Equation of a basic velocity-time graph

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The discussion centers on determining an equation for a velocity-time graph that describes an object's motion involving acceleration, constant velocity, and deceleration. Participants clarify that the graph is piecewise continuous, requiring separate equations for each time interval rather than a single function. Suggestions include using Fourier series or special functions to approximate the graph's shape, but a straightforward single equation is deemed impossible. The conversation highlights the complexities of representing such motion mathematically, particularly regarding continuity and discontinuities in acceleration. Overall, the consensus is that while approximations can be made, a single equation cannot accurately represent the entire graph.
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Okay, so say we had the following situation: An object accelerates from rest to 3 m/s in 1 second. It then maintains this velocity for 2 seconds before decelerating to a velocity of -3 m/s over the span of 2s. It maintains this velocity for a further 2 seconds. The object then accelerates back up to a velocity of 3m/s in 2 seconds, which it then travels constantly at for a final second.

I want to know if it would be possible to come up with an equation which would sketch a graph which would look like this. I would prefer the equation to be this exact graph, but an equation which sketches the same general shape of this graph would also be fine. Also, if it is possible to do, does the graph which follows this general shape have a specific name?

Thank you all for your time.
 
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Hi,

This looks so much like homework that I invite you to fill in more of the template: (reason: see guidelines)

1. Homework Statement
See above. this part is OK.​
2. Homework Equations
See e.g. in your textbook, or else here
3. The Attempt at a Solution
You don't want to leave this empty: it enables us to assist at an effective level.​
 
Last edited:
BvU said:
Hi,

This looks so much like hoework that I invite you to fill in more of the template: (reason: see guidelines)

1. Homework Statement
See above. this part is OK.​
2. Homework Equations
See e.g. in your textbook, or else here
3. The Attempt at a Solution
You don't want to leave this empty: it enables us to assist at an effective level.​

It wasn't a homework question. I am just wondering if it would be possible to determine the equation of this graph or not.
 
Yes it is. A straightforward application of the equations in the link under 2.
I do make the assumption that the various accelerations are constants during the time they act.
 
BvU said:
Yes it is. A straightforward application of the equations in the link under 2.
I do make the assumption that the various accelerations are constants during the time they act.

I'm sorry, I don't think I made myself clear. I didn't mean I wanted to use the physics formulas to solve for unknowns in the equation; I meant I wanted to know if you could represent the entire graph using one equation (like a base parabola can be expressed as y=x^2, or variants of a cubic can be represented as y=x^3, y=x^3 + x^2 - 3x, etc)
 
No. You can't do it with a single function like this, because the velocity vs time is "piecewise" continuous, and you need a separate equation for each time interval.
 
Bill_Nye_Fan said:
I'm sorry, I don't think I made myself clear. I didn't mean I wanted to use the physics formulas to solve for unknowns in the equation; I meant I wanted to know if you could represent the entire graph using one equation (like a base parabola can be expressed as y=x^2, or variants of a cubic can be represented as y=x^3, y=x^3 + x^2 - 3x, etc)
I see. That would end up as some infinitely long series: If you change from e.g. y = x2 to y = 8 - (x-4)2 at x =2 there is a discontinuity in the second derivative that makes things dificult.
You could try a power series or a Fourier series or some other series and break off at some point, but may end up with the equivalent of the Gibbs phenomenon (which google) at the points where the acceleration changes.

Contrary to what Chet concludes, I would venture that the velocity is continuous, but the acceleration is not (that is piecewise continuous).

