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Let the position of a particle be given by the vector:

$$\vec{r}=(2t-t^2) \vec{e_{x}} +(8+6t)\vec{e_{y}}$$

where $t$ is the time, starting from $0$.

When solving for the distance the particle has traveled from $t=[0.0 ; 2.0] s$, my teacher said it was equal to the displacement since:

$$s= \parallel\vec{r}\parallel = \sqrt{(\Delta x^2+\Delta y^2)}$$

Upon solving that would give: $$\vec{r_{0}}=(0) \vec{e_{x}} +(8)\vec{e_{y}}=8\vec{e_{y}}$$ $$\vec{r_{2}}=(2\times2-2^2) \vec{e_{x}} +(8+6\times 2)\vec{e_{y}}=(0) \vec{e_{x}} +(20)\vec{e_{y}}=20\vec{e_{y}}$$ $$\Delta\vec{r}=(0-0) \vec{e_{x}} +(20-8)\vec{e_{y}}=0 \vec{e_{x}} +12\vec{e_{y}}$$ $$\mathbf{s=\sqrt{0^2+12^2}=12} meters$$

However, using this method, aren't we forgetting to have in count the distance the particle has traveled in the x-axis during this time?

I also drew the function of the distance traveled in the x-axis by the particle according to time : $$\Delta x=-t^2-2t$$ And concluded that it is a parabola with a vertice on the point of coordinates (1,1) and intersections on the points (0,0) and (2,0). I then assumed that the distance the particle traveled in the xx-axis was 2 meters (1 to the left + 1 to the right).

That would give us a total distance of 14 meters instead of 12 meters.

My question is whether any of these solutions are right or whether we need integrals to solve for the distance?

My apologies if this is not a relevant question, but I am still new to movements in 2D.

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  • $\begingroup$ from the parabola you can see that the movement on x between 0 and 2 is forward and then backwards, so it returns to x=0 when x=2. So displacement has not changed in the x direction. a different think is the length of the path you traveled, that can be calculated integrating tutorial.math.lamar.edu/classes/calcii/arclength.aspx $\endgroup$ – Wolphram jonny Sep 28 at 16:52
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    $\begingroup$ Distance and displacement are not the same thing. Which are you wanting to calculate? $\endgroup$ – Not_Einstein Sep 28 at 19:22
  • $\begingroup$ @Not_Einstein Trying to solve for distance. $\endgroup$ – Diogo Sousa Sep 28 at 20:14
  • $\begingroup$ Your teacher may be defining distance as the displacement. Someone else might define distance differently. A non-ambiguous term for something other than displacement would be path length, which would be the sum of all the small steps along the path. $\endgroup$ – Bill N Sep 30 at 13:48
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The displacement of the particle from $t=0$ to $t=2$ is indeed $12$ units because the particle starts at $(0,8)$ and ends at $(0,20)$.

But, as you say, the distance travelled by the particle between $t=0$ and $t=2$ is greater than $12$ units because it does not travel in a straight line between the start and end points, but follows a parabola with its vertex at $(1,14)$.

The find the distance travelled you would have to find the arc length of the parabola between $(0,8)$ and $(0,20)$. This will be more than $12$ units but less than $14$ units.

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  • $\begingroup$ Thanks! I think I ended up figuring it out using Wolfram (wolframalpha.com/input/…). The result was ~12,2187. (I don't think so because it is a parabola), but is there any way we can solve it without using integrals? $\endgroup$ – Diogo Sousa Sep 29 at 12:57
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There is a homework rule that means I cannot give you a worked answer. However as you ask a conceptual question and are asking for clarification this is how it works.

You have some path the object travels on.

You work out the position at each time you are given. Hint : have you done this correctly ?

The distance between these two positions is what you are looking for.

Your confusion is :

However, using this method, aren't we forgetting to have in count the distance the particle has traveled in the xx-axis during this time?

This is already accounted for in the equation for position. It has a time dependency.

As a maybe useful exercise, try and work out what initial velocity, position and acceleration that function represents. It may help clarify what is going on. You can find these using differentiation.

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  • $\begingroup$ If it helps, this wans't homework. In class we worked out that the correct answer was 12 meters, however the only thing that I don't understand is how the displacement and the distance traveled can be the same in this scenario. As you suggested I tried to work out the position of the particle in each moment (t=0 -> (0,0) ; t=1s -> (1,14); t=2s -> (0,20)) and I concluded that it had to travel atleast 2 meters in the x-axis apart from the 12 meters it travelled in the y-axis. I also know that the initial velocity is 2 and 6 (for x, y), accelaration=-2 (x) and position=8 (y). $\endgroup$ – Diogo Sousa Sep 28 at 17:21
  • $\begingroup$ @DiogoSousa The definition of homework on this site includes self-study. We give conceptual help even to homework problems but only in exceptional circumstances are we supposed to give worked or detailed answers to these questions. In general people are expected find e.g. duplicate or similar questions on this site and elsewhere rather than have what will essentially be very similar questions being repeated answered. $\endgroup$ – StephenG Sep 28 at 17:30
  • $\begingroup$ I used Wolfram to see the path of the particle. Is the arc length of parametric curve equal to the distance travelled by the particle? imgur.com/a/NdtRDIn $\endgroup$ – Diogo Sousa Sep 28 at 17:42
  • $\begingroup$ @DiogoSousa This depends on your definition of distance traveled. If I go from point A to point B through point C, it is perfectly reasonable to say that the distance traveled is just the distance between A and B, but equally valid to say it is the length of the path I followed via C. I would suggest you play around with the equations to see what the different approaches mean. $\endgroup$ – StephenG Sep 28 at 17:54
  • $\begingroup$ I consider the first scenario to be the displacement and the second to be the distance travelled by the particle. Nevertheless, now I don't understand why Wolfram gives the path lenght to be ~12,22 meters and not 14. $\endgroup$ – Diogo Sousa Sep 28 at 18:01

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