# Vector expressions - equivalence

## Homework Statement

A car is travelling at 12ms^-1. To the passenger in the car the wind appears to be blowing at 8.0ms^-1 at right angles to the road. What is the magnitude and direction of the velocity of the wind with respect to the ground.

## Homework Equations

I can think of two possible vector expressions which should be equivalent. I am clearly doing something wrong because they are not.
Using:
vwg = velocity of wind wrt ground
vcg = velocity of car wrt ground
vwc = velocity of wind wrt car
Expression 1
vwg = vwc + vcg

Expression 2 (following the rule of subtracting the observer's movement)
vwc= vwg-vcg

Yes - I'm aware I should add the negative vector and that the negative sign means the reverse direction

## The Attempt at a Solution

Expression 1:
The solution vector diagram is (excuse the dots so I don't lose the spaces):

^----------->vcg
|..............^
|vwc ..... /
|........ / vwg
|....../
|../
|/
It gives a vwg magnitude of c. 14ms^-1 and a direction of tan^-1(8/12) = c. 34 degrees with the wind coming from behind the car

Expression 2:
The solution vector diagram is:

<----------------- -vcg
.^.......................^
....\....................|
........\ ...............| vwc
....vwg..\.............|
..............\..........|
..................\......|
......................\..|
.........................\|
Of course the magnitude is the same as in the previous example but the direction is
tan^-1 (8/-12) = -34 degrees.

My concern is that the wind appears to be coming from the front!

Gloom - what have I misunderstood?