SOLUTION: So I had a kinda hard test today...well it was easy until this hard head wind/tail wind problem came up. The problem went like this Johnny went head wind (east to west I believe) f
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Question 571025: So I had a kinda hard test today...well it was easy until this hard head wind/tail wind problem came up. The problem went like this Johnny went head wind (east to west I believe) from Miami to San Francisco 2100 miles at 5.6 lies per hour. When this other person (a girl) went tail wind (west to east) from San Francisco to Miami 2100 miles too but only took 4.8 miles per hour...what is the air speed and the wind speed?
Help??? Please...has anybody heard of this problem before or know where I could find it?
Thanks!!! :) Found 2 solutions by ankor@dixie-net.com, Edwin McCravy:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! I will rewrite this so it makes sense:
;
Johnny went head wind (east to west I believe) from Miami to San Francisco 2100 miles in 5.6 hours.
When this other person (a girl) went tail wind (west to east) from San Francisco to Miami 2100 miles too but only took 4.8 hours...
:
what is the air speed and the wind speed?
:
Let s = speed in still air
let w = rate of the wind
then
(s-w) = effective speed against the wind
and
(s+w) = effective speed with the wind
;
Write a distance equation for each trip; dist = time * speed
:
5.6(s-w) = 2100
4.8(s+w) = 2100
Simplify this, divide the 1st equation by 5.6 and the 2nd equation by 4.5, resulting in:
s - w = 375
s + w = 437.5
--------------- adding eliminates w, find s
2s = 812.5
s = 812.5/2
s = 406.25 mph is the speed in still air
then
406.25 + w = 437.5
w = 437.5 - 406.25
w = 31.25 mph is the speed of the wind
;
:
You can confirm this by replacing s and w in the original equations and doing the math
You can put this solution on YOUR website! So I had a kinda hard test today...well it was easy until this hard head wind/tail wind problem came up. The problem went like this Johnny went head wind (east to west I believe) from Miami to San Francisco 2100 miles at 5.6 lies per hour. When this other person (a girl) went tail wind (west to east) from San Francisco to Miami 2100 miles too but only took 4.8 miles per hour...what is the air speed and the wind speed?
The way you have it doesn't make sense. All the planes I've seen fly a little
faster than 4.8 miles per hour, because you can walk that fast.
Here's a way it could have been:
Johnny flew 2100 miles Miami to SF against the wind while his girl friend flew
from SF to Miama with the wind in 4.8 hours. They communicated by cell phone
and kept their air speed indicators exactly the same speed the whole way. It
took him 5.6 hours and her 4.8 hours.
x = air speed (how fast they would be going if the air was still.
distance rate time
Johnny 2100 x-w 5.6
girl 2100 x+w 4.8
2100 = (x-w)5.6
2100 = (x+w)4.8
Multiply the first by 5.6 and the second by 4.8
375 = x - w
437.5 = x + w
Add the equations get
812.5 = 2x
406.25 = x
Multiply the first by -1 and add
-375 = -x + w
437.5 = x + w
52.5 = 2w
25.25 = w
So the airspeed they kept their planes at was 406.25 mph and
the wind was blowing 25.25 miles per hour toward the west.
That's probably not the way it was stated but at least that
makes a little more sense.
Edwin
