# 2-D Relative Velocity Problem

1. Aug 25, 2004

### NoPhysicsGenius

I am having difficulty with Problem 49 of Chapter 3 from Physics for Scientists and Engineers by Paul A. Tipler, 4th Edition:

Airports A and B are on the same meridian, with B 624 km south of A. Plane P departs airport A for B at the same time that an identical plane, Q, departs airport B for A. A steady 60 km/h wind is blowing from the south 30' east of north. Plane Q arrives at airport A 1 h before plane P arrives at airport B. Determine the airspeeds of the two planes (assuming that they are the same) and the heading of each plane.

In drawing the vector diagram for plane Q, we use the following equation:

$$\vec{v}_{Qg} = \vec{v}_{Qw} + \vec{v}_{wg}$$

where $\vec{v}_{Qg}$ is the velocity of Plane Q relative to the ground, $\vec{v}_{Qw}$ is the velocity of Plane Q relative to the wind, and $\vec{v}_{wg}$ is the velocity of the wind relative to the ground.

Since I am unable to include a picture of the vector diagram, I will instead describe it as follows (and you might wish to draw it on a sheet of scratch paper for future reference):

$\vec{v}_{Qg}$ points from due south to due north;
$\vec{v}_{wg}$ points from the tail of $\vec{v}_{Qg}$ toward the northeast;
$\theta$ is used to denote the angle between $\vec{v}_{Qg}$ and $\vec{v}_{wg}$, which is 30' east of north;
$\vec{v}_{Qw}$ points from the head of $\vec{v}_{wg}$ to the head of $\vec{v}_{Qg}$, in a direction west of north;
$\theta_Q$ is used to denote the angle of the heading of Plane Q. It is located between $\vec{v}_{Qw}$ and a line drawn from the tail of $\vec{v}_{Qw}$ parallel to $\vec{v}_{Qg}$. Note that, by interior angles, $\theta_Q$ is also the angle between $\vec{v}_{Qg}$ and $\vec{v}_{Qw}$.

In drawing the vector diagram for plane P, we use the following equation:

$$\vec{v}_{Pg} = \vec{v}_{Pw} + \vec{v}_{wg}$$

where $\vec{v}_{Pg}$ is the velocity of Plane P relative to the ground, $\vec{v}_{Pw}$ is the velocity of Plane P relative to the wind, and $\vec{v}_{wg}$ is once again the velocity of the wind relative to the ground.

A description of the vector diagram for Plane P is as follows:

$\vec{v}_{Pg}$ points from due north to due south;
$\vec{v}_{wg}$ points from the tail of $\vec{v}_{Pg}$ toward the northeast;
$\theta$ is used to denote the angle between $\vec{v}_{wg}$ and a line drawn parallel to $\vec{v}_{Pg}$, which is 30' east of north;
$\vec{v}_{Pw}$ points from the head of $\vec{v}_{wg}$ to the head of $\vec{v}_{Pg}$, in a direction west of south;
$\theta_P$ is used to denote the angle of the heading of Plane P. It is located between $\vec{v}_{Pw}$ and a line drawn from the tail of $\vec{v}_{Pw}$ parallel to $\vec{v}_{Pg}$. Note that, by interior angles, $\theta_P$ is also the angle between $\vec{v}_{Pg}$ and $\vec{v}_{Pw}$.

We note the following:

$$v_{wg} = 60 km/h$$
$$\theta = 30'(\frac{1^\circ}{60'}) = 0.5^\circ$$
$$v_{Qw} = v_{Pw}$$
$t_Q = t_P - 1 h$, where $t_Q$ is the time of arrival of Plane Q and $t_P$ is the time of arrival of Plane P.

$\vec{v}_{wg}$ is computed as follows:

$$v_{wg,y} = v_{wg} \cos \theta = (60 km/h) \cos 0.5^\circ = 60 km/h$$
$$v_{wg,x} = v_{wg} \sin \theta = (60 km/h) \sin 0.5^\circ = 0.524 km/h$$
Therefore, $\vec{v}_{wg} = v_{wg,x} \widehat{i} + v_{wg,y} \widehat{j} = (0.524 \widehat{i} + 60 \widehat {j}) km/h$.

So far, so good.

Here's where things start to get potentially troublesome ...