:smile:
 
Here's a function for the acceleration at time t:

##a(t) = 3cos(\lfloor{\frac{t+1}{2}} \rfloor \frac{\pi}{2})##
 
And, for velocity:

##v(t) = 3[cos(\lfloor{\frac{t+1}{2}} \rfloor \frac{\pi}{2})(t - 2\lfloor{\frac{t+1}{2}} \rfloor) + sin(\lfloor{\frac{t+1}{2}} \rfloor \frac{\pi}{2})]##
 
  • #10
Bill_Nye_Fan said:
I'm sorry, I don't think I made myself clear. I didn't mean I wanted to use the physics formulas to solve for unknowns in the equation; I meant I wanted to know if you could represent the entire graph using one equation (like a base parabola can be expressed as y=x^2, or variants of a cubic can be represented as y=x^3, y=x^3 + x^2 - 3x, etc)
It does not matter what you want to do with the equations. The will be the same either way.
Assuming that the acceleration is constant for the portions with changing speed, there is only one general equation for the velocity:
v(t)=vo+a*t.
The values of vo and a will be different for different portions of the motion described.

Or you can use perok's trick with the integer part function.
 
  • #11
Coming back to this series approximation: your function looks like a cosine, so a Fourier expansion should be appropriate. Depending on how accurately you want to reproduce the function, you need a good number of terms. I played with excel Fourier analysis and found that coefficients for ##\cos(nx)## with n = 0,1,3,5,7,9,11 namely

4.5 -4.37861 -0.16217 0.035028 0.012765 -0.00601 -0.00329
give a maximum error of 0.02 (function range 0-9)

The Gibbs phenomenon isn't distinguishable, probably thanks to the first derivative being continuous :wink:

Fourier1.jpg
 
  • #12
What do you mean by "actual"?
His velocity function does not look like this.
It has flat portions at both maximum and minimum velocity.
 
  • #13
His velocity function does not look like this.
I agree, I assumed the poster was asking for the position function.

Fourier1a.jpg
 
  • #14
Chestermiller said:
No. You can't do it with a single function like this, because the velocity vs time is "piecewise" continuous, and you need a separate equation for each time interval.
Even though Bill's v-t graph is full of discontinuities, it could be written using special functions which are discontinuous by nature. Such functions are often used to write equations for shear and bending moment functions for loaded beams, using the singularity function method or the very similar Macaulay's method.

https://en.wikipedia.org/wiki/Macaulay's_method

https://en.wikipedia.org/wiki/Singularity_function

The method could be extended to any situation where piecewise functions would normally be used.
 
  • #15
I only see discontinuities in the a(t) graph - must admit the excel picture doesn't do them justice (took steps of 1/16 sec to get 128 steps from 0 to 8. The 128 is inspired by a queer restriction in excel Fourier analysis: it wants powers of 2).
 
  • #16
Curious to hear from Bill if such an approach satisfies his needs ...
Bill_Nye_Fan said:
an equation which sketches the same general shape of this graph
 
  • #17
No i don't think that the locus you are telling would have single figure which has specific equation, but it would be mixture of two locus which we already know, parabola and straight line. a single equation is not possible. I can explain the reason that generally in a locus which as by you in an equation, we apply certain condition on the motion of moving point, generally(x,y) like we apply in circle thet it has to always have a fixed distance from a fixed point. but the graph you are telling we cannot apply a single condition.

Hope you understand !
 
  • #18
In the OP there is no mention of a graph of position versus time. It's all about velocity.
All these mentions of parabolas may be confusing. The graph he is talking about has just straight lines.
 
  • #19
nasu said:
In the OP there is no mention of a graph of position versus time. It's all about velocity.
All these mentions of parabolas may be confusing. The graph he is talking about has just straight lines.
sorry i thought it was position-time graph. Still I think what I said about the locus, I was right.
 
  • #20
I did not refer specifically to your post. There are several post with parabolas. :)
But it is not very clear what the OP actually expects. He did not come back for a while.
 
  • #21
Not easily discouraged: If Bill maintains radiosilence we can philosophize ad infinitum about any of the functions as much as we like :smile:.

How about describing the function in a single expression as a convolution ? This time I'll target the velocity function (reddish in post #13) and I think it's the autoconvolution of the blue guy. Which itself is a convolution of a delta function with alternating sign ( ##(-1)^n\, \delta(t-2n) ## or something like that ) and a square wave.
 
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