Question

An aircraft has to fly between two cities, one of which is 600.0 km north of the other. The pilot starts from the southern city and encounters a steady 100.0 km/h wind that blows from the northeast. The plane has a cruising speed of 236.0 km/h in still air. In what direction (relative to east) must the pilot head her plane

Answer 1

72.57° North of east

Step-by-step explanation:

From the given information:

We can compute the velocity plane that is related to the ground in air in the North direction as;

`v^(\to) _(PG) = v \\ \\ v^(\to) _(PG,x) = 0 \\ \\ v^(\to) _(PG,y) = v`

However, the velocity of the wind-related to the ground from the NorthEast direction is;

`v^(\to)_(wG)=100 \ km/h \\ \\ \text{from North East} \\ \\ v_(wG,x) = (-100 \ km/h ) cos 45 = -70.7 km/h \\ \\ v_(wG,y) = (-100 \ km/h ) sin 45 = -70.7 km/h`

Now,

Since the plane is moving with a 236 km/h speed in the Northeast direction;

Then;

`v^(\to) _(pw) = 236 \ km/h \\ \\ v^(\to) _(pw.x) = (236 m/s) cos \theta \\ \\ v^(\to) _(pw,y) = (236\ m/s) sin \theta \\ \\ v_(pG,x) = v_(pw,x) + v_(w G,x) \\ \\ \implies 0 = (236 \ km/h) sin \theta -( 70.7 \ km/h) \\ \\ \implies cos \theta = (70.7 \ km/h)/(236 \ km/h) \\ \\ \theta = cos^(-1) (0.2996) \\ \\ \mathbf{\theta = 72.57}`