an airplane is heading due west at a velocity of 500 miles/hr. There is a cross wind due North at a velocity of 50 miles/hr. find the airplanes velocity relative to the ground (magnitude and direction)
well, sketch it.
YOu see a right triangle:
magnitude= sqrt(500^2+50^2)
direction N of W by the angle whose tangent is 50/500
To find the airplane's velocity relative to the ground, we need to combine the velocities of the airplane in the west direction and the crosswind in the north direction. We can use vector addition to do this.
Step 1: Draw a diagram to visualize the situation. Draw a north (up) direction and a west (left) direction. Label the airplane's velocity as 500 miles/hr toward the west and the crosswind's velocity as 50 miles/hr toward the north.
Step 2: Convert the velocities into vector form. The airplane's velocity (Va) can be represented as -500i (since it is in the west direction) and the crosswind's velocity (Vc) can be represented as +50j (since it is in the north direction). Here, i and j represent the unit vectors in the x and y direction, respectively.
Step 3: Add the two velocities together. Vg = Va + Vc.
Vg = -500i + 50j.
The magnitude of the airplane's velocity relative to the ground, |Vg|, can be found using the Pythagorean theorem:
|Vg| = sqrt((-500)^2 + 50^2).
Simplifying this calculation, we get:
|Vg| = sqrt(250000 + 2500).
|Vg| = sqrt(252500).
|Vg| ≈ 502.49 miles/hr.
Step 4: Determine the direction of the airplane's velocity relative to the ground. To find the direction, we can use trigonometry. The direction can be determined by calculating the angle (θ) between the vector (Vg) and the west direction.
θ = arctan(Vg_y / Vg_x),
where Vg_x represents the x-component (-500) and Vg_y represents the y-component (+50) of the vector Vg.
θ = arctan(50 / -500).
Using a calculator, we find:
θ ≈ -05.71 degrees.
The direction of the airplane's velocity relative to the ground is approximately 05.71 degrees west of the north direction.
Therefore, the airplane's velocity relative to the ground is approximately 502.49 miles/hr in a direction of 05.71 degrees west of north.