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.

Respuesta :

Answer:

984.8 mph

Explanation:

The initial velocity of the jet in terms of components is

[tex]v_x = 852 mph\\v_y = 0[/tex]

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

The velocity of the wind is

[tex]v'_x = (206)(cos 55^{\circ})=118.2 mph\\v'_y = (206)(sin 55^{\circ})=168.7 mph[/tex]

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

[tex]V_x = v_x + v'_x = 852+118.2=970.2 mph\\V_y = v_y + v'_y = 0 + 168.7 =168.7 mph[/tex]

And so the new speed of the jet is

[tex]V=\sqrt{V_x^2+V_y^2}=\sqrt{(970.2)^2+(168.7)^2}=984.8 mph[/tex]