A plane flies 1.3 hr at 100mph on a bearing of N40degreesE. It then turns and flies 1.5hr at the same speed on a bearing of S50degreesE. How far is the plane from its starting point?
130 at 60 deg up from x axis
150 at 40 deg down from x axis
x component = 130 cos 60 + 150 cos 40
y component = 130 sin 60 - 150 sin 40
magnitude = sqrt (x^2+y^2)
To find the distance of the plane from its starting point, we can use the concept of vector addition. Let's break down the two legs of the flight into their respective components.
First leg:
Speed = 100 mph
Time = 1.3 hours
Bearing = N40°E
The N40°E bearing can be decomposed into its northward and eastward components. To do this, we need to find the sin and cos of the angle involved.
sin(40°) = Northward component / Speed
cos(40°) = Eastward component / Speed
Now, we can solve for the northward and eastward components:
Northward component = Speed * sin(40°)
Eastward component = Speed * cos(40°)
Second leg:
Speed = 100 mph
Time = 1.5 hours
Bearing = S50°E
The S50°E bearing can also be decomposed into its northward and eastward components using similar trigonometric calculations:
Northward component = Speed * sin(180° - 50°)
Eastward component = Speed * cos(180° - 50°)
The angles for the second leg are calculated differently because the bearing is counter-clockwise from the negative y-axis.
Now that we have the northward and eastward components for both legs of the flight, we can calculate the overall northward and eastward displacement of the plane:
Northward displacement = Northward component of leg 1 + Northward component of leg 2
Eastward displacement = Eastward component of leg 1 + Eastward component of leg 2
To find the overall displacement, we can use the Pythagorean theorem:
Overall displacement = sqrt(Northward displacement^2 + Eastward displacement^2)
Calculating these values, we find that the distance of the plane from its starting point is approximately 224.6 miles.