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?

 

[baba89]

[13.12 degrees W of N]

I would like to clarify that the calculation provided in the "Attempt at a Solution" section is incorrect. The correct way to calculate the relative velocity of the helicopter would be to use vector addition, taking into account both the wind velocity and the desired velocity of the helicopter.

First, we can break down the wind velocity and the desired velocity of the helicopter into their respective x and y components. The wind velocity of 38km/h at 25 degrees N of E can be broken down into 38cos25 km/h in the x direction and 38sin25 km/h in the y direction. Similarly, the desired velocity of the helicopter of 91km/h at 17 degrees W of N can be broken down into 91cos17 km/h in the x direction and 91sin17 km/h in the y direction.

Next, we can use vector addition to find the resultant velocity of the helicopter. This can be done by adding the x components and the y components separately. The resultant x component would be 38cos25 + 91cos17 = 115.86 km/h and the resultant y component would be 38sin25 + 91sin17 = 103.92 km/h.

Finally, we can use the Pythagorean theorem to find the magnitude of the resultant velocity, which would be the relative velocity of the helicopter. This can be done by taking the square root of the sum of the squares of the x and y components. Therefore, the relative velocity of the helicopter would be sqrt((115.86)^2 + (103.92)^2) = 152.02 km/h.

In terms of direction, we can use trigonometry to find the angle of the resultant velocity. This can be done by taking the inverse tangent of the y component over the x component. Therefore, the direction of the relative velocity of the helicopter would be tan^-1(103.92/115.86) = 41.74 degrees W of N.

Therefore, the correct answer would be that the helicopter needs to achieve a relative velocity of 152.02 km/h at an angle of 41.74 degrees W of N.