Two planes leave simultaneously from Chicago's O'Hare Airport, one flying due north and the other due east. The northbound plane is flying 50 miles per hour faster than the eastbound plane. After 3 hours the planes are 2840 miles apart. Find the speed of each plane. (Round your answers to the nearest whole number.)
let speed of slower plane be x mph
speed of faster plane = x+50 mph
After 3 hours they will have gone 3x and 3x+150 miles respectively,
and their flightpaths will form a right-angled triangle so that
(3x)^2 + (3x+150)^2 = 2840^2
Simplify and solve this quadratic using your favourite method.
Make sure to dismiss the negative answer and use only the positive one.
Let me know what you get
i used the quadratic formula and i got
eastbound plane is 643.93
northbound plane is -693.93
Let's denote the speed of the eastbound plane as 'x' miles per hour.
Since the northbound plane is traveling 50 miles per hour faster, its speed can be represented as 'x + 50' miles per hour.
To find the distance each plane travels, we can use the formula: distance = speed * time.
For the eastbound plane:
distance = speed * time
distance = x * 3
distance = 3x miles
For the northbound plane:
distance = speed * time
distance = (x + 50) * 3
distance = 3x + 150 miles
Since the total distance between the two planes after 3 hours is 2840 miles:
total distance = distance of eastbound plane + distance of northbound plane
2840 = 3x + 3x + 150
Simplifying the equation:
2840 = 6x + 150
2840 - 150 = 6x
2690 = 6x
Dividing both sides by 6:
2690/6 = x
448.3 = x
So the speed of the eastbound plane is approximately 448 miles per hour.
To find the speed of the northbound plane:
speed = x + 50
speed = 448 + 50
speed = 498
Therefore, the speed of the northbound plane is approximately 498 miles per hour.
To find the speed of each plane, we can use the distance formula: distance = speed × time.
Let's assume the speed of the eastbound plane is x mph. Since the northbound plane is flying 50 mph faster, its speed will be (x + 50) mph.
In 3 hours, the eastbound plane would have traveled a distance of 3x miles, and the northbound plane would have traveled a distance of 3(x + 50) miles.
Since the two planes are flying at right angles to each other, we can use the Pythagorean theorem to find the distance between them. According to the theorem, the square of the hypotenuse (distance between the planes) is equal to the sum of the squares of the other two sides (distances traveled by each plane).
So, using the Pythagorean theorem, we have:
(3x)^2 + (3(x + 50))^2 = 2840^2
Simplifying the equation:
9x^2 + 9(x + 50)^2 = 2840^2
Expanding and combining like terms:
9x^2 + 9(x^2 + 100x + 2500) = 2840^2
9x^2 + 9x^2 + 900x + 22500 = 2840^2
18x^2 + 900x + 22500 = 2840^2
Now, we can solve this quadratic equation to find the value of x.
18x^2 + 900x + 22500 - 2840^2 = 0
Simplifying:
18x^2 + 900x - 7,969,600 = 0
To find the values of x, we can either factor this quadratic equation or use the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 18, b = 900, and c = -7,969,600.
Plugging the values into the formula:
x = (-900 ± √(900^2 - 4(18)(-7,969,600))) / (2(18))
Simplifying:
x = (-900 ± √(810,000,000 + 574,630,400)) / 36
x = (-900 ± √1,384,630,400) / 36
Now, we can calculate the two possible values for x and determine the speeds of each plane.
x ≈ (-900 ± 37,200.67) / 36
For the positive value of x:
x ≈ (-900 + 37,200.67) / 36
x ≈ 36,300.67 / 36
x ≈ 1008.35
Now, let's find the speed of the northbound plane:
Northbound plane's speed ≈ x + 50
≈ 1008.35 + 50
≈ 1058.35
Therefore, the estimated speed of the eastbound plane is approximately 1008 mph, and the estimated speed of the northbound plane is approximately 1058 mph.