# Math Help - relative velocities

1. ## relative velocities

A man was walking with a velocity of 4 i km/h when he obvserved that the wind was blowing from the north . When the man increased his velocity to 6 i km/h , he noticed that the wind seemed to be blowing from N 45 W . Find the speed and the direction of the wind .
(Take the unit vector i to be in the direction due east.)

2. Originally Posted by hooke
A man was walking with a velocity of 4 i km/h when he obvserved that the wind was blowing from the north . When the man increased his velocity to 6 i km/h , he noticed that the wind seemed to be blowing from N 45 W . Find the speed and the direction of the wind .
(Take the unit vector i to be in the direction due east.)
If v is the man's velocity and w is the winds velocity, both relative to teh ground, then the wind's velocity, relative to him, is w- [/b]v[/b].

Let the velocity of the wind (relative to the ground) be xi+ yj. The wind velocity relative to the man, when he is walking at 4i is xi+ yj- i= (x- 1)i+ j. The fact that the wind is from the north relative to him means x- 1= 0. When he is walking at 6i, the wind velocity, relative to him, is xi + y[/b]j[/b]- 6i= (x- 6)i+ yi. The fact that the wind is at N 45 W relative to him means that x- 6= y (legs of a isosceles right triangle).

3. Originally Posted by HallsofIvy
If v is the man's velocity and w is the winds velocity, both relative to teh ground, then the wind's velocity, relative to him, is w- [/b]v[/b].

Let the velocity of the wind (relative to the ground) be xi+ yj. The wind velocity relative to the man, when he is walking at 4i is xi+ yj- 4i= (x- 1)i+ j. The fact that the wind is from the north relative to him means x- 1= 0. When he is walking at 6i, the wind velocity, relative to him, is xi + y[/b]j[/b]- 6i= (x- 6)i+ yi. The fact that the wind is at N 45 W relative to him means that x- 6= y (legs of a isosceles right triangle).
thanks , is the speed of wind the same for both cases ? And also , did you miss a 4 (in red) ?

4. Hello, hooke!

A fascinating problem . . . hope I got it right.

A man was walking with a velocity of 4i km/h
when he obvserved that the wind was blowing from the north.
When the man increased his velocity to 6i km/h ,
he noticed that the wind seemed to be blowing from N 45° W.
Find the speed and the direction of the wind.
Evidently, the wind is blowing from the northeast.

The first phase looks like this:
Code:
                  C
*
* |
w *   |
*     |
* θ     |
A * - - - - * B
4

The man walks from $A$ to $B\!:\;\;|\overrightarrow{AB}| \,=\,4$
The wind has vector $\vec w \,=\,\overrightarrow{CA}\!:\;\;w \,=\,|\vec w|,\;\;\theta \,=\,\angle A$
The apparent wind vector is: . $\overrightarrow{CB}$

We have: . $\cos\theta \,=\,\frac{4}{w} \quad\Rightarrow\quad w\cos\theta \,=\,4 \quad\Rightarrow\quad w\sin\theta \,=\,\sqrt{w^2-16}$ .[1]

The second phase looks like this:
Code:
                C
*
*   *
w *       *
*           *
* θ         45° *
A * - - - - - - - - - * D
6

The man walks from $A$ to $D\!:\;|\overrightarrow{AD}| \,=\,4$
The wind vector $\vec w$ is exactly the same.
The apparent wind vector is: . $\overrightarrow{CD},\;\angle D \,=\,45^o$

We have: . $\angle C \:=\:180^o - \theta - 45^o \:=\:135^o - \theta$

Law of Sines: . $\frac{w}{\sin45^o} \:=\:\frac{6}{\sin(135^o-\theta)} \quad\Rightarrow\quad w\sin(135^o-\theta) \:=\:6\sin45^o$ .[2]

The left side is: . $w\bigg[\sin135^o\cos\theta - \cos135^o\sin\theta\bigg] \;=\;w\bigg[\frac{1}{\sqrt{2}}\cos\theta - \left(-\frac{1}{\sqrt{2}}\right)\sin\theta\bigg]$

. . . . . . . . . . $=\; w\bigg[\frac{1}{\sqrt{2}}\cos\theta + \frac{1}{\sqrt{2}}\sin\theta\bigg] \;=\; \frac{1}{\sqrt{2}}\,\bigg[w\cos\theta + w\sin\theta\bigg]$

. . Substitute [1]: . $\frac{1}{\sqrt{2}}\left(4 + \sqrt{w^2-16}\right)$

Then [2] becomes: . $\frac{1}{\sqrt{2}}\left(4 + \sqrt{w^2-16}\right) \:=\:6\left(\frac{1}{\sqrt{2}}\right) \quad\Rightarrow\quad 4 + \sqrt{w^2-16} \:=\:6$

. . . . . $\sqrt{w^2-16} \:=\:2\quad\Rightarrow\quad w^2-16 \:=\:4 \quad\Rightarrow\quad w^2 \:=\:20 \quad\Rightarrow\quad w \:=\:2\sqrt{5}$

The wind speed is about: . $\boxed{4.47\text{ km/hr}}$

Substitute into [1]: . $w\cos\theta \:=\:4 \quad\Rightarrow\quad 2\sqrt{5}\cos\theta \:=\:4 \quad\Rightarrow\quad \cos\theta \:=\:\frac{2}{\sqrt{5}}$

. . Hence: . $\theta \:\approx\:26.6^o$

The direction of the wind is: . $\boxed{S\,63.4^o\,W}$