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 November 4th, 2019, 03:06 AM #1 Senior Member   Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 Math Focus: Calculus How can I find the relative velocity given a fixed observer and another who's moving? The problem is as follows: In a race track an orange car $A$ is moving to the north ($\textrm{Y-axis}$) at a rate of $20\frac{m}{s}$ with respect to that a tv cameraman who is at a point labeled $O_{1}$ in the ground. Simultaneously another car (blue) $B$ is moving to the direction known as $N53^{\circ}O$ at $25\frac{m}{s}$ with respect to that another cameraman $O_{2}$. If the second cameraman $O_{2}$ is holding a camera in a dolly moving to the same direction of the blue car $B$ and the velocity $v_{o_{2}o}$ is $5.0\,\frac{m}{s}$. Find the velocity of the blue car $B$ with respect to that $A$ in $\frac{m}{s}$.The given alternatives on my book are: $\begin{array}{ll} 1.&-15\hat{i}+4\hat{j}\frac{m}{s}\\ 2.&-24\hat{i}-2\hat{j}\frac{m}{s}\\ 3.&-15\hat{i}-2\hat{j}\frac{m}{s}\\ 4.&+26\hat{i}+2\hat{j}\frac{m}{s}\\ 5.&-26\hat{i}+4\hat{j}\frac{m}{s}\\ \end{array}$ Okay I'm lost with this problem. Essentially my source of confusion is how should I understand $v_{o_{2}o}= 5.0\,\frac{m}{s}$ My only guess here is that what the author intended to explain was that the velocity of the blue car with respect to that the first cameraman is $5\frac{m}{s}$. But other than that. I don't know exactly what else should I do with the given information to work with the given bearing angles. Can somebody help me here?.  November 4th, 2019, 06:49 AM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 The 5 m/s is the speed of cameraman $O_2$ relative to cameraman $O_1$. Let's take the ground ($O_1$) to be our reference point (i.e., $v_{O_1}=0$). Then $v_{O_2}=5.0~m/s$ and $(v_B - v_{O_2}) = 25~m/s$, both in the same direction. This means $v_B=30~m/s$ relative to our reference. You are looking for $v_B - v_A$. The only thing confusing me is the direction of the velocity of car B. Is that supposed to be nord-53°-ouest (or the Spanish equivalent)? As in, 53° west of north? November 4th, 2019, 08:28 AM   #3
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Quote:
 Originally Posted by DarnItJimImAnEngineer The 5 m/s is the speed of cameraman $O_2$ relative to cameraman $O_1$. Let's take the ground ($O_1$) to be our reference point (i.e., $v_{O_1}=0$). Then $v_{O_2}=5.0~m/s$ and $(v_B - v_{O_2}) = 25~m/s$, both in the same direction. This means $v_B=30~m/s$ relative to our reference. You are looking for $v_B - v_A$. The only thing confusing me is the direction of the velocity of car B. Is that supposed to be nord-53°-ouest (or the Spanish equivalent)? As in, 53° west of north?
if your hunch regarding the direction of $v_B$ is correct ...
Attached Images velocity_vector_rel.jpg (29.0 KB, 4 views) November 5th, 2019, 02:48 AM   #4
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 Originally Posted by DarnItJimImAnEngineer The 5 m/s is the speed of cameraman $O_2$ relative to cameraman $O_1$. Let's take the ground ($O_1$) to be our reference point (i.e., $v_{O_1}=0$). Then $v_{O_2}=5.0~m/s$ and $(v_B - v_{O_2}) = 25~m/s$, both in the same direction. This means $v_B=30~m/s$ relative to our reference. You are looking for $v_B - v_A$. The only thing confusing me is the direction of the velocity of car B. Is that supposed to be nord-53°-ouest (or the Spanish equivalent)? As in, 53° west of north?
I'm so sorry. When I transcribed from the original source I forgot the translation of $O$ should had been $W$ for west.

But I read your explanation and I still don't get. Can you tell me exactly how to use the reference frame?.

The only way that I could understand why you did a straight subtraction between $v_{B}-v_{O_{2}}$ is that both share the same orientation?

I roughly understand what you mean so should I subtract

$v_{B}-v_{A}$ But this cannot be done directly. I have to consider the orientation of $A$ in other words, a vectorial subtraction must be done.

