two airplanes depart simultaneously from an airport.One flies due south, the other flies due east at a rate 20 mi/h faster than that of the other airplane. After one hour , radar indicates that the airplanes are approximately 450 mi apart. What is the ground speed of each airplane?
This is a right triangle.
The legs are (x+20)1 and 1x
(x+20)^2 +x^2 = 450^2
2x^2 +40x + 400 = 202500
2(x^2+20 x + 200) =202500
x^2 + 20x + 200 =101250
x^2 + 20x -101050 = 0
Can you finish from here?
To find the ground speed of each airplane, we need to break down the problem into components.
Let's call the speed of the slower airplane 'x' mi/h. Since the faster airplane is flying 20 mi/h faster than the slower one, its speed can be written as 'x + 20' mi/h.
The slower airplane flies due south, so its velocity in the south direction is 'x' mi/h.
The faster airplane flies due east, so its velocity in the east direction is 'x + 20' mi/h.
Now, after one hour, the distance between the airplanes is 450 miles. We can use the Pythagorean theorem to relate the distances in the north-south (y) and east-west (x) directions.
According to the Pythagorean theorem, the sum of the squares of the legs is equal to the square of the hypotenuse. In this case, the square of the hypotenuse (450 mi) is equal to the square of the distance traveled in the north-south direction plus the square of the distance traveled in the east-west direction.
Mathematically, this can be written as:
(450)^2 = x^2 + (x + 20)^2
Let's solve this equation to find 'x', which represents the speed of the slower airplane:
202,500 = x^2 + (x^2 + 40x + 400)
Combine like terms:
2x^2 + 40x - 202,100 = 0
We can now solve this quadratic equation for 'x'. Using the quadratic formula (-b ± √(b^2 - 4ac))/(2a), where a = 2, b = 40, and c = -202,100, we get:
x = (-40 ± √(40^2 - 4 * 2 * -202100))/(2 * 2)
x = (-40 ± √(1600 + 1616800))/(4)
x = (-40 ± √(1618400))/(4)
x = (-40 ± 1271.57)/(4)
To find x, we have two possible solutions:
1) x = (-40 + 1271.57)/4 = 313.89
2) x = (-40 - 1271.57)/4 = -327.89
Since speed cannot be negative, we can disregard the negative value.
Therefore, the speed of the slower airplane (going due south) is approximately 313.89 mi/h.
The speed of the faster airplane (going due east) is x + 20 = 313.89 + 20 = 333.89 mi/h.
Hence, the ground speed of the slower airplane is approximately 313.89 mi/h, and the ground speed of the faster airplane is approximately 333.89 mi/h.