Basic Kinematics Problem -- A plane flies North with a crosswind

[rr96]

1. A plane is aimed north and is traveling 851 km/h. A wind blows the plane from 40° S of E at 36 km/h. What is the plane’s resultant velocity?

2. I think the answer is 828.3 km/h at 88.1° East of North but I am not sure.

Ay= 851

Bx= cosθ
cos40=x/36
x=27.6

By=sinθ
sin40=-y/36
y=-23.1

Rx=27.7

Ry=851-23.1
=827.9

a^2 + b^2 = c^2
(27.6)^2 + (827.9)^2 = c^2
c= 828.3

Using tan, θ=88.1

 

[Greg Bernhardt]

1. A plane is aimed north and is traveling 851 km/h. A wind blows the plane from 400 S of E at 36 km/h. What is the plane’s resultant velocity?

2. I think the answer is 828.32 km/h at 88.09° East of North but I am not sure.

How did you reach your answer? We can then see where you might have gone wrong.

 

[SteamKing]

According to your answer, a plane originally traveling north is blown so far off course by a modest SEasterly wind that it winds up traveling almost due east. Who needs hurricanes?

If the wind is coming from the SE and the plane is traveling north, is the plane experiencing a headwind or a tailwind?

In these problems, drawing a simple sketch helps to clarify the direction of the various vector components.

 

[haruspex]

We cannot see inside your head. When you make up symbols for variables, unless the meanings are blindingly obvious, please state what they represent.

 

[ciubba]

The numbers look good, but I do not believe that you have interpreted the location of theta properly. Are you certain that theta is made between the y-axis and the Resultant vector i.e. east of north?