Respuesta :
Answer:
[tex]\Sigma v_{res} = 93.35 km/h[/tex]
[tex]\theta = 88.17^{\circ}[/tex]
Given:
Air speed of plane, [tex]\v_{a} = 109 km/hr[/tex]
bearing angle, [tex]\theta_{1} = 83^{\circ}[/tex]
wind speed, [tex]\v_{w} = 21 km/hr[/tex]
bearing angle, [tex]\theta_{2} = 225^{\circ}[/tex]
Step-by-step explanation:
To calculate the ground speed we take the resultant of the vector sum of air speed and wind speed.
Also, head wind speed is added to ground speed and tail wind speed is subtracted from ground speed.
Now,
[tex]\Sigma v_{x} = v_{a}sin\theta_{1} - v_{w}sin\theta_{2}[/tex]
[tex]\Sigma v_{x} = 109sin83^{\circ} - 21sin45^{\circ}[/tex]
= 93.34 km/h
[tex]\Sigma v_{y} = v_{a}cos\theta_{1} - v_{w}cos\theta_{2}[/tex]
[tex]\Sigma v_{y} = 109cos83^{\circ} - 21cos45^{\circ}[/tex]
= - 1.565
Now, resultant of two vectors is given by:
[tex]\Sigma v_{res} = \sqrt{(\Sigma v_{x})^{2} + (\Sigma v_{y})^{2}}[/tex]
[tex]\Sigma v_{res} = \sqrt{( 93.34)^{2} + (- 1.565)^{2}}[/tex]
[tex]\Sigma v_{res} = 93.35 km/h[/tex]
Bearing angle,
[tex]\theta = tan^{-1}(\frac{\Sigma v_{y}}{\Sigma v_{x}})[/tex]
[tex]\theta =tan^{-1}(\frac{- 1.5656}{93.34})= -1.828^{\circ}[/tex]
[tex]\theta = -1.828^{\circ} + 90^{\circ} = 88.17^{\circ}[/tex]
