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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 ($\textrm{Yaxis}$) 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 nord53°ouest (or the Spanish equivalent)? As in, 53° west of north? 
November 4th, 2019, 08:28 AM  #3  
Math Team Joined: Jul 2011 From: Texas Posts: 3,093 Thanks: 1675  Quote:
 
November 5th, 2019, 02:48 AM  #4  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 Math Focus: Calculus  Quote:
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  
Senior Member Joined: Jun 2019 From: USA Posts: 376 Thanks: 202  Quote:
Quote:
$\vec{v}_B  \vec{v}_{O_2} = 25~m/s ~[sin(53°)\hat{i}+cos(53°)\hat{j}]$ 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  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 Math Focus: Calculus  Quote:
Last edited by Chemist116; November 7th, 2019 at 01:26 AM. Reason: added compliment  

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