Resultant Velocity


by sweet877
Tags: resultant, velocity
sweet877
sweet877 is offline
#1
Nov9-05, 10:25 PM
P: 30
A plane flying at 125 m/s [south] encounters a storm that changes its course due to the wind blowing 25 m/s 30 degrees N of E. What is the plane's resultant velocity?

How would you draw the wind's component?
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Diane_
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#2
Nov9-05, 10:30 PM
HW Helper
P: 402
In mathematics, angles are generally measured anti-clockwise from the positive x-axis. If you superimpose a compass rose on the standard Cartesian coordinate axes, the positive x-axis becomes east. So 30 degrees north of east would be a 30 degree angle as typically measured.

Does that do it?
sweet877
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#3
Nov9-05, 10:33 PM
P: 30
So the diagram would roughly be:
..../
.../ 25 m/s
./
| 30 degrees
|
| 125 m/s
|
?
Then would I find the resultant velocity using cosine law?

Diane_
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#4
Nov9-05, 10:38 PM
HW Helper
P: 402

Resultant Velocity


Not quite. When you're adding vectors graphically, you place the tail of the next one at the head of the last one. If you draw the wind speed first, it would make a 30 degree angle with the positive x-axis. When you draw in the airspeed of the plane, it would head due south from the head of the wind speed vector. A little geometry will give you the angle between the vectors - then go for the Law of Cosines.
sweet877
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#5
Nov9-05, 10:57 PM
P: 30
So the angle between the two vectors is 180-30-5.74 = 144.26.
Then, using Cosine Law:
Result V= sqrt[25^2 + 125^2 - 2(25)(125)cos (144.26)]= 146.02 m/s at 5.74 degrees W of N

Was what I did correct?
Diane_
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#6
Nov9-05, 11:01 PM
HW Helper
P: 402
Not quite. I'm not sure where you're getting the 5.74.
sweet877
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#7
Nov9-05, 11:07 PM
P: 30
Oh...I made a mistake in thinking that the angle opposite the 125 m/s is 30 degrees. Let me try this again...
sweet877
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#8
Nov9-05, 11:23 PM
P: 30
So the angle in between the vectors is 60 degrees. Then, using Cosine Law,
resultant vector: sqrt[25^2 + 125^2 - 2(25)(125)(cos 60)] = 115 m/s at 60 degrees E of S?
Diane_
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#9
Nov9-05, 11:29 PM
HW Helper
P: 402
Yes, the angle is 60 degrees, and I'm getting the same answer as you for the speed. The direction is another matter - you'll need to use the Law of Sines to get the angle. Be careful - the angle in the triangle is obtuse, so you'll have to interpret the results the Law of Sines gives you. Then, you'll have to remember that the angle you're getting is with respect to the wind speed - you'll need to subtract 30 degrees from it to get the correct direction.
sweet877
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#10
Nov9-05, 11:51 PM
P: 30
Hmm...would it be 10.9 W of S?
sin 60/115 = sin x/25
Diane_
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#11
Nov10-05, 12:03 AM
HW Helper
P: 402
Right idea, but what you're finding there is the obtuse angle between the resultant and the wind. The Law of Sines - well, your calculator, actually - will always give you the acute angle. That's what I meant by "you'll need to interpret the result." So - find the obtuse angle that has that same sine. Lastly, remember that that angle is not what you want. You want the angle it makes with one of the axes.

Look at your sketch again - I think you'll see it.
sweet877
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#12
Nov10-05, 12:26 AM
P: 30
Is it sin 60/115 = sin x/125 to get 70.2765. Then, subtract it from 180, which equals 109.7235. Then, subtract 30 from that number, which equals 79.7.

So the direction would be 79.7 degrees E of S?
Diane_
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#13
Nov10-05, 12:55 AM
HW Helper
P: 402
I ended up with 76 degrees south of east, but that may be because I used the full length calculated instead of just the 115. Remember - there is a difference between "south of east" and "east of south". "South of east" is measured clockwise from the positive x-axis. "East of south" would be measured clockwise from the negative y-axis.

Outside of that, I think you've got it. :)
sweet877
sweet877 is offline
#14
Nov10-05, 07:02 PM
P: 30
Thanks very much! I appreciated your help.


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