Trying to calculate relative speed between two moving pointsby hkBattousai Tags: derivative, moving, relative, speed, velocity 

#1
Jul1411, 02:53 PM

P: 58

There are two points on the 2D rectangular coordinate system, namely P and M.
Their positions are function of time and are: [tex]Position \, of \, P: \, (p_x(t), \, p_y(t))[/tex] [tex]Position \, of \, M: \, (m_x(t), \, m_y(t))[/tex] Distance between them is: [tex]R(t) \, = \, \sqrt{(p_x  m_x)^2 \, + \, (p_y  m_y)^2}[/tex] And the relative speed (magnitude of relative velocity) between them is: [tex]V_{pm}(t) \, = \,  \frac{dR}{dt}[/tex] Is it correct up to this step? If so, can you please help me take this derivative? If not, how do I calculate this relative speed? Any help will be appreciated. 



#2
Jul1411, 03:39 PM

Sci Advisor
P: 5,942

To get the derivative you need explicit expressions for the components of P(t) and M(t). Based on the information presented, all you can get is an expression involving the derivatives of the components.
Aside: why the minus sign for the relative speed? 



#3
Jul1411, 03:45 PM

P: 58

Think of it a little, correct me if I'm wrong... 



#4
Jul1511, 01:55 PM

P: 56

Trying to calculate relative speed between two moving points
Try something like y_{1}= x^2 and y_{2} = x
let the x component velocities be the same for both functions then the y component velocities of the functions at any point as respects x would be the derivative of the functions at those points (slope) and the relative velocities of those point are the difference of the y component velocites. I don't think the distance between the 2 points has anything to do with relative velocities since the 2 function above cross at x = 1. I think that is correct but correct me if Im wrong. 



#5
Jul1511, 02:32 PM

P: 56

jim pohl
dont see how vector addition of derivatives can find relative velocites between points could you give more details? My understanding is that vector addiiton gives a new vector but then that new vector only represents a direction and magnitude but as I already pointed out the distance beween two points has nothing to do with relative velocity of those points  maybe im not understanding your post i will study further. later: do you mean the vector addition of the y and x component velocites? OK maybe thatll work  need to go over this 



#6
Jul1811, 05:54 PM

P: 56

(If you have trouble reading this try holding down CTRL key while you move the scroll wheel on your mouse.) A particle travels along vector A in the direction shown and crosses vector C at point a Another particle travels along vector B in the direction shown and crosses vector C at point b What is the relative velocity of points a, b if the particle along vector A has a velocity of 2 feet per second and the particle along vector B has a velocity of 3 feet per second? (using degrees in this example) V = VA cos 110 + VB cos 40 = 2 cos 110 + 3 cos 40 = 1.61409 feet per second The relative velocity of points a,b can also be zero. For the above triangle, any positive velocity along vector A will have a velocity along vector B such that the relative velocity of a,b will equal zero. Example The particle along vector A has a velocity of 2 feet per second as it did before. What is the velocity of the particle along vector B so relative velocity of points a, b equal zero? 2 cos 110 / cos 40 = .89295 feet per second. so relative velocity of 2 moving points are dependant on the velocity of a particle along the vector and the angles between vectors. Possible implications. An observer at point b looks along vector C and notices that a certain star at point a does not have a red shift. But that star could be traveling at .2C along vector A while the observer is traveling at .089295C along vector B. Einstein did indicate that there was no absolute motion. The reasoning is that the relative velocity of objects are dependant on their vector directions as well as the velocity along the vectors. 


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