Relative Velocity of a Helicopter

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SUMMARY

The discussion focuses on calculating the relative velocity of a helicopter in the presence of wind. The average wind velocity is given as 38 km/h at 25 degrees North of East, while the helicopter must achieve a velocity of 91 km/h at 17 degrees West of North. The initial calculations for the resultant velocity, using trigonometric functions, incorrectly add the wind velocity to the helicopter's velocity instead of determining the necessary additional velocity to achieve the desired direction and speed. The correct resultant velocity should be calculated using vector addition principles.

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vsharma88
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Homework Statement


Average Wind Velocity 38km/h [25 degrees N of E]
Helicopter needs to achieve 91km/h [17 degrees W of N]


Homework Equations


r=sqrt(x^2+y^2)


The Attempt at a Solution



y=38sin25 + 91sin98 = 106.17
x=38cos25 - 91cos98 = 47.1

r= 116.15

answer should be 94km/h
 
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Hi vsharma88,

vsharma88 said:

Homework Statement


Average Wind Velocity 38km/h [25 degrees N of E]
Helicopter needs to achieve 91km/h [17 degrees W of N]


Homework Equations


r=sqrt(x^2+y^2)


The Attempt at a Solution



y=38sin25 + 91sin98 = 106.17
x=38cos25 - 91cos98 = 47.1

You have a bit of an error with the trig functions, but the most important thing here is how you have written down your equations. It appears that you are adding the 38km/h wind velocity to the 91km/h velocity to find a new resultant.

However, the way I read the problem indicates that the 91km/h is the resultant. So the question here is what do you have to add to the wind velocity so that the helicopter goes 91km/h in the specified direction?
 

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