Question 1033225: An airplane flying against the wind travels 125 miles in the same amount of time it would take the same plane to travel 155 miles with the wind. If the wind speed is a constant 30 miles per hour, how fast would the plane travel in still air?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! rate * time = distance.
rate is composed of the speed of the plane plus or minus the speed of the wind.
if we let p = the speed of the plane and w = the speed of the wind and t = the time and d = the distance, you get:
when the plane is flying against the wind, you get (p-w)*t = d.
when the plane is flying with the wind, you get (p+w)*t = d.
when the plane is flying against the wind, you are given that d = 125.
when the plane is flying with the wind, you are given that d = 155.
you get:
(p-w)*t = 125
(p+w)*t = 255
you are given that w = 30, so these formulas become:
(p-30)*t = 125
(p+30)*t = 155
these are two equations that need to be solved simultaneously because the same value of p and the same value of t applies to both of them.
solve for t in both equation to get:
t = 125/(p-30)
t = 155/(p+30)
since both expressions on the right side of these equations are equal to t, you can set those expressions equal to each other to get:
125/(p-30) = 155/(p+30)
multiply both sides of this equation by (p-30)*(p+30) to get:
125 * (p+30) = 155 * (p-30)
simplify to get 125 * p + 125*30 = 155 * p - 155 * 30
simplify to get 125 * p + 3750 = 155 * p - 4650
subtract 125 * p from both sides of the equation and add 4650 to both sides of the equation to get 3750 + 4650 = 155 * p - 125 * p
simplify to get 8400 = 30 * p
divide both sides of this equation by 30 to get 280 = p
that's the speed of the plane in still air (without any wind).
this one is a little more difficult than the usual rate * time problem because you can get into trouble when eliminating variables.
the way above was one way.
another way would be:
start with:
(p-30)*t = 125
(p+30)*t = 155
simplify to get:
pt - 30t = 125
pt + 30t = 155
add the second equation to the first equation to get:
2pt = 280
divide both equations by 2 to get pt = 140.
solve for t to get t = 140/p.
you still don't have a solution, but you do have a relationship that you can use.
go back to one of your equations that you started with.
the one i chose is (p-30)*t = 125
replace t with 140/p to get:
(p-30)*140/p = 125
multiply both sides of this equation by p to get (p-30)*140 = 125*p
simplify to get 140*p - 30*140 = 125*p
simplify to get 140*p - 4200 = 125*p
subtract 125*p from both sides of the equation and add 4200 to both sides of the equation to get:
140*p - 125*p = 4200
simplify to get 15*p - 4200
solve for p to get p = 4200 / 15 = 280.
you get the plane's speed in still air is 280.
that's the same as we got above using a different method.
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