Question

14. (16) Greg

Answer 1

a)

velocit of the plane relative to the ground

v(plane-ground) = v(plane-air) + v(air-ground)
= 420+ 120 = 540km/h

b)
the velocity of the plane relative to the air

v(plane-air) =420km/h

c)

For the purposes of getting the proper signs, we'll designate East and North as the 'positive' directions, and West and

south as negative East and North, respectively.
This is the desired, resultant vector: a velocity completely in the North direction
V = V N-direction (+ 0 E=direction)

Wind velocity, East component, Vw E = - (120km/h)cos23= - 110.46km/h E-direction
Vw N = - 120sin23 = - 46.88km/h N-direction

For this part, make sure you've drawn out a diagram to illustrate the situation.

Look at it, and realize that in order to end up heading North while the wind blows Southwest,

the plane will have to be headed in some kind of Northeast direction. This means that breaking down the velocity of the plane into its components gives:
Vp E = (420km/h)sin? E-direction
Vp N = (420km/h)cos? N-direction

So we have the following vector addition:
Vp + Vw = V
...split into two components, but we only need one of them to solve for ?:
Vp E + Vw E = V E = 0 (the East component of the resulting vector is 0)
Vp E = - Vw E
(420km/h)sin? E-dir. = - (- 110.46km/h E-dir.)
sin? = 110.46km/h E-dir /420km/h E-dir.
sin? = 0.263
Use the inverse sine function (secant = sin^-1) to isolate ?
? = sin^-1(0.263)
? = 15.24 degrees