Respuesta :
Answer:
[tex]v=42.66\ m.s^{-1}[/tex]
[tex]\beta=6.75^{\circ}[/tex] clockwise from the south.
Explanation:
Given:
- velocity of the plane southwards, [tex]v_p=35\ m.s^{-1}[/tex]
- velocity of the wind in south-west, [tex]v_w=10\ m.s^{-1}[/tex]
- ∴Angle between the plane and wind velocities, [tex]\theta=45^{\circ}[/tex]
According to the vector addition rule, magnitude of the resultant velocity is given as:
[tex]v=\sqrt{v_p^2+v_w^2+2\times v_p.v_w\ cos\ \theta}[/tex]
[tex]v=\sqrt{35^2+10^2+2\times 35\times 10\ cos\ \45}[/tex]
[tex]v=42.66\ m.s^{-1}[/tex] is the plane's speed with respect to ground.
Direction of this resultant with respect to south:
[tex]tan\ \beta=\frac{v_w\ sin\theta}{(v_p+v_p\ cos\theta)}[/tex]
[tex]tan\ \beta=\frac{10\ sin\ 45}{(35+35\ cos\ 45)}[/tex]
[tex]\beta=6.75^{\circ}[/tex] clockwise from the south.
Answer:
The plane's velocity is [tex](v_{x},v_{y})=(-7.07,-42.07)[/tex].
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
[tex]v_{x}=-v'\cos\theta[/tex]
Put the value into the formula
[tex]v_{x}=-10\cos45[/tex]
[tex]v_{x}=-7.07\ m/s[/tex]
We need to calculate the velocity along the y -direction
Using velocity component
[tex]v_{y}=-(v'\sin\theta+v)[/tex]
Put the value into the formula
[tex]v_{y}=-(10\sin45+35)[/tex]
[tex]v_{y}= -42.07\ m/s[/tex]
Hence, The plane's velocity is [tex](v_{x},v_{y})=(-7.07,-42.07)[/tex].
