Question 67071This question is from textbook Algebra 1
: Against a head wind, Jeff computes his flight time for a trip of 2900 miles at 5 hours. The flight would take 4 hours and 50 minutes if the head wind were half as much. Find the head wind and the plane's air speed.
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This question is from textbook Algebra 1
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Against a head wind, Jeff computes his flight time for a trip of 2900 miles at 5 hours. The flight would take 4 hours and 50 minutes if the head wind were half as much. Find the head wind and the plane's air speed.
:
You must mean find the plane's speed in still air speed.
:
Change 50 min to hrs: 50/60 = 5/6 hr; + 4 hr = 29/6 hrs
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Let s = the planes still air speed
Let w = wind speed.
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Write two equations using the distance: dist = time * speed
:
The 1st equation:
5(s - w) = 2900
Simplify divide equation by 5
s - w = 580
s = (w + 580)
:
The 2nd equation, (wind is half as much)
(29/6)(s - .5w) = 2900
Get rid of this annoying fraction, mult eq by 6 and you have:
29(s-.5w) = 2900 * 6
29s - 14.5w = 17400
:
Substitute (w+580) for s in the 2nd equation, solve for w:
29(w+580) - 14.5w = 17400
29w + 16820 - 14.5w = 17400
29w -14.5w = 17400 - 16820
14.5w = 580
w = 580/14.5
w = 40 mph speed of the wind
:
Find the speed of the plane
s = w + 580
s = 40 + 580
s = 620 mph is the speed of the plane
:
Check our solution:
Wind speed 40 mph: 2900/580 = 5 hrs
Wind speed 20 mph: 2900/600 = 4.8333 hrs (which is 4 & 5/6 hrs)
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