Vector addition, trying to find the angle (resulting velocity of plane with respect to ground)

[coooool222]

Is angle Vag 180 since the vector is a straight line pointing left?
Also you can add 30 degrees with 150 which will be 180?

Relative Velocity: An airplane is flying at an air speed of 100 mph at a heading of 30 degrees southeast. If the velocity of the wind is 20 mph due west, determine the resultant velocity of the plane with respect to the ground.

 

[Steven G]

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A straight angle is 180 regardless of its direction (horizontal, vertical or otherwise).

 

[coooool222]

The link is not working. It is also best to include the image in the post. Please try again.

A straight angle is 180 regardless of its direction (horizontal, vertical or otherwise).

 

[topsquark]

Relative Velocity: An airplane is flying at an air speed of 100 mph at a heading of 30 degrees southeast. If the velocity of the wind is 20 mph due west, determine the resultant velocity of the plane with respect to the ground.

Yes, the angle of vector [imath]\textbf{V}_{ag}[/imath] is 180 degrees (measured from the +x axis, or East if you need to label that.)

I don't know if you've gotten this far yet, but I personally prefer to write it as [imath]\textbf{V}_{ag} = - 20 \, \hat{i}[/imath]. This "component notation."

-Dan

Addendum: And whereas 30 + 150 = 180, it has nothing to do with this problem.

 

[Steven G]

Is angle Vag 180 since the vector is a straight line pointing left?
Also you can add 30 degrees with 150 which will be 180?

Vag is a vector! How can you ask what its angle equal? I see at least two angles associated with Vag.

 

[khansaheb]

Relative Velocity: An airplane is flying at an air speed of 100 mph at a heading of 30 degrees southeast. If the velocity of the wind is 20 mph due west, determine the resultant velocity of the plane with respect to the ground.

If I write these vectors in unit vector notation, I find it easier to "handle".

V_pa = 100*cos(-30°) i + 100*sin(-30°) j = 50√3 i - 50 j

V_ag = 20*cos(180°) i = -20 i

V_pg = V_pa + V_ag = (50√3 i - 50 j ) + (-20 i ) = (50√3 - 20) i + (-50) j mph

 

[khansaheb]

If I write these vectors in unit vector notation, I find it easier to "handle".

Vpa = 100*cos(-30°) i + 100*sin(-30°) j = 50√3 i - 50 j

Vag = 20*cos(180°) i = -20 i

Vpg = Vpa + Vag = (50√3 i - 50 j ) + (-20 i ) = (50√3 - 20) i + (-50) j mph

The angle related to the vector V_pg - with the positive x-axis is:

Θ = arctan[(-50)/(50√3 - 20)]........................ negative sign indicates that Θ is either in second-quadrant or fourth-quadrant

 

[Harry_the_cat]

Of course the other way to approach this problem is using trigonometry.
The cosine rule will enable you to find the magnitude of Vpg.
Then the sine rule will enable you to find the upper angle in the triangle. (Then add 30 degrees to get the direction.)