Question

How do i solve this? An airplane heads N60°E at 600 mi/h in still air. A tail wind begins to blow in the direction N30°E at 50 mi/h. Find the resulting speed of the airplane and its resulting direction of travel. Draw a diagram

Answer 1

Let us first understand what does `"N"60^@"E"` mean.
It is: We go `60^@` towards east from reference north direction.

Now lets take east as `+x`-axis and north as `+y`-axis.
As such airplane's `v_s` speed in still air makes an angle of `30^@` with the `+x`-axis and tail wind `v_w` makes an angle of `60^@` with it.
Let `R` be resultant of both. Resolving all there along the `xand y`-axes and equating both we get, for components along `x`-axis

`R_x=v_(sx)+v_(wx)`
`=>R_x=600\ cos 30^@+50\ cos 60^@`
`=>R_x=600sqrt3/2+50xx1/2`
`=>R_x=519.62\ mph`

Similarly, for components along `y`-axis

`R_y=v_(sy)+v_(wy)`
`=>R_y=600\ sin 30^@+50\ sin 60^@`
`=>R_y=600xx1/2+50xxsqrt3/2`
`=>R_y=343.30\ mph`

Now `R=sqrt(R_x^2+R_y^2)`

`=>R=sqrt((519.62)^2+(343.30)^2)`
`=>R=622.8\ mph`

If `theta` is angle made by resultant with `x`-axis then

`theta=tan^-1(R_y/R_x)`
`theta=tan^-1(343.30/519.62)`
`theta=33.5^@`
This can also be written as `"N"56.5^@"E"`

Please make a diagram and post here.