Question

An airplane flies with a velocity v1=120 m/s to the north, but there is a crosswind of v2=20 m/s to the east.

Answer 1

The resulting velocity of the airplane, considering both its original velocity to the north
`(\(v_1 = 120 \, \text{m/s}\))` and the crosswind to the east
`(\(v_2 = 20 \, \text{m/s}\))`, is approximately
`\(v = 122 \, \text{m/s}\)` at an angle of approximately
`(5.7^\circ\)`) north of east.

Step-by-step explanation:

When an object, like an airplane, moves in the presence of a crosswind, the resulting velocity is the vector sum of its original velocity and the crosswind velocity. This can be represented using vector addition. For this scenario, the Pythagorean theorem and trigonometric functions come into play.

The resultant velocity (v) can be found using the Pythagorean theorem:
`\(v = √(v_1^2 + v_2^2)\)`. Substituting the given values,
`\(v = √(120^2 + 20^2) \approx 122 \, \text{m/s}\)`. The angle
`(\(\theta\))` of the resultant velocity with respect to the original direction can be determined using the arctangent function:
`\(\theta = \arctan{(v_2)/(v_1)}\). Substituting the values, \(\theta = \arctan{(20)/(120)} \approx 5.7^\circ\)`.

In conclusion, the airplane's resulting velocity is approximately
`\(122 \, \text{m/s}\)` at an angle of
`\(5.7^\circ\)` north of east. Understanding vector addition is essential for calculating the overall motion of an object in the presence of multiple velocities.