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notsponge [240]
3 years ago
9

A jet airliner moving initially at 406 mph (with respect to the ground) to the east moves into a region where the wind is blowin

g at 568 mph in a direction 15◦ north of east. What is the new speed of the aircraft with respect to the ground?
Physics
1 answer:
astraxan [27]3 years ago
6 0

Answer:

966 mph

Explanation:

Using as convention:

- East --> positive x-direction

- North --> Positive y-direction

The x- and y- components of the initial velocity of the jet can be written as

v_{1x} = 406 mph\\v_{1y} = 0

While the components of the velocity of the wind are

v_{2x} = (568)(cos 15^{\circ})=548.6 mph\\v_{2y} = (568)(sin 15^{\circ})=147.0 mph

So the components of the resultant velocity of the jet are

v_x = v_{1x}+v_{2x}=406+548.6=954.6 mph\\v_y = v_{1y}+v_{2y}=0+147.0=147.0 mph

And the new speed is the magnitude of the resultant velocity:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(954.6)^2+(147.0)^2}=965.8 mph \sim 966 mph

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