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.

Respuesta :

Answer

given,

intial velocity = 504 mph

wind speed = 219 mph

at an angle of 29◦

from the data

 The resultant velocity =

[tex]V = V_x \hat{i} + V_y\hat{j}[/tex]

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

 [tex]V = 695.54\hat{i} + 106.17 \hat{j}[/tex]

the magnitude of velocity

[tex]V = \sqrt{695.54^2+106.17^2}[/tex]

V = 703.59 m/s

direction

tan θ =  [tex]\dfrac{106.17}{695.54}[/tex]

θ = 8.676°