stan H.
================
(