# motion distance-time graph

It has been a while since I have touched on physics, but reading the material presented in my course, I recall displacement and distance being two different things. Distance is the total length something travels from A to B and back to A, whereas displacement would be 0 when returning to its origin. So, if A to B were 2 metres, back to A would be 2 metres, a total of 4 metres. The displacement would be zero.

See the attachment. The writers might be confused about which they are referring to, or I might be confused.

They have referred to coming back to the starting origin as the distance being zero and displacement in brackets, suggesting they're the same.
(edited 2 months ago)
I guess in their graph theyre measuring distance from starting position to the object rather than distance travelled by object? which is pretty weird - I agree with you in that it would make much more sense to use displacement rather than distance.
KingRich
It has been a while since I have touched on physics, but reading the material presented in my course, I recall displacement and distance being two different things. Distance is the total length something travels from A to B and back to A, whereas displacement would be 0 when returning to its origin.

From https://en.wikipedia.org/wiki/Distance
Distance is a numerical or occasionally qualitative measurement of how far apart objects or points are

Although this isn't quite as informative as displacement, it is still consistent with the graph given.

If they said "distance travelled" I feel you'd have more of a case.

I know there seems to be a recent trend to insist it should be displacement, but

indicates distance/time is still the most common terminology. [And in particular, from other things you've posted about your course, I'm guessing it will tend to use older terminolgy].
Original post by dfranklin
From https://en.wikipedia.org/wiki/Distance
Although this isn't quite as informative as displacement, it is still consistent with the graph given.
If they said "distance travelled" I feel you'd have more of a case.
I know there seems to be a recent trend to insist it should be displacement, but
indicates distance/time is still the most common terminology. [And in particular, from other things you've posted about your course, I'm guessing it will tend to use older terminolgy].

I see. So, distance/time is becoming more used in textbooks for teachings rather than displacement/time?

Moreover, what is more confusing is that it talks about distance/time graph but then jumps to velocity/time graph.

It’s going from a scalar quantity graph to a vector quantity graph.

So, it’s stating that I find velocity from the distance/time graph, referring to the gradient but as I recall this is only true to a displacement/time graph.

If we’re talking distance/time, we’d be considering speed and not velocity.

Also, the area under the line on the velocity/time graph, I recall being displacement but they’re referring it as distance?

My textbook teaching from Cambridge are contradicting this course material.

From my book

From the course.

After such, it then jumps to SUVAT formulas lol

The reason why I recall this is because I remember integrating to find displacement under the graph.

They call this graph distance-time graph but then refer to the journey traveled as displacement.

In this short text, they're not referring to velocity and speed to be the same.

It isn't just me, right? It seems the material itself has no idea.
(edited 2 months ago)
Original post by KingRich
I see. So, distance/time is becoming more used in text books for teachings rather than displacement/time?
Moreover, what is more confusing is that it talks about distance/time graph but then jumps to velocity/time graph.
It’s going from a scalar quantity graph to a vector quantity graph.
So, it’s stating that I find velocity from the distance/time graph, referring to the gradient but as I recall this is only true to a displacement/time graph.
If we’re talking distance/time, we’d be considering speed and not velocity.
Also, the area under the line on the velocity/time graph, I recall being displacement but they’re referring it as distance?
My textbook teaching from Cambridge are contradicting this course material.

You seem to be overthinking it a bit in this case. Distance from the origin (a point) is different from distance travelled and the former is the same as displacement when displacement is not negative (your scalar/vector distinction and you have to choose your positive direction for the later), which is the case here. As Dfranklin notes, they have been frequently interchanged and I wouldnt say that either are more/less commonly used these days, just that distance a more familiar term than displacement. In certain cases, distinguishing between them is important, but not here.

Note that if there was a second "bump" after the first one, then youd not be able to say whether it was in the same direction as the first one, though if the y axis was displacement then youd be able to determine that.

Mixing speed and velocity here may be more problematic as the speed of the last section would be 5m/s (say), but the velocity would be -5m/s (with the usual reason about which direction is positive). However, for simple trajectories inferring that displacement is the same as distance from the origin and hence using that to determine velocity is common ... for simple trajectories.
Original post by mqb2766
You seem to be overthinking it a bit in this case. Distance from the origin (a point) is different from distance travelled and the former is the same as displacement when displacement is not negative (your scalar/vector distinction and you have to choose your positive direction for the later), which is the case here. As Dfranklin notes, they have been frequently interchanged and I wouldnt say that either are more/less commonly used these days, just that distance a more familiar term than displacement. In certain cases, distinguishing between them is important, but not here.
Note that if there was a second "bump" after the first one, then youd not be able to say whether it was in the same direction as the first one, though if the y axis was displacement then youd be able to determine that.
Mixing speed and velocity here may be more problematic as the speed of the last section would be 5m/s (say), but the velocity would be -5m/s (with the usual reason about which direction is positive). However, for simple trajectories inferring that displacement is the same as distance from the origin and hence using that to determine velocity is common ... for simple trajectories.

