(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant 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

3. 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?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Vector expressions - equivalence

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