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?

Answer 1

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°