From the vector diagram for Plane P, we have:

$$v_{Pw,x} = -v_{wg,x} = -0.524 km/h$$

From the vector diagram for Plane Q, we have:

$$v_{Qw,x} = -v_{wg,x} = -0.524 km/h$$

Now ...

$$\sin \theta_Q = \frac{opposite}{hypotenuse} = \frac{|v_{wg,x}|}{v_{Qw}}$$
$$\Rightarrow v_{Qw} = \frac{|v_{wg,x}|}{\sin \theta_Q} = \frac{0.524 km/h}{\sin \theta_Q}$$

Also ...

$$\sin \theta_P = \frac{opposite}{hypotenuse} = \frac{|v_{wg,x}|}{v_{Pw}}$$
$$\Rightarrow v_{Pw} = \frac{|v_{wg,x}|}{\sin \theta_P} = \frac{0.524 km/h}{\sin \theta_P}$$

The condition $v_{Pw} = v_{Qw}$ implies:

$$\frac{0.524 km/h}{\sin \theta_P} = \frac{0.524 km/h}{\sin \theta_Q}$$
$\Rightarrow \theta_P = \theta_Q$ (for our purposes, at least--though the angles can actually differ by certain fractional multiples of $\pi$, though I forgot the exact formula ...)

With $\vec{r}_{Qg}$ denoting the position vector of Plane Q relative to the ground and $\vec{r}_{Pg}$ denoting the position vector of Plane P relative to the ground, we have the following conditions:

$$r_{Qg,y} = v_{Qg,y}t_Q = 624 km$$
$$r_{Pg,y} = v_{Pg,y}t_P = -624 km$$

The condition $r_{Pg,y} = -r_{Qg,y}$ yields:

$$v_{Pg,y}t_P = -v_{Qg,y}t_Q$$

The relation $t_Q = t_P - 1$ then yields:

$$v_{Pg,y}t_P = -v_{Qg,y}(t_P - 1)$$
$$\Rightarrow (v_{Pg,y} + v_{Qg,y})t_P = v_{Qg,y}$$
$$\Rightarrow t_P = \frac{v_{Qg,y}}{v_{Pg,y} + v_{Qg,y}}$$

From the vector diagram for Plane Q ...

$$v_{Qg,y} = v_{Qw,y} + v_{wg,y} = v_{Qw,y} + 60$$

Now ...

$$\cos \theta_Q = cos \theta_P = \frac{adjacent}{hypotenuse} = \frac{|v_{Qw,y}|}{v_{Qw}}$$
$$\Rightarrow |v_{Qw,y}| = v_{Qw} \cos \theta_Q = v_{Pw} \cos \theta_P$$

Therefore ...

$$v_{Qg,y} = v_{Pw} \cos \theta_P + 60$$

Now ...

$$v_{Pg,y} = v_{wg,y} + v_{Pw,y}$$

Since $v_{Pw,y} = -v_{Qw,y}$, we then have:

$$v_{Pg,y} = v_{wg,y} - v_{Qw,y} = 60 - v_{Pw} \cos \theta_P$$

Therefore ...

$$t_P = \frac{v_{Qg,y}}{v_{Pg,y} + v_{Qg,y}} = \frac{v_{Pw} \cos \theta_P + 60}{(60 - v_{Pw} \cos \theta_P) + (v_{Pw} \cos \theta_P + 60)}$$
$$\Rightarrow t_P = \frac{v_{Pw} \cos \theta_P + 60}{120}$$

Then $v_{Pw} = \frac{0.524}{\sin \theta_P}$ implies:

$$t_P = \frac{0.524 \cot \theta_P + 60}{120}$$

I haven't the slightest clue where to go from here.

The answer in the back of the book is as follows:

$261.7 km/h$, $6.58^\circ$ west of north

There is at least one problem with this answer ... Obviously, although the two planes might well have the same angle of heading (as I have shown above), it is obvious from the vector diagrams that the heading of Plane Q would be $6.58^\circ$ west of north, whereas the heading of Plane P would be $6.58^\circ$ west of south.

Also, I should note the following:

$v_{Pw} = \frac{0.524 km/h}{\sin \theta_P} = \frac{0.524 km/h}{\sin 6.58^\circ} = 4.57 km/h \ll$ the correct answer of $261.7 km/h$.

$$t_P = \frac{v_{Pw} \cos \theta_P + 60}{120} = \frac{261.7 \cos 6.58^\circ + 60}{120} = 2.67 h \neq \frac{0.524 \cot \theta_P + 60}{120} = \frac{0.524 \cot 6.58^\circ + 60}{120} = 0.538 h$$

Indeed, if the value on the right (namely, $0.538 h$) were correct, then $t_Q = t_P - 1 h = 0.538 h - 1 h = -0.462 h$!

Clearly, I have done a great deal wrong here. Can someone please spot the flaw(s) in my reasoning?

