A plane flies due north at 225 km/h. A wind carries it due east at 55 km/h. What are the magnitude and direction of the plane's resultant velocity?
Add the two vectors. Since they are perpendicular, you can get the magnitude from the Pythagorean theorem.
The ratio 55/225 is the tangent of the resultant ground velocity, angle east of north
How do you add the vectors?
To find the magnitude and direction of the plane's resultant velocity, we can use vector addition. The plane's velocity can be broken down into its northward component (along the y-axis) and its eastward component (along the x-axis).
Given that the plane flies due north at 225 km/h, its northward component is 225 km/h in the positive y-direction. Since the wind carries it due east at 55 km/h, the eastward component is 55 km/h in the positive x-direction.
To find the resultant velocity, we can use the Pythagorean theorem. The magnitude of the resultant velocity vector (V) is given by:
|V| = √(Vx^2 + Vy^2)
where Vx and Vy are the components of the velocity vector.
In this case, Vx = 55 km/h (eastward component) and Vy = 225 km/h (northward component).
|V| = √(55^2 + 225^2)
|V| = √(3025 + 50625)
|V| = √53650
|V| ≈ 231.6 km/h
So, the magnitude of the plane's resultant velocity is approximately 231.6 km/h.
To find the direction, we can use trigonometry. The plane's resultant velocity vector forms a right-angled triangle with the x-axis and y-axis.
The direction (θ) of the resultant velocity vector can be found using the following trigonometric relationship:
θ = arctan(Vy / Vx)
In this case, Vy = 225 km/h (northward component) and Vx = 55 km/h (eastward component).
θ = arctan(225/55)
θ ≈ 74.2 degrees
So, the direction of the plane's resultant velocity is approximately 74.2 degrees north of east.