Question

A plane has an airspeed of 109 km/h. It is flying on a bearing of 83° while there is a 21 km/h wind out of the northeast (bearing 225*). What are the ground speed and the bearing of the plane? The ground speed is km/h. Round to the nearest tenth as needed.) : 0

Answer 1

`\Sigma v_(res) =  93.35 km/h`

`\theta = 88.17^(\circ)`

Given:

Air speed of plane,
`\v_(a) = 109 km/hr`

bearing angle,
`\theta_(1) = 83^(\circ)`

wind speed,
`\v_(w) = 21 km/hr`

bearing angle,
`\theta_(2) = 225^(\circ)`

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,

`\Sigma v_(x) = v_(a)sin\theta_(1) - v_(w)sin\theta_(2)`

`\Sigma v_(x) = 109sin83^(\circ) - 21sin45^(\circ)`

= 93.34 km/h

`\Sigma v_(y) = v_(a)cos\theta_(1) - v_(w)cos\theta_(2)`

`\Sigma v_(y) = 109cos83^(\circ) - 21cos45^(\circ)`

= - 1.565

Now, resultant of two vectors is given by:

`\Sigma v_(res) = \sqrt{(\Sigma v_(x))^(2) + (\Sigma v_(y))^(2)}`

`\Sigma v_(res) = \sqrt{( 93.34)^(2) + (- 1.565)^(2)}`

`\Sigma v_(res) = 93.35 km/h`

Bearing angle,

`\theta = tan^(-1)((\Sigma v_(y))/(\Sigma v_(x)))`

`\theta =tan^(-1)((- 1.5656)/(93.34))= -1.828^(\circ)`

`\theta = -1.828^(\circ) + 90^(\circ) = 88.17^(\circ)`