A jet airliner moving initially at 889 mph

(with respect to the ground) to the east moves

into a region where the wind is blowing at

830 mph in a direction 59° north of east.

What is the new speed of the aircraft with

respect to the ground?

Answer in units of mph.

1500 mph

Explanation:

Take east to be + x and north to be + y.

The x component of the velocity is:

vₓ = 889 cos 0° + 830 cos 59°

vₓ = 1316.5 mph

The y component of the velocity is:

vᵧ = 889 sin 0° + 830 sin 59°

vᵧ = 711.4 mph

The speed is found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1316.5 mph) ² + (711.4 mph) ²

v = 1496 mph

Rounded to two significant figures, the jet's speed relative to the ground is 1500 mph.