Plane speed

Question

An airplane takes 3 hours to travel a distance of 1680 miles with the wind. The return trip takes 4 hours against the wind. Find the speed of the plane in still air and the speed of the wind.

Answer 1

Dear Ryan,

 

    Let x= rate of plane in still air, and let y=rate of wind.

 

._____3 hours____________>  1680 miles      Plane is flying with the wind.

 

.<____4 hours____________.    1680 miles     Return trip was flying against the wind.

 

Note:  Both distances are equal.       Formula:  rate * time= distance 

 

Set up a table like below:

       

 

Plane          Rate     *       time      =      Distance 

 

                  (x+y)                  3              3(x+y)=1,680     Rate of wind is increasing rate of plane.

 

return trip (X-y)                   4              4(x-y)=1,680        Rate of wind is decreasing rate of plane.

 

 

Since the two equations equal the same distance, make them equal each other. Use the Distributive Property, and then solve for "y" and "x."

 

 

3(x+y)=4(x-y)             3x +3y= 4x-4y                     Distributive Property      

 

                                                                             Get the "x's" on one side of equal sign, and the "y's" on the  other side of equal sign.

 

3x-3x+3y+4y=4x-3x-4y+4Y

 

3y+4y= 4x-3x

 

7y=X             

 

3(x+y)= 1,680                  Substitute for "x=7y."

 

3(7y +y)=3(8y)=1,680                    24y=1,680    Now, solve for "y," and find "x=7y" through substitution.

 

Remember    x= rate of plane in still air=______miles/hour     and y=rate of wind=-______miles/hour

 

 

If you have further questions, please let me know.  If I have helped you, please give me a thumbs up.

                                                                             

Susan C.

                                                  

Answer 2

d = r * t , distance = (rate = speed) * time

We need to use this formula twice since the plane is traveling back and forth

distance = 1680 we know that distance will be same for both equations

notice we get information on the time and influence with the wind, substitute known values

first equation --- 1680 = (r+h) * 3 notice r is now r+h because the speed is faster when traveling with the wind

second equation (the return trip) --- 1680 = (r-h) * 4 and here r is now r-h since speed is slower when traveling against the wind

1680 = (r+h)3

1680 = (r-h)4

divide first equation by 3 then we get 560 = r + h

divide second equation by 4 then we get 420 = 4- h

you can use a substitution or elimination method to solve these (unless you are allowed to use a graphing calculator) which would result in r = 490 and h = 70

speed of the plane in still air = 490 and speed of the wind = 70