Two planes take off from the same airport at the same time. One plane travels on a bearing S 12degreeW at 400 miles per hour. The other plane travels on a bearing N 75degreeE at 300 miles per hour. How far are the planes from each other 3 hours after takeoff?
To find the distance between the two planes after 3 hours, we can use the concept of vector addition.
First, let's find the displacement vector for each plane after 3 hours of travel. The displacement vector can be calculated by multiplying the velocity vector (speed and direction) by the time traveled.
For the first plane traveling on a bearing S 12° W at 400 miles per hour, the displacement vector can be calculated as follows:
Velocity vector = 400 miles/hour (S 12° W)
Displacement vector (plane 1) = 400 miles/hour * 3 hours = 1200 miles (S 12° W)
For the second plane traveling on a bearing N 75° E at 300 miles per hour, the displacement vector can be calculated as follows:
Velocity vector = 300 miles/hour (N 75° E)
Displacement vector (plane 2) = 300 miles/hour * 3 hours = 900 miles (N 75° E)
Now, we have the displacement vectors for both planes. To find the distance between the planes, we can find the magnitude of the difference vector between the two displacement vectors.
Difference vector = Displacement vector (plane 1) - Displacement vector (plane 2)
To calculate the difference vector, we need to convert the displacement vectors to a common coordinate system. We can use a Cartesian coordinate system, where the x-axis represents the east-west direction and the y-axis represents the north-south direction.
For plane 1:
Displacement vector x-component = 1200 miles * sin(12° W) = -243.58 miles (westward)
Displacement vector y-component = 1200 miles * cos(12° W) = 1180 miles (southward)
For plane 2:
Displacement vector x-component = 900 miles * sin(75° E) = 873.08 miles (eastward)
Displacement vector y-component = 900 miles * cos(75° E) = 222.73 miles (northward)
Now, we can calculate the x and y components of the difference vector by subtracting the corresponding components of the displacement vectors:
Difference vector x-component = (Displacement vector x-component of plane 1) - (Displacement vector x-component of plane 2)
= -243.58 miles - 873.08 miles = -1116.66 miles
Difference vector y-component = (Displacement vector y-component of plane 1) - (Displacement vector y-component of plane 2)
= 1180 miles - 222.73 miles = 957.27 miles
Finally, let's calculate the magnitude of the difference vector using the Pythagorean theorem:
Distance between the planes = sqrt((Difference vector x-component)^2 + (Difference vector y-component)^2)
= sqrt((-1116.66 miles)^2 + (957.27 miles)^2)
= sqrt(1250145.56 + 915914.74) miles
≈ sqrt(2166060.3) miles
≈ 1472.19 miles
Therefore, after 3 hours, the two planes are approximately 1472.19 miles apart.
d1=3h * 400mi/h @ 258Deg=1200mi@258Deg.
d2 = 3h * 300mi/h @ 15deg=900km@15deg.
Da = d1 - d2 = Distance apart.
Da = 1200km@258 = 900km@15deg. =
X = hor.=1200cos258 - 900cos15=-1119mi.
Y=ver.= 1200sin258 - 900sin15=-1407mi
tanA = Y/X = -1407/-1119 = 1.2574.
A = 51.5 deg.
A = 51.5 + 180 = 231.5 Deg.
Da = -1119 / cos231.5 = 1798mi apart.