Two airplanes leave a airport at the same time , one going northwest ( bearing 315degrees) at 422 mph and the other going east at 341 mph. How far are the planes after 4 hours ?
Please help have no idea of how to solve?
Any change you could clear the question? Is the question how far are the planes from each other?
If my assumption is correct the way to solve it would be via geometry.
East is 90 degrees.
The plane going 90 has cathetus with the lenght of 4*341.
The other one has 4*422.
The angle between these 2 cathetuses is 90+45=135 degrees.
Solve hypotenus.
(Cosine theorem, A^2+B^2-2ABcosx=C^2)
Still the same guy here: ansver is 13055857/5186 miles.
To find the distance between the two airplanes after 4 hours, we need to calculate the distance each airplane has traveled.
First, let's find the distance traveled by the northwest-bound airplane.
Distance = Speed × Time
Distance = 422 mph × 4 hours
Distance = 1688 miles
Now, let's find the distance traveled by the eastbound airplane.
Distance = Speed × Time
Distance = 341 mph × 4 hours
Distance = 1364 miles
To find the total distance between the two airplanes after 4 hours, we need to find the diagonal of a right triangle formed by the distances traveled by each airplane.
Using the Pythagorean theorem, we can find the diagonal distance.
Diagonal Distance² = Northwest Distance² + East Distance²
Diagonal Distance² = (1688 miles)² + (1364 miles)²
Diagonal Distance² = 2,846,144 + 1,860,496
Diagonal Distance² = 4,706,640
Diagonal Distance = √4,706,640
Diagonal Distance ≈ 2,169.6 miles
Therefore, after 4 hours, the two airplanes are approximately 2,169.6 miles apart.
To solve this problem, we can use the concept of vectors and the Pythagorean theorem.
First, let's break down the velocities of the two airplanes into their horizontal (east-west) and vertical (north-south) components.
For the airplane going northwest, the bearing is given as 315 degrees. To find the horizontal and vertical components, we can use trigonometry.
The horizontal component (x-component) can be found by multiplying the velocity (422 mph) by the cosine of the angle:
Horizontal component = 422 mph * cos(315 degrees) = 422 mph * cos(-45 degrees) = 422 mph * (√2 / 2) = 422 * 0.7071 ≈ 298.85 mph (rounded to two decimal places).
The vertical component (y-component) can be found by multiplying the velocity (422 mph) by the sine of the angle:
Vertical component = 422 mph * sin(315 degrees) = 422 mph * sin(-45 degrees) = 422 mph * (√2 / 2) = 422 * 0.7071 ≈ 298.85 mph (rounded to two decimal places).
For the airplane going east, the bearing is 90 degrees. Its velocity is 341 mph.
Since the airplanes left the airport at the same time, we can assume that their positions relative to the airport are the same, and the distance between them after 4 hours is the magnitude of their displacement vectors.
The displacement vector for the northwest airplane is given by (298.85 mph * 4 hours, 298.85 mph * 4 hours), which simplifies to (1195.4 miles, 1195.4 miles).
The displacement vector for the eastward airplane is given by (341 mph * 4 hours, 0 miles), which simplifies to (1364 miles, 0 miles).
To find the distance between the two airplanes after 4 hours, we apply the Pythagorean theorem:
Distance = √((1195.4 miles - 1364 miles)^2 + (1195.4 miles - 0 miles)^2)
= √((-168.6 miles)^2 + (1195.4 miles)^2)
= √(28384.96 miles^2 + 1425870.16 miles^2)
= √(1454255.12 miles^2)
≈ 1205.76 miles (rounded to two decimal places).
Therefore, the airplanes are approximately 1205.76 miles apart after 4 hours.