I see.

I’m more familiar with displacement when dealing with suvat and that’s likely why I’m a little confused.

In the case that the velocity time graph had a curve though, the use of distance=speed x time would no longer apply here like they suggest to use but as their example shows steady acceleration, what they’re saying isn’t technically wrong.
Original post by kingrich
I see.
I’m more familiar with displacement when dealing with suvat and that’s likely why I’m a little confused.
In the case that the velocity time graph had a curve though, the use of distance=speed x time would no longer apply here like they suggest to use but as their example shows steady acceleration, what they’re saying isn’t technically wrong.

The distance=speed*time is the very simple case kids are taught. "Time" here is really a time interval (as its usually assumed to start at time 0) and hence "distance" is the distance moved from the initial point and "speed" is assumed positive constant (zero acceleration). So when you use the suvat
s = ut + 1/2 at^2
the only thing thats different (1/2 at^2 would be zero) is the sign of u and s (velocity and displacement) to denote direction. If theyre positive, then its equivalent to the distance=speed*time, as would s=(u+v)/2 t when the u=v and are positive as the average velocity is the constant speed.

I presume youre talking about the epearl, blue shaded area example? As velocity is always positive, you could call it speed and hence displacement or distance are equivalent. Indeed as velocity is positiv, distance from a point and distance travelled are also equivalent (this is not the case in the OP, due to the last section with negative velocity where the direction of motion is reversed). The distance=speed*time is the area of the dark blue block, and the distance or displacement of the first phase (light blue triangle) is similar (s = (u+v)/2 t, or displacement = average velocity * time).

As before, the terms can be used interchangeably in simple situations and Im sure the writers do this to make it sound more friendly, but obviously you need to understand them.
(edited 2 months ago)
KingRing
I see. So, distance/time is becoming more used in textbooks for teachings rather than displacement/time?
No, the opposite. When I was at school/university, I'd say "distance/time" was far more common usage than displacement.

If I'm honest, I find the distinctions fairly pointless pedantry. If I say a rocket is traveling at a speed of -10m/s, everyone knows what that means. (Or at least they are no more confused than if I say it's traveling at a velocity of -10m/s). Similarly for distance. This seems particularly so when you're saying things like "velocity is a vector, speed is a scalar", but your handling of the problem means you're treating velocity as a one dimensional quantity. In real problems, you're far more likely to get bitten by that than by confusions about what a negative speed or distance means.
Original post by DFranklin
No, the opposite. When I was at school/university, I'd say "distance/time" was far more common usage than displacement.
If I'm honest, I find the distinctions fairly pointless pedantry. If I say a rocket is traveling at a speed of -10m/s, everyone knows what that means. (Or at least they are no more confused than if I say it's traveling at a velocity of -10m/s). Similarly for distance. This seems particularly so when you're saying things like "velocity is a vector, speed is a scalar", but your handling of the problem means you're treating velocity as a one dimensional quantity. In real problems, you're far more likely to get bitten by that than by confusions about what a negative speed or distance means.

I understand what you mean. I would have considered velocity and speed as interchangeable in reality. But as I’m still learning the mechanics side of it, when I learn velocity as being different to speed and same with distance and displacement, it leaves me with questions, more than answers when I see them used as the same within a context and extremely confused.
Original post by KingRich
I understand what you mean. I would have considered velocity and speed as interchangeable in reality. But as I’m still learning the mechanics side of it, when I learn velocity as being different to speed and same with distance and displacement, it leaves me with questions, more than answers when I see them used as the same within a context and extremely confused.

Without being prescriptive, just think about whats the same / whats different / what the motion actually is for some simple cases like

When the velocity is always positive. How is it related to speed, how are displacement, distance from, distance travelled curves related?

For the bump in the OP, how is speed and velocity related? How is the distance from curve related to displacement and distance travelled curves?

How do you (graphically) calculate the distance travelled and distance from from a velocity curve that includes both positive and negative sections/areas?

Really you should understand clearly the (calculus/suvat) relationships for
acceleration <-> velocity <-> displacement
then pick examples like above to think about how it relates to the simpler concepts of speed, distance travelled, ...