Help me! Math Vectors! A helicopter wishes to fly directly East at 100 km/h.

Question

A helicopter wishes to fly directly East at 100 km/h. If a constant wind blows directly from the North-East at 40 km/h, find the direction and velocity of the wind speed that must be maintained in order to achieve this result (i.e. the speed of the helicopter with respect to the air). (Hint: Construct a diagram first)

Answer 1

A sketch would look like this:

You want vector v₂ to add to vector v₃ to get vector v₁

Vector v₁ is only in the x direction. Vector v₂ can be broken into 2 components v_2x and v_2y using trig ratios.

v_2x = 40cos45 but we need to remember its in the negative x direction so it equals -28.28. And v_2y (also in the negative y direction) is 40sin45 = -28.28

So, summing in the y direction: v_2y + v_3y = v_1y or v_3y = v_1y - v_2y or v_3y = 0 - -28.28 so v_3y = 28.28.

Summing in the x direction v_2x + v_3x = v_1x or v_3x = v_1x - v_2x or v_3x = 100 - -28.28 = 128.28

Putting v₃ back together again, using the Pythagorean Theorem,

The magnitude of v₃ = sqrt(v_3x² + v_3y²) = sqrt(128.28² + 28.28²) = 131.365 km/hr. This would need to be adjusted for proper sig figs if your teacher has taught you that.

The direction of v₃ comes from taking the inverse tangent. θ = tan^-1(v_3y/v_3x) = tan^-1(28.28/128.28) = 12.43° (so 12.43° north of east)