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I wrote the equation for the time it takes for an object to travel a specific distance at a specific speed. This equation is defined as $t(s) = \frac cs$ where $d$ is a constant for the distance the object will travel where $s$ is the speed at which the object is traveling.

Likewise, this is a simple rational function where the faster the object is traveling, the less time it will take for that object to reach the distance defined by $d$. However, I took the indefinite integral of this function which was equal to $d\ln|s|+C$.

I was wondering what the interpretation of this integral would be in regard to physics. I believe it could be the sum of the time that the object is moving at that specific speed, but I'm not sure if there is a physical representation of this integral. Also, sorry for my inexperience in MathJax.

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  • $\begingroup$ The title talks about integrating a "function of time", but in the body you integrate a function of distance $s$. The interpretation of the integral in "physical terms" is possible but will depend on assumptions like the number of dimensions. $\endgroup$
    – hardmath
    Oct 18, 2018 at 2:10
  • $\begingroup$ @hardmath could you explain what you mean by assumptions about the number of dimensions? $\endgroup$ Oct 18, 2018 at 15:18
  • $\begingroup$ Your function depends on a one-dimensional variable "distance" $s$. Starting (apparently) with the rational function $1/s$ you obtain by integration $\ln |s| + C$. Such a result could be connected to the Newtonian potential in dimension $d=2$. $\endgroup$
    – hardmath
    Oct 18, 2018 at 15:34
  • $\begingroup$ @hardmath $s$ is the speed, not the distance. $c$ is the distance. $\endgroup$ Oct 18, 2018 at 15:38

2 Answers 2

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I was wondering what the interpretation of this integral would be in regard to physics.

I don't think your integral has any physical significance. This is because when you use the equation $t=\frac cs$ where $t$ is the time it takes to travel a distance of $c$ at a speed $s$, you are already assuming a constant $s$. Therefore, if you decide to integrate this equation with respect to $s$, you are considering a varying speed mathematically, which then takes you out of the physical interpretation of the equation.

I believe it could be the sum of the time that the object is moving at that specific speed

If you are integrating with respect to the speed $s$, then your integral has units of distance, not of time. Therefore this cannot be the interpretation of the integral, even if it did have physical significance.

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We have

$$\int_{s_0}^{s} t(s)\,ds=\int_{t_0}^{t} t\frac{ds(t)}{dt}dt$$

and by parts

$$\int_{t_0}^{t} t\frac{ds(t)}{dt}dt=\left.ts\right|_{t_0}^t-\int_{t_0}^t s(t)\,dt=ts(t)-t_0s_0-x(t)+x_0$$ where $x$ is the distance traversed.

The terms have the dimension of a distance. The first one doesn't have a common interpretation. It can be seen as "the traversed distance, had the speed kept the current value all the time".

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  • $\begingroup$ If you want $s$ to depend on time $t$, then technically you would want to define your function differently as something like $T(s(t))$ right? Since the OP is specifically interested in a set time interval $T$ it takes to travel some distance at some speed. $\endgroup$ Oct 18, 2018 at 17:45
  • $\begingroup$ @AaronStevens: $T(s(t))=t$. $\endgroup$
    – user65203
    Oct 18, 2018 at 17:53
  • $\begingroup$ If the speed is time dependent then it can't be... If that were the case then by changing our speed we would be changing time itself. You have to distinguish between time in a general sense $t$ and the time interval in question $T$. $\endgroup$ Oct 18, 2018 at 18:00
  • $\begingroup$ @AaronStevens: no, there is no $T$. Just a functional relation between $t$ and $s$. $\endgroup$
    – user65203
    Oct 18, 2018 at 18:12
  • $\begingroup$ Find. You can do that mathematically, but physically it doesn't make sense. $\endgroup$ Oct 18, 2018 at 18:21

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