2. Aug 25, 2004

### wisky40

try to use 30 degrees instead of 30 minutes because If I used the the data of your book's solution, see what happends (V_p)sin6.58= 261.7sin6.58=29.988...=60sintheta
=> sin(theta)=29.9988.../60=.4998..~.5 =>theta~30degrees.
By the way the solution that I had is V={[(2dv'cos30)/1h]+(v')^2}^(1/2).
where (d) is the distance between the airports and (v') the wind's speed.

3. Aug 25, 2004

### JasonRox

I tried it before looking at the solution, and this is what I got.

I started with:

Plane P - $$v_p = v_1 - 52$$ Against the wind.
Plane Q - $$v_q = V_1 + 52$$ With the wind.

Note: It says that the velocities are the same.

I went on and calculated the vertical velocity of the wind.

Assuming this is what it meant.

l---/
l--/
lx/
l/ -> x=30 degrees

So, naturally using trig, we get $$cos30^o*60=51.9$$. I rounded to 52.

We know that the $$v=d/t$$.

So, we get for plane P $$v_1 - 52 = \frac{624}{t_1 + 1}$$.

With a little work we get $$v_1 - 52t_1 + v_1 t_1 = 676$$. Remember it took P an extra hour to reach its destination, so t + 1.

For plane Q, $$v_1 +52 = \frac{624}{t_1}$$.
$$t_1 = \frac{624}{v_1+52}$$, and we isolate t, so we can put it in the equation of P. We get:
$$v^2_1 - 52v_1 - 67548 = 0$$

Use the quadratic formula, and we shall get:

$$v_1 = -235.2 or 287.2$$

Using trigonometry or the pythagorean theorem to balance it back against the wind.

I got 285.2km/h, not bothering with the 235, which is 233km/h.

T_1 should be the same for both equations.

They are about .2 hours off, so it's possible that 51.9 could have fixed some of it.

I hope I got it.

Last edited: Aug 25, 2004
4. Aug 26, 2004

### NoPhysicsGenius

I agree ... This is a typo in the book that I had missed. Thanks.

5. Aug 27, 2004

### NoPhysicsGenius

Your solution method is correct ... and that is what counts.

It is here that things went wrong for you. Let me perform the substitution in detail ...

$$v_1 - 52 = \frac{624}{\frac{624}{v_1+52} + 1}$$

Multiply the expression on the RHS by $\frac{v_1 + 52}{v_1 + 52}$ to get:

$$v_1 - 52 = \frac{(624)(v_1 + 52)}{(624) + (v_1 + 52)} = \frac{624 v_1 + (52)(624)}{v_1 + 676} = \frac{624 v_1 + 32448}{v_1 + 676}$$
$$\Rightarrow (v_1 - 52)(v_1 + 676) = 624 v_1 + 32448$$
$$\Rightarrow {v_1}^2 - 52 v_1 + 676 v_1 - (52)(676) = 624 v_1 + 32448$$
$$\Rightarrow {v_1}^2 + 624 v_1 - 35152 = 624 v_1 + 32448$$
$$\Rightarrow {v_1}^2 = 32448 + 35152 = 67600$$
$$\Rightarrow v_1 = 260 km/h$$
$$\Rightarrow v_Q = v_1 + 52 = 260 + 52 = 312 km/h$$

Now $v_Q$ corresponds to $v_{Qg}$ using the notation from my original post.

Applying the Pythagorean Theorem then gives:

$${v_{Qw}}^2 = {v_{Qg}}^2 + {v_{wg}}^2 - 2v_{Qg}v_{wg}\cos \theta$$
$$\Rightarrow {v_{Qw}}^2 = (312 km/h)^2 + (60 km/h)^2 - 2(312 km/h)(60 km/h)\cos 30^\circ = 68520 \frac{km^2}{h^2}$$
$$\Rightarrow v_{Qw} = 261.8 km/h$$

This compares very favorably to the book's answer of $261.7 km/h$.

To compute the angle of each plane's heading, we do the following:

$$v_{Qw} = \frac{|v_{wg,x}|}{\sin \theta_Q} = \frac{v_{wg} \sin \theta}{\sin \theta_Q} = \frac{(60 km/h) \sin 30^\circ}{\sin \theta_Q} = \frac{30 km/h}{\sin \theta_Q}$$
$$\Rightarrow \sin \theta_Q = \frac{30 km/h}{v_{Qw}}$$
$$\Rightarrow \theta_Q = \sin^{-1} \frac{30 km/h}{261.8 km/h} = 6.58^\circ$$

This agrees precisely with the book's answer.

Last edited: Aug 27, 2004
6. Aug 27, 2004

### JasonRox

It happens.

You pointed out something else I should have done, too. Answers wouldn't change, but simply a different method that can be useful one day.

Thanks.