Is displacement the sum of speed times time?

[PrudensOptimus]

Hi,

Is the exact displacement = sum of ( speed * time ) ?

thanks.

 

[whozum]

Exact displacement is the area between a velocity curve and the x-axis.

 

[dextercioby]

It's not a sum,but an integral...

$$\int d\vec{r} =\int \vec{v} \ dt$$

Choose two moments of time $t_{1} < t_{2}$ and u'll find

$$\vec{r}\left(t_{2}\right)-\vec{r}\left(t_{1}\right) =\int_{t_{1}}^{t_{2}} \vec{v} \ dt$$

Daniel.

 

[PrudensOptimus]

integral is sum...

 

[dextercioby]

No.The integral is a limit of a Riemann/Darboux/... sum...

Daniel.

 

[PrudensOptimus]

i forgot to mention that the time is very very very small... it is in measured in miliseconds

 

[dextercioby]

The process of taking a limit is independent of the unit chosen for a physical quantity...

Daniel.

 

[PrudensOptimus]

so how do I make a good estimation of the area under the curve given the speed and time in miliseconds?

 

[dextercioby]

By evaluating that integral...Formulate the problem with its original text.Then we'll see what it needs to be done.

Daniel.

 

[kleinwolf]

Just try to apply a basic discrete integration scheme, like rectangles :

$$d=\sum_{i=1}^n v_i*\Delta t_i$$

I suppose in your case, the intervals $$\Delta t_i$$ are all the same.
You can also use trapezoidal scheme.

But you can apply better integration scheme, like Simpson (2n order interpolation), or even higher splines stuff, aso...

 

[PrudensOptimus]

what is the Simpson's method?

 

[kleinwolf]

I think in the Simpson method you interpolate 3 points with a parabola and integrate the obtained curve. This is equivalent to the ponderation : 2/3*rectangle+1/3*trapezes and gives :

$$d=\sum_{i=1}^{N-1}(v(t_{i+1})+4v(\frac{t_i+t_{i+1}}{2})+v(t_{i})) \frac{\Delta t_i}{6}$$

It's of double order than rectangles...

 

[whozum]

The easiest way is to find a fit to your curve, integrate that wrt time, and evaluate your endpoints. Best estimation you could get.

 

[Data]

If you are given intervals in time and speeds that remain constant (piecewise, presumably) over those intervals, then taking a sum over all the intervals of the time on the interval times the speed on the interval will give you the exact displacement, yes.

 

[PrudensOptimus]

If you are given intervals in time and speeds that remain constant (piecewise, presumably) over those intervals, then taking a sum over all the intervals of the time on the interval times the speed on the interval will give you the exact displacement, yes.

Thank you.