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November 4th, 2019, 02:21 AM   #1
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Question How can I find the relative velocity of yacht respect to a passenger ship?

The problem is as follows:
A passenger liner sails in the direction $N23^{\circ}E$ following a rate of $\textrm{21 knots}$ and a yacht sails in the direction of $S67^{\circ}E$ at a rate of $\textrm{28 knots}$. Find the relative velocity in knots of the yacht with respect to that of the passenger ship.
$\begin{array}{ll}
1.\,35\,\textrm{knots with a bearing angle of}\,N60^{\circ}E\\
2.\,45\,\textrm{knots with a bearing angle of}\,S60^{\circ}E\\
3.\,35\,\textrm{knots with a bearing angle of}\,N30^{\circ}W\\
4.\,25\,\textrm{knots with a bearing angle of}\,N60^{\circ}W\\
5.\,35\,\textrm{knots with a bearing angle of}\,S30^{\circ}E\\
\end{array}$

For this particular problem, I'm a confused at how to use the bearing angles, since I don't know whether I should assume that both ships are in the same spot, which should be physically impossible (although a submarine could obey this assumption). The other assumption is that the yacht is at a certain distance from the ship. But from its relative position, how can I tell whether it is behind or ahead of the passenger liner? Does this matter?

For this particular situation, I assumed what is described in the drawing from below:



The best thing that I could come up with was that relative might mean that I should take the difference of such vectors. But to do so, what I did was to rotate in the counterclockwise direction the axis so I could encounter an angle which could have been easier to work with as shown in the picture.



From then, what I did was to establish the coordinates for the vector $\vec{v}$ for the passenger liner (in blueberry color) and $\vec{u}$ in orange for the yacht.

$\vec{v}=21\cos60^{\circ}\hat{i}+21\sin60^{\circ} \hat{j}$

$\vec{v}=\frac{21}{2}\hat{i}+\frac{21\sqrt{3}}{2} \hat{j}$

$\vec{u}=28\cos30^{\circ}\hat{i}+28\sin30^{\circ} \hat{j}$

$\vec{u}=14\sqrt{3}\hat{i}+14\hat{j}$

From then I took the difference between those vectors as indicated in the figure:

$\vec{v}-\vec{u}=\left(\frac{21}{2}-14\sqrt{3}\right)\hat{i}+\left(\frac{21\sqrt{3}}{2 }-14\right)\hat{j}$

Now all that would be left is to obtain the angle which can be obtained from the inverse tangent function:

$\tan\omega=\frac{\frac{21\sqrt{3}}{2}-14}{\frac{21}{2}-14\sqrt{3}}$

$\tan\omega=\frac{\frac{21\sqrt{3}}{2}-14}{\frac{21}{2}-14\sqrt{3}}$

$\tan\omega=\frac{\frac{3\sqrt{3}}{2}-2}{\frac{3}{2}-2\sqrt{3}}$

$\tan\omega=\frac{3\sqrt{3}-4}{3-4\sqrt{3}}$

$\tan\omega=\frac{3\sqrt{3}-4}{3-4\sqrt{3}}\times \frac{3+4\sqrt{3}}{3+4\sqrt{3}}$

$\tan\omega=-\frac{24-7\sqrt{3}}{39}$

However the last result did not yield any desired result. Can somebody help me here? Did I do something wrong?

Last edited by skipjack; November 4th, 2019 at 10:26 PM.
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November 4th, 2019, 06:17 AM   #2
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Almost correct method.

1) Velocity of the yacht relative to the liner would be u - v, not v - u.

2) Very clever noticing you could turn the coordinate system 7°. Just don't forget to turn it back afterwards.

3) Speaking of clever, did you notice you have a right triangle? That should make the magnitude of the relative velocity easy to find.

4) Something helpful when working with vector maths like this would be to draw things approximately to scale when you can.

Let's morphin'!
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November 4th, 2019, 07:10 PM   #3
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Originally Posted by DarnItJimImAnEngineer View Post
Almost correct method.

1) Velocity of the yacht relative to the liner would be u - v, not v - u.

2) Very clever noticing you could turn the coordinate system 7°. Just don't forget to turn it back afterwards.

3) Speaking of clever, did you notice you have a right triangle? That should make the magnitude of the relative velocity easy to find.

4) Something helpful when working with vector maths like this would be to draw things approximately to scale when you can.

Let's morphin'!
Ryōkai! 了解

This problem took me sometime to draw. I hope it is accordingly to your suggestions. I did corrected the orientation of the subtraction of the vectors. In this case it should indicate the velocity of the yacht minus the velocity of the passenger liner as seen in the figure from below.



Indeed it helped much to identify that the angle between those two was $90^{\circ}$ and as a consecuence I could obtain the hypotenuse of the triangle. Initially I had some problems at trying to find this by "asuming" that such right triangle was $30^{\circ}-60^{\circ}-90^{\circ}$ but it was not the case but instead a $37^{\circ}-53^{\circ}-90^{\circ}$. From it I could find the modulus of the resultant.

I attempted to do it by the "formal" approach by taking the norm of the vector subtraction, which it was by no means an easy arithmetical task. But I could obtain $35$ as well.

I did "restored" the original orientation of the axis by turning in the clockwise direction those $7^{\circ}$ degrees which I had intentionally turned in the beginning.



And finally the answer would be:

$S35^{\circ}E$ which corresponds to the fifth choice in the alternatives given and it checks with the answer sheet.
This was rather fun to draw by practicing my artistic abilities but I did noticed that the most important thing here was to spot how to put those vectors.

I'm getting the idea that relative speed means that we have to subtract one from the other. But in this case was specifically stated.

I have more questions regarding this matter but those are in another question with similar situation.

To this problem. Shutdown, 完了 kanryou!.
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