How Do Wind Velocity and Direction Affect Aircraft Travel Time and Direction?

[Benzoate]

Homework Statement

An aircraft is to fly from a from point A to an airfield B 600km due North of A. If a stead wind of 90 km/hr is blowing from the north-west, find the direction of the plane should be pointing and the time to reach B if the cruising speed of the aircraft in still air is 200 km/hr.

Homework Equations

The Attempt at a Solution

let u(A) be the velocity of the airplane and let u(W) be the velocity of wind.

u(A)+u(W)=200

u(A)-u(W)=90

u(A)= 145 km/hr and u(W)= 55

u(W)*sin(theta)=u(A) ==> sin(theta)=145/55 => theta=55.5 degrees. my velocities are wrong.

time= 600 km/145 km/hr= 4.13 hr

 

[Benzoate]

why hasn't anyone answer this basic physics question yet?

 

[vega_1992]

i`ll try
not difficult

 

[vega_1992]

speed of wind --u
speed of ship --- v
ucos45=vcosx
we found angle x relative x axis.
speed of ship in y-axis is v sinx
l(length)/vsinx= time

 

[Benzoate]

speed of wind --u
speed of ship --- v
ucos45=vcosx
we found angle x relative x axis.
speed of ship in y-axis is v sinx
l(length)/vsinx= time

I didn't think you could automatically assumed that the just because the wind is heading from the North west direction the angle the wind automatically makes is a 45. in the book, it says the direction of the wind is 18 degrees and the time is 4 hr 46 min .

 

[vega_1992]

it is a simple calculation

 

[vega_1992]

you can change some values

 

[Benzoate]

I'm still lost. I want someone to tell me what's wrong with my calculations and what alternative approach should I used to tackle this problem.

 

[tiny-tim]

vector addition

Hi Benzoate!

u(A)+u(W)=200

u(A)-u(W)=90

u(A)= 145 km/hr and u(W)= 55

Nooo … velocities don't work like that …

velocities are vectors …

so you must use vector addition …

draw a triangle, and then use either trig or components to find the unknown angle and side.

 

[Benzoate]

Hi Benzoate!


Nooo … velocities don't work like that …

velocities are vectors …

so you must use vector addition …

draw a triangle, and then use either trig or components to find the unknown angle and side.

I drew my given velocity vectors , on a coordinate axis , that travels in the North ,West, East and South directions. The velocity of the plane would be headed in the North direction , so its velocity will only have a y-component. The velocity of the wind would have a northern component and a western component.So I would write the velocity of the wind , using trigonometry. Here are my new equations:

velocity of plane relative to still wind

y component: 200 km/hr

x component: 0 km/hr

velocity of the wind relative to the ground:

y component(northern direction): 90 km/hr *sin(theta)

x component(western direction): 90 km/hr*cos(theta)

velocity of the airplane relative to the ground(resultant velocity):

y:component : 200 km/hr + 90 km/hr*sin(theta)= ?

x: component: 0+90km/hr*cos(theta)=90km/hr*cos(theta)?

Now I am lost and not sure how to calculate the direction or resultant velocity.

 

[tiny-tim]

Hi Benzoate!

ooh … sorry … but you really have got this completely wrong …

I drew my given velocity vectors , on a coordinate axis:

Before we go any further …

did you draw a trangle, or did you just draw vectors all coming out of the origin?

you must draw a triangle.

velocity of plane relative to still wind

y component: 200 km/hr

x component: 0 km/hr

Nooo … you need to find the velocity of the plane relative to the ground … you know its magnitude, but not its directon.

velocity of the wind relative to the ground:

y component(northern direction): 90 km/hr *sin(theta)

x component(western direction): 90 km/hr*cos(theta)

Yes … except why not say 45º? …

and it's a north-west wind, which means it blows south and east!

velocity of the airplane relative to the ground(resultant velocity):

y:component : 200 km/hr + 90 km/hr*sin(theta)= ?

x: component: 0+90km/hr*cos(theta)=90km/hr*cos(theta)?

Nooo … you know the direction of this velocity is due north …

but the magnitude is unknown.

Draw a triangle, and start again. ​

 

[Benzoate]

Hi Benzoate!

ooh … sorry … but you really have got this completely wrong …


Before we go any further …

did you draw a trangle, or did you just draw vectors all coming out of the origin?

you must draw a triangle.



Nooo … you need to find the velocity of the plane relative to the ground … you know its magnitude, but not its directon.



Yes … except why not say 45º? …

and it's a north-west wind, which means it blows south and east!

velocity of the airplane relative to the ground(resultant velocity):


Nooo … you know the direction of this velocity is due north …

but the magnitude is unknown.

Draw a triangle, and start again. ​

Yes I drew the triangle on my Coordinate axis is yes both vectors start from the origin; the northwest vector starts from the origin and the velocity vector for the airplane entirely lies on the y-axis and begins at the origin . The direction is unknown and its one of the things that I am asked to find. My book says the direction is 18 degrees not 45 degrees. I did not think you could automatically assumed that the direction of the vector of the wind was 45 degrees , just because a problem stated that the velocity vector was pointed in the northwest direction.

 

[tiny-tim]

Yes I drew the triangle on my Coordinate axis is yes both vectors start from the origin; the northwest vector starts from the origin and the velocity vector for the airplane entirely lies on the y-axis and begins at the origin . The direction is unknown and its one of the things that I am asked to find. My book says the direction is 18 degrees not 45 degrees. I did not think you could automatically assumed that the direction of the vector of the wind was 45 degrees , just because a problem stated that the velocity vector was pointed in the northwest direction.

ok … start again, like this …

draw a vertical line from the origin,

and a 45º line from the end of that line,

then draw another line from the origin to meet the 45º line.

Adjust the lengths so that it looks roughly as if that third line is about twice the 45º line.

Now you have a triangle with two known sides and one known angle. Find the angle at the origin.

 

[Benzoate]

ok … start again, like this …

draw a vertical line from the origin,

and a 45º line from the end of that line,

then draw another line from the origin to meet the 45º line.

Adjust the lengths so that it looks roughly as if that third line is about twice the 45º line.

Now you have a triangle with two known sides and one known angle. Find the angle at the origin.

x: component: 90km/hr*cos(theta)+200km/hr*cos(alpha) = 0
>

y: -90km/hr*sin(theta) + 200km/hr*sin(alpha) = vy...
theta=45 degrees in this case. I can solve for alpha and once I solved for alpha, I can calculated to vy and now I can use vy to find t=d/vy

 

[tiny-tim]

x: component: 90km/hr*cos(theta)+200km/hr*cos(alpha) = 0
>

y: -90km/hr*sin(theta) + 200km/hr*sin(alpha) = vy...
theta=45 degrees in this case. I can solve for alpha and once I solved for alpha, I can calculated to vy and now I can use vy to find t=d/vy

Yes that's fine!

(except … shouldn't that be a minus … 90km/hr*cos(theta) - 200km/hr*cos(alpha) = 0?)

Are you happy with the general technique now, or do you want some more help on this?

 

[Benzoate]

Yes that's fine!

(except … shouldn't that be a minus … 90km/hr*cos(theta) - 200km/hr*cos(alpha) = 0?)

Are you happy with the general technique now, or do you want some more help on this?

Yes it should be a minus sign; thanks for pointing that out to me . You been very helpful. I appreciate it.