Therefore:

$\vec{v_{B}}=\left\langle -30\sin 53^{\circ}, 30 \cos 53^{\circ} \right\rangle$

$\vec{v_{B}}=\left\langle -30\left(\frac{4}{5}, 30 \left(\frac{3}{5} \right ) \right )\right\rangle$

$\vec{v_{B}}=\left\langle -24, 18 \right\rangle$

Therefore since:

$\vec{v_{A}}=\left\langle 0,20\right\rangle$

Thus to make a subtraction from both will be

$\vec{v_{B}}-\vec{v_{A}}= \left\langle -24, -2 \right\rangle$

Which would be what skeeter described accurately with its graph.

But my major question is those $25$ meters per second is the velocity as seen by the driver? or as seen by the cameraman in the dolly? I'm confused with the reference frame.

This subtraction thing is what is makind me confused.  November 5th, 2019, 10:03 AM   #5
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 Originally Posted by Chemist116 Car $A$ is moving to the north ($\textrm{Y-axis}$) at a rate of $20\frac{m}{s}$ with respect to point $O_{1}$
This means cameraman $O_1$ would see the car moving north at this speed. Mathematically, this says $\vec{v}_A - \vec{v}_{O_1} = 20~m/s~\hat{j}$.

Quote:
 Originally Posted by Chemist116 Car $B$ is moving $N53^{\circ}W$ at $25\frac{m}{s}$ with respect to cameraman $O_{2}$.
So, cameraman $O_2$ would see car B moving at this speed.
$\vec{v}_B - \vec{v}_{O_2} = 25~m/s ~[-sin(53°)\hat{i}+cos(53°)\hat{j}]$

From cameraman $O_1$'s point of view on the ground, the dolly is already moving NW $(-sin(53°)\hat{i}+cos(53°)\hat{j})$ at $5~m/s$, so car B is moving at $30~m/s$ relative to the ground.
$\vec{v}_B - \vec{v}_{O_1} = (\vec{v}_B - \vec{v}_{O_2}) + (\vec{v}_{O_2} - \vec{v}_{O_1})$

If the driver of car B looked down at the ground, it would appear to be moving behind him at 30 m/s. If he looked at the dolly chasing him, it would appear to be moving backwards at 25 m/s.

Position/velocity/acceleration "with respect to" or "relative to" a reference "O" means subtract the reference position/velocity/acceleration. This is the same as defining a reference frame with "O" at the origin. November 7th, 2019, 01:23 AM   #6
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Quote:
 Originally Posted by DarnItJimImAnEngineer This means cameraman $O_1$ would see the car moving north at this speed. Mathematically, this says $\vec{v}_A - \vec{v}_{O_1} = 20~m/s~\hat{j}$. So, cameraman $O_2$ would see car B moving at this speed. $\vec{v}_B - \vec{v}_{O_2} = 25~m/s ~[-sin(53°)\hat{i}+cos(53°)\hat{j}]$ From cameraman $O_1$'s point of view on the ground, the dolly is already moving NW $(-sin(53°)\hat{i}+cos(53°)\hat{j})$ at $5~m/s$, so car B is moving at $30~m/s$ relative to the ground. $\vec{v}_B - \vec{v}_{O_1} = (\vec{v}_B - \vec{v}_{O_2}) + (\vec{v}_{O_2} - \vec{v}_{O_1})$ If the driver of car B looked down at the ground, it would appear to be moving behind him at 30 m/s. If he looked at the dolly chasing him, it would appear to be moving backwards at 25 m/s. Position/velocity/acceleration "with respect to" or "relative to" a reference "O" means subtract the reference position/velocity/acceleration. This is the same as defining a reference frame with "O" at the origin.
Just as a practical conclusion can I say that each time I shift from one reference to another I could swap their signs and with that I'll get what that person will see in his reference frame relative to the first reference?. Btw I felt that the thanks button was not enough to tell my gratitude to your help in translating the words into math and explaining the nitty gritty details of reference frames as these concepts are not easy for me to swallow. Last edited by Chemist116; November 7th, 2019 at 01:26 AM. Reason: added compliment Tags find, fixed, moving, observer, relative, velocity Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Chemist116 Physics 2 November 4th, 2019 07:10 PM arybhatta01 Algebra 2 June 9th, 2018 11:02 PM asifrahman1988 Calculus 1 March 17th, 2013 04:24 AM joker Calculus 5 October 31st, 2010 02:30 AM vlamers Calculus 1 January 27th, 2010 08:12 PM

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