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An aeroplane can travel at a speed of 120 km/h there's no wind. Today there's a wind velocity of 3i + 2j km/h. Find:
a) the speed of the wind
b) the poisition vector if the aeroplane is to end up due north of its starting point after 1 hour
c) the actual bearing required.
a) $\displaystyle |v_w| = \sqrt{3^2 + 2^2}$Quote:
Originally Posted by Mr Rayon
a bit easier to do part (c) first ...
to travel due north relative to the ground, the airplane will have to steer west of north because of the wind blowing to the northeast.
let $\displaystyle \theta$ = air velocity vector, $\displaystyle \vec{v_a}$ , angle relative to east.
$\displaystyle \vec{v_a} = 120(\cos{\theta} \vec{i} + \sin{\theta}\vec{j})$
wind velocity vector ...
$\displaystyle \vec{v_w} = 3\vec{i} + 2\vec{j}$
ground velocity vector ...
$\displaystyle \vec{v_g} = |v_g|\vec{j}$
now, since $\displaystyle \vec{v_a} + \vec{v_w} = \vec{v_g}$ ...
in the x-direction, $\displaystyle 120\cos{\theta} + 3 = 0$
$\displaystyle \cos{\theta} = -\frac{1}{40}$
$\displaystyle \theta \approx 91.4^\circ$ relative to east
the airplane needs to steer $\displaystyle 1.4 ^\circ$ west of north.
in the y-direction ...
$\displaystyle 120\sin{\theta} + 2 = |v_g| \approx 122$ km/hr
and for part (b) ... the position vector of the airplane after 1 hr is
$\displaystyle \vec{r_g} = 122\vec{j}$ km
