# Is this correct regarding the displacement?

by LegendF
Tags: correct, displacement
 P: 29 Hello there, I'm feeling really confused regarding this subject. If I place my zero-level at the water, and the positive direction upwards, that mean that the cliff is at +20m. If I then want to drop something off this cliff, then it is also simply travelling -20m downwards? Is this correct? If I then define positive direction as down instead, does that mean that the cliff is at -20m, and if I drop something from here then it has to travell +20m? If I place my zero-level at the cliff which is 20m instead, and assume that the positive direction is down, then it has to travell 20m to end into the water? If I place my zero-level at the cliff and state that negative direction is down, then I can say that it has to travell -20m? So it's all about what you put your positive and negative direction as?, if you assume that the positive direction is +50m up from the zero-level, then it's also -50m down to the zero-level from the same point? Did I get this correct? Happy New Year! :D
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Hello LegendF! Welcome to PF!
 Quote by LegendF If I place my zero-level at the water, and the positive direction upwards, that mean that the cliff is at +20m. If I then want to drop something off this cliff, then it is also simply travelling -20m downwards? Is this correct?
Yes, except you should omit the word "downwards".
 If I then define positive direction as down instead, does that mean that the cliff is at -20m, and if I drop something from here then it has to travell +20m? If I place my zero-level at the cliff which is 20m instead, and assume that the positive direction is down, then it has to travell 20m to end into the water? If I place my zero-level at the cliff and state that negative direction is down, then I can say that it has to travell -20m?
Yes.

But i think you're confusing travelling with position: if you say it travels 1 metre, that means that the displacement of its finish point is 1 meter more than the displacement of its start point: the zero-level does not matter.

The zero-level is only relevant to position (of one point): for the difference between positions of two points, changing the zero-level obviously makes no difference .
Merry Fishmas and Happy New Year!
 P: 29 Hello there! Firstly, Thank you! :D I've just some questions based on your reponse. Why should I omit the word ''downwards''?, is it because the ''-'' sign already indicates that the stone is falling downwards? However, for example if I wish to calculate the time it takes from a stone to travel from me dropping it from the cliff, until it falls into the water, then I always thought that I have to show respect to the direction of the displacement. For example; The cliff is 30 metres high, if I stand on the cliff and decide to drop my stone wanting to calculate the time of the displacement, then I have to show respect to the directions of the vectors? If I define negative direction as being down, then I have that the acceleration is -9,82m/s, and that the direction of travel is -20m(Displacement). What I was most insecure of is if for example a height is 30m, and I decide to drop something of it, is it then the displacement -30m from the cliff to the water if the stone is going ''downwards''(the word I should omit) :( :p Or did I understand everything wrong? Best regards, LegendF :D
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Is this correct regarding the displacement?

What he said. I just want to be a bit more specific... JIC:
 Quote by LegendF Hello there, I'm feeling really confused regarding this subject.
I suspect the confusion is between displacement and position - as TinyTim said.

 If I place my zero-level at the water, and the positive direction upwards, that mean that the cliff is at +20m. If I then want to drop something off this cliff, then it is also simply travelling -20m downwards? Is this correct?
... the zero-level is just the origin of your coordinate system, it's position is arbitrary.
So if we indicate height (position) by the variable h, then the zero-level is h=0.
The top of the cliff has position h=+20m
When the object falls from h=+20 to h=0, it changes position by ##\Delta h = h_f-h_i=0\text{m}-20\text{m}=-20\text{m}##
so we can say that it has displaced -20m from it's initial position.
Displacement is change in position.

 If I then define positive direction as down instead, does that mean that the cliff is at -20m, and if I drop something from here then it has to travel +20m?
That is correct - we do not have to define "up" to be positive: this is completely arbitrary.
Be careful, however, about the difference between the distance traveled and the displacement.
If, in this last example, you lifted the object to a height h and let it fall back to zero, then it would have traveled a total of 40m, but it's displacement would be 0m because it ended up where it started.

 If I place my zero-level at the cliff which is 20m instead, and assume that the positive direction is down, then it has to travell 20m to end into the water?
Yes.

 If I place my zero-level at the cliff and state that negative direction is down, then I can say that it has to travell -20m?
No.
It still travels a distance of 20m ... it has a displacement of -20m.
The minus sign indicates the direction of the displacement from the start point.

 So it's all about what you put your positive and negative direction as?, if you assume that the positive direction is +50m up from the zero-level, then it's also -50m down to the zero-level from the same point?
The direction of "positive" is arbitrary ... so +10m north is -10m south. You pick the direction to call positive according to what makes the math easier.

But each of your examples either start or finish at the zero position.
Let's define + to be upwards.
If the object started at hi=+50m and finished at hf=+20m, then it's displacement is d=hf-hi=-30m and the distance travelled is d=30m.

