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

In summary, the difference between distance and displacement is that displacement is a vector and distance is a scalar. Displacement is the magnitude of the change in position, while distance is the length of the path between two points.
  • #1
natsuki91xp
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Homework Statement


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 isn't displacement the shortest length between two point ?because Xinitial - Xfinal couldn't 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

Homework 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 physics 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.

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 don't 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 don't even know how to phase my question correctly due to my bad english vocab grammar etc.[/B]
 

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  • #2
natsuki91xp said:
where Xi^2 + Xf^2=displacement^2
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).
 
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  • #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 ?
 
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  • #4
problem that i don't 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 doesn't know. imagine that even someone with a dual master's degree in physics doesn't know why . . .
 
  • #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.
 
  • #6
hmmm27 said:
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.

hi , the problem is most book are stating that the change of X is equal to the displacement. how can this be? shouldn't the displacement be the length of the straight line from point A to B? shouldn't (30^2)+(0^2)=(displacement^2)? why is displacement equals to final position-initial position(this makes no sense)?
 
  • #7
natsuki91xp said:
most book are stating that the change of X is equal to the displacement.

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.
 
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  • #8
hmmm27 said:
The graph you are referring to, has
x - time
y - length
Pythagorean theorem means nothing.

If your graph was instead a map, then
x - length
y - length
Pythagorean theorem is good.

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 don't understand though i manage to score 87% of my physics final but i feel empty as i only memorize everything like a robot and do substitution based on the formula .
 
  • #9
natsuki91xp said:
what kind of graph that has x as length y also as length?
Usually they're called "maps" or "charts" or "diagrams".
natsuki91xp said:
i feel empty as i only memorize everything like a robot and do substitution based on the formula
Then I most humbly apologize for not giving you the opportunity to expand your knowledge.
 
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  • #10
hmmm27 said:
Usually they're called "maps" or "charts" or "diagrams".
Then I most humbly apologize for not giving you the opportunity to expand your knowledge.
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 .
 
  • #11
:smile:
 

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

1. What is displacement?

Displacement is a measure of an object's change in position, or the distance between its final and initial positions.

2. How is displacement calculated?

Displacement is calculated by subtracting an object's initial position from its final position. The result is the shortest path between the two positions.

3. Does displacement always equal the shortest length of position?

Yes, displacement always equals the shortest length of position between an object's initial and final positions. This is because displacement is a vector quantity, meaning it includes both magnitude and direction, and the shortest path between two points is always a straight line.

4. Can displacement be negative?

Yes, displacement can be negative. This occurs when an object moves in the opposite direction of its initial position. The negative sign indicates a change in direction, but the displacement value is still the shortest length of position between the initial and final positions.

5. How is displacement different from distance?

Displacement and distance are often confused, but they are different concepts. Distance is a scalar quantity that measures the total length of the path an object travels, while displacement is a vector quantity that only measures the change in position from an object's initial to final position.

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