Question

An airplane heads due north at v₁ = 10m/s through a v₂ =7m/s cross wind blowing from the east to the west. What is the magnitude of the resultant velocity of the airplane (relative to due north)? Write your answer with one decimal place. What is the direction of the resultant velocity of the airplane above (relative to due north)? Note that your interactive lab shows the angle from +x-axis (relative to due east).

Answer 1

The magnitude of the airplane's resultant velocity is approximately 12.2 m/s, and the direction of this velocity is 35.0 degrees west of north.

Step-by-step explanation:

The problem at hand involves vector addition to find the resultant velocity of an airplane experiencing a crosswind. Given that the airplane heads due north with a velocity v1 = 10 m/s and there is a crosswind from the east, with a velocity v2 = 7 m/s to the west, we can calculate the resultant magnitude using the Pythagorean theorem since the velocities are perpendicular to each other.

The magnitude of the resultant velocity (Vr) is

Vr = √(v1² + v2²)
= √(10² + 7²)
= √(100 + 49)
= √149
= 12.2 m/s (rounded to one decimal place).

Next, to find the direction relative to due north, we use the arctan function:

θ = arctan(v2/v1)
= arctan(7/10)
= arctan(0.7)
≈ 35.0 degrees west of north.

So, the airplane's resultant velocity relative to the ground is approximately 12.2 m/s at an angle of 35.0 degrees west of north.