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Is this correct regarding the displacement? 
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#1
Jan214, 04:15 PM

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Hello there, I'm feeling really confused regarding this subject.
If I place my zerolevel 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 zerolevel 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 zerolevel 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 zerolevel, then it's also 50m down to the zerolevel from the same point? Did I get this correct? Happy New Year! :D 


#2
Jan214, 04:48 PM

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Hello LegendF! Welcome to PF!
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 zerolevel does not matter. The zerolevel is only relevant to position (of one point): for the difference between positions of two points, changing the zerolevel obviously makes no difference . Merry Fishmas and Happy New Year! 


#3
Jan214, 05:07 PM

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 


#4
Jan214, 05:09 PM

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Is this correct regarding the displacement?
What he said. I just want to be a bit more specific... JIC:
So if we indicate height (position) by the variable h, then the zerolevel 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_fh_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. 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. 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. But each of your examples either start or finish at the zero position. Let's define + to be upwards. If the object started at h_{i}=+50m and finished at h_{f}=+20m, then it's displacement is d=h_{f}h_{i}=30m and the distance travelled is d=30m. Notice that I was being more careful here  the boldface indicates a vector quantity. Position and displacement are vectors, so they have a direction as well as a size and we write them as boldface 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.phyastr.gsu.edu/hbase/posit.html 


#5
Jan214, 05:26 PM

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=hfhi=30m and the distance travelled is d=20m.) Regards, 


#6
Jan214, 05:50 PM

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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. 


#7
Jan214, 05:57 PM

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, 


#8
Jan214, 06:26 PM

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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 (y1y0)<down> would be the displacement. ##\vec y_0 = y_0\hat{\jmath}## would be the initial position vector. (do you know ijk notation? ##\hat{\jmath}## just means "in the direction of the positive yaxis) 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: $$y_1y_0=\frac{1}{2}gT^2$$ ... 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:$$\vec y = \vec v T + \frac{1}{2}\vec a T^2 \\ \Leftrightarrow (y_1y_0)\hat{\jmath}=(v_0\hat{\jmath})T+\frac{1}{2}(g\hat{\jmath})T^2$$ ... 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. 


#9
Jan214, 06:40 PM

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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?
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, 


#10
Jan214, 06:50 PM

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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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