If a golden eagle is flying due west with a speed of 32 mph, and encounters a wind that blows it north at 11 mph, what is the resultant velocity of the eagle in m/s? At what angle must the eagle be heading in this direction?
To solve this problem, we can break it down into two components: the east-west component and the north-south component of velocity.
1. East-West Component:
Since the golden eagle is flying due west with a speed of 32 mph, its velocity in the east-west direction is simply -32 mph (negative sign indicating westward direction).
2. North-South Component:
The wind blows the eagle north at 11 mph, so its velocity in the north-south direction is +11 mph (positive sign indicating northward direction).
Now, let's find the resultant velocity by combining these two components using vector addition:
Resultant velocity (v) = √(v_east-west^2 + v_north-south^2)
v = √((-32)^2 + (11)^2)
v ≈ √(1024 + 121)
v ≈ √(1145)
v ≈ 33.85 mph
To convert this velocity from miles per hour to meters per second, we need to multiply by a conversion factor:
1 mph = 0.44704 m/s
Resultant velocity in m/s = 33.85 mph * 0.44704 m/s
Resultant velocity ≈ 15.13 m/s
So, the resultant velocity of the eagle is approximately 15.13 m/s.
To find the angle at which the eagle must be heading in this direction, we can use trigonometry. The angle (θ) can be found using the tangent function:
θ = tan^(-1)(v_north-south / v_east-west)
θ = tan^(-1)(11 / -32)
θ ≈ -19.4 degrees
Note that the negative sign indicates the direction is west of north. Therefore, the eagle must be heading in a direction approximately 19.4 degrees west of north.
32@W = <-32,0>
11@N = <0,11>
Resultant is <-32,11> = 33.8@W19°N