Question

The nose of an ultra-light plane is pointed is pointed South and its airspeed indicator shows 35m/s. The plane is in a 10m/s wind blowing towards the SouthWest relative to the ground. What is the plane's speed with respect to ground?

Answer 1

Answer:

$v=42.66\ m.s^{-1}$

$\beta=6.75^{\circ}$ clockwise from the south.

Explanation:

Given:

• velocity of the plane southwards, $v_p=35\ m.s^{-1}$

• velocity of the wind in south-west, $v_w=10\ m.s^{-1}$

• ∴Angle between the plane and wind velocities, $\theta=45^{\circ}$

According to the vector addition rule, magnitude of the resultant velocity is given as:

$v=\sqrt{v_p^2+v_w^2+2\times v_p.v_w\ cos\ \theta}$

$v=\sqrt{35^2+10^2+2\times 35\times 10\ cos\ \45}$

$v=42.66\ m.s^{-1}$ is the plane's speed with respect to ground.

Direction of this resultant with respect to south:

$tan\ \beta=\frac{v_w\ sin\theta}{(v_p+v_p\ cos\theta)}$

$tan\ \beta=\frac{10\ sin\ 45}{(35+35\ cos\ 45)}$

$\beta=6.75^{\circ}$ clockwise from the south.

Answer 2

Answer:

The plane's velocity is $(v_{x},v_{y})=(-7.07,-42.07)$.

Explanation:

Given that,

Airspeed v= 35 m/s

Speed of wind v'= 10 m/s

Let x be the east and y be the north.

We need to calculate the velocity along the x -direction

Using velocity component

$v_{x}=-v'\cos\theta$

Put the value into the formula

$v_{x}=-10\cos45$

$v_{x}=-7.07\ m/s$

We need to calculate the velocity along the y -direction

Using velocity component

$v_{y}=-(v'\sin\theta+v)$

Put the value into the formula

$v_{y}=-(10\sin45+35)$

$v_{y}= -42.07\ m/s$

Hence, The plane's velocity is $(v_{x},v_{y})=(-7.07,-42.07)$.