Two airplanes leave an airport at the same time. one travels in the direction of 38 degrees at 140mph and the other travels in the direction of 350 degrees at 180mph. After two hours, how far apart are they? Thanks
Plane #1:
d1 = 140mi/h[38o] * 2h = 280mi/h[38o]
Plane #2:
d2 = 180mi/h[350o] * 2h = 360mi[350]
d = d2-d1 = 360[350o] - 280[38o]
X = 360*Cos350 - 280*Cos38 = 134 mi
Y = 360*sin350 - 280*sin38 = -235 mi
D^2 = X^2 + Y^2 = 134^2 + (-235)^2 = 73,181
D = 270.5 Miles apart.
To find the distance between the two airplanes after two hours, we can use the concept of vector addition.
First, we need to find the displacement of each airplane after two hours.
The first airplane is traveling at a speed of 140 mph in the direction of 38 degrees. To find the horizontal and vertical components of its displacement, we can use trigonometry.
Horizontal Component = 140 mph * cos(38 degrees)
Vertical Component = 140 mph * sin(38 degrees)
Similarly, the second airplane is traveling at a speed of 180 mph in the direction of 350 degrees. Again, we can use trigonometry to find its horizontal and vertical components.
Horizontal Component = 180 mph * cos(350 degrees)
Vertical Component = 180 mph * sin(350 degrees)
Now, we can calculate the total horizontal and vertical displacements by adding the respective components of each airplane.
Total Horizontal Displacement = Horizontal Component of the first airplane + Horizontal Component of the second airplane
Total Vertical Displacement = Vertical Component of the first airplane + Vertical Component of the second airplane
Using these displacements, we can apply the Pythagorean theorem to find the distance between the two airplanes:
Distance = √((Total Horizontal Displacement)^2 + (Total Vertical Displacement)^2)
Substituting the values, we can calculate the distance between the two airplanes after two hours.