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Does displacement=shortest length of position (final-initial)?

  1. Sep 25, 2016 #1
    1. The problem statement, all variables and given/known data
    if displacement is the shortest distance between two point then that would also meant that it is equivalent to the hypotenuse when using phytogaras theorem right?. but why isnt displacement the shortest length between two point ?because Xinitial - Xfinal couldnt posibly be the displacement (shortest distance between two point is from phytogaras theorem. where Xi^2 + Xf^2=displacement^2) this is soo confusing. please help

    2. Relevant equations
    pythagoras theorem states that the length of hypotenus^2(a.k.a shortest distance between two position) is equal to the length of (opposite^2)+ (adjacent^2).
    but in physic the displacement is not equal to the length of the hypotenus why is? this so contrary to a lot of books stating that displacement is the shortest distance between two position.

    3. The attempt at a solution
    this is just an understanding problem . i understand 100% the difference between distance and displacement . i just got confuse why some book define displacement as the shortest distance between two position or points in a straight line. But Pythagoras would said NO that is not true as that would be the hypotenuse . who is right ?physicist or pythagoras? or am i totally got the wrong idea all together? if car 400m at time 0sec moves in a straight line and stop 1000m at time 5sec. if we plot this in Cartesian plane then to get the velocity we do rise over run which is the slope of the line. rise over run also means that it can be obtained from forming a triangle this also means that phytagoras theorem should be valid. but why ?this is soo confusing i dont understand . is hypotenus not equivalent to the magnitude of change in position? why is the change in position is the final minus initial ( which makes a lot of sense) but i am confuse. i dont even know how to phase my question correctly due to my bad english vocab grammar etc.

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  3. Sep 25, 2016 #2


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    No, that is simply wrong.
    To explain, we need to consider two dimensions at least. If initial position is (xi, yi) and final is (xf, yf) then the distance between them is the hypotenuse of the displacements in the two coordinate directions. The X displacement is xf-xi, for example. So the whole distance moved is √((xf-xi)2+(yf-yi)2).

    By the way, a displacement is a vector, so the above formula gives the magnitude of the displacement. The displacement is (xf-xi,yf-yi).
  4. Sep 25, 2016 #3
    I think the confusion arises from different types of notation (for different use).

    If on a Cartesian graph we start at say (10,10) and end at (13,14) then we can express the displacement as an x,y vector - ##Displacement=(3,4)## , or we can use Pythagoras to express it as a length - ##Displacement=5## .

    Does this help ?
    Last edited: Sep 25, 2016
  5. Sep 26, 2016 #4
    problem that i dont understand is that, for example, if it takes me 10 sec to run from point A to point B.(assume that at point A=0sec , 0m & at point B= 10sec,30m) and do note that i ran in a straight line . if we were to plot this on a Cartesian plane then wouldn't the (displacement) or shortest distance AB follows phytagoras theorem? displacement or shortest distance should be the length of AB as i ran in a straight line right? then why do we calculate the displacement as Xfinal-Xinitial= 30m-0m= 10m . Why not follow phytagoras theory , where
    displacement^2 = (30-0)^2 +( 10-0)^2 ??????
    its very confusing . i ask my lecturer .and he bluntly said he doesnt know. imagine that even someone with a dual master's degree in physic doesnt know why . . .
  6. Sep 26, 2016 #5
    I still don't see the problem...

    If you graph time on the x axis and distance on the y axis, then in the example that you give:

    (0 seconds, 0 minutes) to (10 seconds, 30 minutes) then

    ##Displacement_{time} = +10 seconds##
    ##Displacement_{distance} = +30 metres##
    ##Displacement_{spacetime}= (10s,30m) =\sqrt{100s^2+900m^2}##

    I'm not sure how useful that last one would be... at least until we can convert seconds into metres.
  7. Sep 26, 2016 #6
    hi , the problem is most book are stating that the change of X is equal to the displacement. how can this be? shouldnt the displacement be the length of the straight line from point A to B? shouldnt (30^2)+(0^2)=(displacement^2)? why is displacement equals to final position-initial position(this makes no sense)?
  8. Sep 26, 2016 #7
    The graph you are referring to, has
    x - time
    y - length
    Pythagorean theorem means nothing for basic physics.

    If your graph was instead a map, then
    x - length
    y - length
    Pythagorean theorem is good.
  9. Sep 26, 2016 #8
    wow thank you it has been 3 years i tried to understand this but fail. u are soo awsome i think you are explaining sumthing really crucial for my understanding here, can you please elaborate on that. as u were saying " a map both length" and "graph x time y length"
    why for Pythagorean theorem is useless for a displacement time graph ? what kind of graph that has x as length y also as length? =D please help me it's been three years i still dont understand though i manage to score 87% of my physic final but i feel empty as i only memorize everything like a robot and do substitution based on the formula .
  10. Sep 26, 2016 #9
    Usually they're called "maps" or "charts" or "diagrams".
    Then I most humbly apologize for not giving you the opportunity to expand your knowledge.
  11. Sep 26, 2016 #10
    i spent much tought on the hint u gave me thanks you.. i can now relate all the chapter and everything makes much more sense now.thank you .
  12. Sep 26, 2016 #11
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