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.