Question

A jet airliner moving initially at 852 mph

(with respect to the ground) to the east moves
into a region where the wind is blowing at
206 mph in a direction 55° north of east.
What is the new speed of the aircraft with
respect to the ground?
Answer in units of mph.

Answer 1

984.8 mph

Step-by-step explanation:

The initial velocity of the jet in terms of components is

`v_x = 852 mph\\v_y = 0`

where we took east as positive x-direction and north as positive y-direction.

The velocity of the wind is

`v'_x = (206)(cos 55^(\circ))=118.2 mph\\v'_y = (206)(sin 55^(\circ))=168.7 mph`

So, the resultant velocity of the jet considering also the wind is

`V_x = v_x + v'_x = 852+118.2=970.2 mph\\V_y = v_y + v'_y = 0 + 168.7 =168.7 mph`

And so the new speed of the jet is

`V=√(V_x^2+V_y^2)=√((970.2)^2+(168.7)^2)=984.8 mph`