2-D Relative Velocity Problem

[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 $261.7 km/h$.

Additionally, note the following discrepancy:

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

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?

 

[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.

 

[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.

 

[NoPhysicsGenius]

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.

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

 

[NoPhysicsGenius]

I hope I got it.

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

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 [we use] t₁ + 1.

For plane Q, $$v_1 +52 = \frac{624}{t_1}$$.
$$t_1 = \frac{624}{v_1+52}$$, and we isolate t₁, so we can put it in the equation of P. We get:
$$v^2_1 - 52v_1 - 67548 = 0$$

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.

Thanks for your help.

 

[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.