Question

An airplane is flying through the air at a speed of 150 mph at a heading of 60 degrees. if the wind is blowing at 20 mph from the south-east, (a) what is its ground speed, and (b) how far and (c) in what direction from its starting point will the plane be located after flying for 30 minutes.

Answer 1

Answer:

Explanation:

Given

speed of plane 150 mph

heading $60^{\circ}$

wind is blowing at 20 mph from south-east

velocity of plane w.r.t. wind

$v_{pw}=150(\cos 60\hat{i}+\sin 60\hat{j})$

$v_{w}=20(\cos 45\hat{i}-\sin 45\hat{j})$

$v_{p}=v_{pw}+v_{w}$

$v_{p}=(150\cos 60+20\cos 45)\hat{i}+(150\sin 60-20\sin 45 )\hat{j}$

$v_{p}=89.14\hat{i}+115.76\hat{j}$

$|v_{p}|=146.103 m/s$

(b)after 30 minutes

Plane traveled in  x direction $=89.14\times 0.5=44.57 miles$

Plane in Y direction $=115.76\times 0.5=57.88 miles$

Total distance$=\sqrt{x^2+y^2}=73.05 miles$

direction $tan\theta =\frac{115.76}{89.14}=1.298$

$\theta =52.38^{\circ}$ w.r.t east