Relative Velocity of a Helicopter

[vsharma88]

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

 

[alphysicist]

Hi vsharma88,

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?