Question

What is the resultant velocity for a plane that is flying due east at 33m/s that is also being carried due north at 12m/s by the wind

Answer 1

v = 35.11 m / s, θ = 20º north of east

Step-by-step explanation:

This is a compound problem of relative velocities, we can use the Pythagorean theorem to find the modulus of the resulting velocity

v² = v₁² + v₂²

where the eastward velocity is v₁=33 m/s and the northward velocity

v₂=12 m / s

v =
`√(33^2 +12^2)`

v = 35.11 m / s

the direction of this velocity can be found with the uses of trigonometry

tan θ = v₂ / v₁

θ = tan⁻¹ v₂ / v₁

θ = tan⁻¹ 12/33

θ = 20º

this angles is north of east