A pilot wishes to fly from Bayfield to Kitchener, a distance of 100 km on a bearing of 105°. The speed of the plane in still air is 240 km/h

Question

A pilot wishes to fly from Bayfield to Kitchener, a distance of 100 km on a bearing of 105°. The speed of the plane in still air is 240 km/h. A 20 km/h wind is blowing on a bearing of 210°.
Remembering that she must fly on a bearing of 105° relative to the ground (i.e. the resultant must be on that bearing), find (6 marks)
a) the heading she should take to reach her destination.
b) how long the trip will take.

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Nguyệt Ánh 6 months 2021-07-18T04:59:34+00:00 1 Answers 37 views 0

Answers ( )

    0
    2021-07-18T05:00:49+00:00

    Here is are by step 🙂 hope this helps

    First things first, you want the resultant to be in the direction of the bearing of 105 °. The distance required is irrelevant for question 1.

    You want the addition of the wind speed and the planes velocity to equal some resultant in the direction one 105 °. Draw a parallelogram (vector addition) of the resultant (the 100km in direction 105°) and the wind speed. The diagonal that results from the vector addition will equal 240. You only know the angle between the resultant and the wind speed right now (105°).

    https://www.geogebra.org/geometry/rbejhvta

    What is the angle between the resultant (the direction we want to travel) and the direction the plane will actually go when wind speed is taken into accound???

    Law of sines.

    You said you have it, so we will continue under the assumption you got 4.617°.

    now, take into account that the plane is accelerating 4.617° off of the 105° bearing in order to travel on the 105° bearing when wind speed is taken into account… what heading is the plane taking? 105° – 4.617° = 100.4°

    There is part 1.

    now you simply need to find the actual speed she is travelling along the bearing 105°. (she is traveling 240km/h with a 100.4° heading, find the correct speed for the resultant).

    EDIT: The representation says 105.1°, it is supposed to say 105°

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