Notice that I was being more careful here - the bold-face indicates a vector quantity.
Position and displacement are vectors, so they have a direction as well as a size and we write them as bold-face or put a little arrow over the top (or a tilde underneith).
Distance just has a size - we call it a scalar.

In the above example, the distance traveled is the same as the size of the displacement - this is not always the case.

Also see:
http://hyperphysics.phy-astr.gsu.edu/hbase/posit.html
 P: 29 Hello! I believe I've understood everything, just confirm my post which I wrote some seconds before yours if you may, so I can be sure that I've understood it correctly. :) However isn't my ''confusement'' leaning more towards me confusing the terms distance with the terms position&displacement? Because displacement&position are vectors, while distance is a scalar (absolute value of the vector quantities?) And, a big thank you for helping me out! I was feeling really confused on this subject. Very good explanation! :) Just one thing, I believe you've accidently wrote 20m instead of 30m at this point, in case anyone else feels as confused as I was in the future :p (Then it's displacement is d=hf-hi=-30m and the distance travelled is d=20m.) Regards,
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 However isn't my ''confusement'' leaning more towards me confusing the terms distance with the terms position&displacement? Because displacement&position are vectors, while distance is a scalar (absolute value of the vector quantities?)
This is likely yes.
Well spotted on the typo.

The word "distance" by itself is not always very clear ... the magnitude (absolute value or size) of the displacement vector is a distance, but not all distances are like that.
"The distance travelled", for instance, can be different.

If you run to the other end of a field and back and I stay put, we each have the same displacement - 0 - the magnitude of the displacement is a distance of 0, but that is not the distance you just ran. This is a rich source of confusion.

It's good to iron them out now before you meet velocity and acceleration.
i.e. if you have a negative velocity and a negative acceleration, then you are speeding up.

This is something that can get clearer when you have to deal with more than one dimension.
 P: 29 Very good example. However, my last question. If I'm using a formula am I then using the displacement, or the distance? For example, like I was into, If I try to calculate the time it takes for a stone to fall into the water from a cliff 20m above the water, I would use the formula: y = -1/2gt^2 Is y then the distance, or the displacement?, In other words would I assume that y is 20m as distance, or use the displacement -20m, (if I define negative direction down..) I've always thought of y as a displacement I suppose. If I then defined the negative direction as down, would I then use y as -20m? Thank you once again, that was my last question, to not bother you any further! :P Very good explanation once again! Best regards,
 Homework Sci Advisor HW Helper Thanks P: 12,463 The formulas don't mean anything without a contaxt ... which is why you should always define/list your variables at the start of a problem. Per your example, stone drops of a 20m cliff, need the time to hit bottom. Define: down = +ve initial position y0 = 0 (at top of cliff) final position y1 = +20m (at bottom of cliff) initial time = t0=0 final time = t1=T (y1-y0) would be the displacement. ##\vec y_0 = y_0\hat{\jmath}## would be the initial position vector. (do you know i-j-k notation? ##\hat{\jmath}## just means "in the direction of the positive y-axis) The relation is: $$y_1=y_0+\frac{1}{2}gT^2$$ ... is for postions. Notice how my equation is a little different from yours? The one you wrote only works for a particular circumstance - re: y0=0 and +ve is upwards. But rearrange: $$y_1-y_0=\frac{1}{2}gT^2$$ ... the LHS is the displacement. Of course ##y_0=0## so, in this case, the change in position is the same as the position ... that's why I was concerned that all you earlier examples involved displacements from the origin - it kinda muddies the waters. The strictly correct vector form of the equation would be:$$\vec y = \vec v T + \frac{1}{2}\vec a T^2 \\ \Leftrightarrow (y_1-y_0)\hat{\jmath}=(v_0\hat{\jmath})T+\frac{1}{2}(g\hat{\jmath})T^2$$ ... and we can just cancel out the ##\hat{\jmath}##'s. This is why we can get away with working in magnitudes while we are in 1D. In your case the initial velocity and position are both 0. Note - the ##\vec y## in the first line is the displacement. Take away lesson: the displacement from the origin is the same as the position.
 P: 29 In this case it seem that it doesn't matter if I take the displacement or the position, but when does it then actually matter? I can't really see it infront of me, do you possibly have any example of this problem, because I've realized that I'm a bit unsure if I'm to use the displacement or the position.. :/ Thank you so much, once again. Best regards,
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 In this case it seem that it doesn't matter if I take the displacement or the position, but when does it then actually matter?
It matters when the displacement is not from the origin.

 I can't really see it infront of me, do you possibly have any example of this problem, because I've realized that I'm a bit unsure if I'm to use the displacement or the position.. :/
The issue is that the position vector is not always the same as the displacement.

An object moves from ##\vec p = (0,4)## to ##\vec q =(3,4)## then the displacement is ##\vec d = \vec q - \vec p = (3,0)## - different from it's position vector.

In fact it is fair to define position as a displacement from the origin... i.e. a position is a special case of a displacement. Usually though, it is good practice to think of displacements as being associated with movement.

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