Question 290339: Flying with the wind, a small plane flew 540 km in 3 hours. Against the wind, the plane could fly only 420 km in the same amount of time. Find:
a) The rate of the plane in calm air and,
b) The rate of the wind
Answer by Grinnell(63)   (Show Source):
You can put this solution on YOUR website! The key to problems that involve a current in water or wind airplane/rate problems is that the initial values for speed will be accelerated or decelerated because of nature--flow of water or flow of air. So, the actual speeds will be the speeds slowed by drag or accelerated by force. When asking for a condition in calm water or in calm air, our answer will never be the first values of Speed or Rate. (Speed and Rate are the same.) With this said let us begin. shall we?
ALWAYS BEGIN WITH THE FORMULA: Rate (TIMES) Time= Distance.
let x be the speed of the plane flying with the wind
x(3)=540
x=180 mph. This is the speed of the plane flying WITH THE WIND. (n.b. NOT the true speed)
let y be the speed of the plane flying against the wind
y(3) (note that the times are the same for both planes)=420
y=140 mph. This is the speed of the plane flying against the wind.
BOTH OF THESE SPEEDS ARE NOT THE TRUE SPEED (WITHOUT NATURE'S WIND)
LET'S TAKE THE WIND OUT COMPLETELY AND SEE WHAT EXACTLY IS IT'S SPEED.
The wind is Konstant. This Konstant is, let's say, 'w'.
180 -w, here we take away the help of the wind from the speed of the plane flying with the wind.
180 -w= 140+w, here we added to make up for the drag caused by the wind against the plane.
40 =2w
20=w NOW WE SEE THE TRUE AFFECT OF THE WIND!
Our first set of speeds was 180 and 140 mph.
Now we know the rate of the plane in calm air is 180-20 (or 140+20) = 160MPH!
The rate of the wind again is 20MPH!
This is a classic Rate/Times/Distance problem. Go over and over this answer. Once you know the concept behind these problems you will have fun solving them.
|
|
|
