SOLUTION: I ihave two problems and need the detailed answers asap 1.Jogging: Two athletes jog 10 miles. One of the athletes jogs 2 miles per hour faster and finishes 10 minutes ahead of the

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Question 486469: I ihave two problems and need the detailed answers asap
1.Jogging: Two athletes jog 10 miles. One of the athletes jogs 2 miles per hour faster and finishes 10 minutes ahead of the other athlete. Find the average speed of each athlete.
2.Strength of a Beam: The strength of a beam varies jointly as its width w and the square of its thickness t. If a beam 8 inches wide and 5 inches thick supports 650 pounds, how much can a similar beam 6 inches wide and 6 inches thick support?
Please if I can get these asap Thank you so much
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
1.Jogging: Two athletes jog 10 miles. One of the athletes jogs 2 miles per hour faster and finishes 10 minutes ahead of the other athlete.
Find the average speed of each athlete.
Let s = jogging speed of the athlete
then
(s+2) = jogging speed of the faster athlete
:
Change 10 min to 1%2F6 hr
:
Write a time equation
:
slower time - faster time = 10 min
10%2Fs - 10%2F%28s%2B2%29 = 1%2F6
multiply by 6s(s+2), resulting in
6(s+2)*10 - 6s(10) = s(s+2)
:
10(6s+12) - 60s = s^2 + 2s
:
60s + 120 - 60s = s^2 + 2s
Arrange to form a quadratic equation
s^2 + 2s - 120 = 0
Factors to
(s+12)(s-10) =
positive solution
s = 10 mph is the speed of the slower
then
10+2 = 12 mph is the speed of the faster
:
:
2.Strength of a Beam: The strength of a beam varies jointly as its width w and the square of its thickness t.
S = w*t^2*k,where k = the variation constant
:
If a beam 8 inches wide and 5 inches thick supports 650 pounds,
Find k
8*5^2*k = 650
200k = 650
k = 650%2F200
k = 3.25
:
how much can a similar beam 6 inches wide and 6 inches thick support?
s = 6 *6^2*3.25
s = 702 lb can be support by a 6 by 6" beam