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A jet airliner moving initially at 504 mph (with respect to the ground) to the east moves into a region where the wind is blowing at 219 mph in a direction 29◦ north of east. What is the new speed of the aircraft with respect to the ground? Answer in units of mph.

User Parrots
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1 Answer

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Answer

given,

intial velocity = 504 mph

wind speed = 219 mph

at an angle of 29◦

from the data

The resultant velocity =


V = V_x \hat{i} + V_y\hat{j}


V = (504 + 219 cos 29^0 ) \hat{i} +(219 sin 29^0 )\hat{j}


V = 695.54\hat{i} + 106.17 \hat{j}

the magnitude of velocity


V = √(695.54^2+106.17^2)

V = 703.59 m/s

direction

tan θ =
(106.17)/(695.54)

θ = 8.676°

User Hbogert
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