SOLUTION: John and Brian leave Williston at the same time. JOhn drives north and Brian drives east. John's average speed is 10 miles per hour slower than Brian's. At the end of one hour t

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Question 834303: John and Brian leave Williston at the same time. JOhn drives north and Brian drives east. John's average speed is 10 miles per hour slower than Brian's. At the end of one hour they are 50 miiles apart. Find Brian's average speed.
50+10 =60
Brian was driving 50 and John 60
Found 2 solutions by richwmiller, ankor@dixie-net.com:
Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website!
Your solution has many problems
John's average speed is 10 miles per hour slower than Brian's
but you have
Brian was driving 50 and John 60
50^2+60^2=x^2
that doesn't put them 50 miles apart.
x^2=6100
x=10*sqrt(61)
x=78.102 miles apart
(x+10)^2+(x)^2=50^2
x=30 x+10=40
check
40^2+30^2=50^2
1600+900=2500
ok

Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
John and Brian leave Williston at the same time.
JOhn drives north and Brian drives east.
John's average speed is 10 miles per hour slower than Brian's.
At the end of one hour they are 50 miles apart.
Find Brian's average speed.
:
b = Brian's speed
then
(b-10) = John's speed (he drives 10 mph slower than Brian)
:
They are driving in directions which are 90 degrees apart, therefore
The distance between them will be the hypotenuse of right triangle (50 mi)
:
Since they drive 1 hr, they will have traveled the same as their speed
:
b^2 + (b-10)^2 = 50^2
FOIL (b-10)(b-10)
b^2 + b^2 - 20b + 100 - 2500
Combine to forma quadratic equation
2b^2 - 20b + 100 - 2500 = 0
2b^2 - 20b - 2400 = 0
simplify, divide by 2
b^2 - 10b - 1200 = 0
Factors to
(b-40)(b+30) = 0
the positive solution is all we want here
b = 40 mph is Brian's speed
:
:
Check this a on a calc: enter