4. An airplane has an airspeed of 500 kph bearing N40E. The wind velocity is 60 kph in the direction of N30W. Find the resultant vector repr

Question

4. An airplane has an airspeed of 500 kph bearing N40E. The wind velocity is 60 kph in the direction of N30W. Find the resultant vector representing the path of the plane relative to the ground. What is the ground speed of the plane? What is the direction?

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Delwyn 5 months 2021-08-21T22:06:39+00:00 1 Answers 1 views 0

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    2021-08-21T22:08:31+00:00

    Answer:

    The ground speed of the plane is 513.6 km/hr.

    The direction is 46.59°

    Explanation:

    Given that,

    Air speed of plane = 500 kph

    Wind speed = 60 kph

    The velocity of plane is

    v_{p}=(v\cos\theta)i+(v\sin\theta)j

    Put the value into the formula

    v_{p}=(500\cos40)i+(500\sin\40)

    v_{p}=383i+321.3j

    The velocity of wind is

    v_{w}=v\cos(90+30)i+v\sin(90+30)j

    Put the value into the formula

    v_{w}=(60\cos120)i+(60\sin120)j

    v_{w}=-30i+51.9j

    We need to calculate the ground speed of plane

    v_{a}=v_{w}+v_{p}

    v_{a}=-30i+51.9j+383i+321.3j

    v_{a}=(383-30)i+(321.3+51.9)j

    v_{a}=353i+373.2j

    The ground speed is

    |v_{a}|=\sqrt{(353)^2+(373.2)^2}

    |v_{a}|=513.6\ km/h

    We need to calculate the direction

    Using formula of direction

    \tan\theta=\dfrac{373.2}{353}

    \theta=\tan^{-1}(\dfrac{373.2}{353})

    \theta=46.59^{\circ}

    Hence, The ground speed of the plane is 513.6 km/hr.

    The direction is 46.59°

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