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1) Is position a function of time only or also velocity? Likewise, is velocity a function of time only or also the position?

2) The following are functions of time:
$s(t)$ = distance a particle travels from time $0$ to $t$.
$v(t)$ = velocity of a particle at time $t$.
$a(t)$ = acceleration of a particle at time $t$.

If we want to see how the position of a particle changes with respect to time only, then its velocity must remain constant with time. Likewise, if we want to see how velocity varies with time, then the distance between the former position of the particle and the current position should remain constant with time. Similarly, if we want to see how acceleration varies with time, then the difference between the initial velocity U and final velocity V should remain constant with time. Is this what the above functions of time tell us?

3) If we say, $s(t)$ then I think it implies that everything has to be constant but time. Otherwise, if displacement $s$ is a function of more than time, for example if its a function of both 'time' and 'velocity' then we should write $s(v,t)$. I would like to given another example: $p(y)$ = water pressure at depth $y$ below the surface. Water pressure is given by: $p=ρgh$. Here the density $ρ$ has to be constant if pressure is only the function of depth $y$.

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  • $\begingroup$ Suggestion to post (v3): Replace everywhere the word (and concept) distance with position to focus discussion. $\endgroup$
    – Qmechanic
    Commented Apr 24, 2013 at 13:12

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The answer to this question depends very much on what field you're studying. For instance, in many areas of physics, being time derivatives of position, most would take the velocity and acceleration equations and treat the whole system as a differential equation, then solve for distance as a function of time only. Similarly, they would then differentiate the distance to get a velocity equation as a function of time only.

However, in some areas of study like robotics and certain fields in engineering, velocity may not only vary with time, but it may vary differently according to specific position. Thus, in those circumstances, velocity is made a function of time and position. Also, because the velocity has a different time dependence at every position, the position function becomes dependent on the path traveled. This means that in cases where position/velocity/acceleration are discontinuous and/or path-dependent, both distance and velocity must be functions of one another.

ADD version
Sometimes they're only functions of time, sometimes they're functions of time and each other. Depends on the situation.

Edit
It's true that in many cases where velocity is taken as a function of position that it CAN be written as just a function of time; however, this can be very impractical. So, the fact remains that in those circumstances we DO write them as functions of position and time.

Edit 2
Velocity and distance can also be functions of more than just time. Temperature and mass are just some examples.

Edit 3
To answer the new part of your question, no this does not imply that anything is constant. This just means that these three things are functions of time. However, you do not need to hold velocity constant to see how position changes with time. Rather $v(t)$ should be the time derivative of $s(t)$ and similarly for velocity -> acceleration.

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  • $\begingroup$ But, if we say, $s(t)$ then I think it implies that everything has to be constant but time. Otherwise, if displacement $s$ is a function of more than time, for example if its a function of both 'time' and 'velocity' then we should write $s(v, t)$. I would like to given another example: $p(y)$ = water pressure at depth $y$ below the surface. Water pressure is given by: $p = \rho gh$. Here the density $\rho$ has to be constant if pressure is only the function of depth $y$. $\endgroup$ Commented Apr 25, 2013 at 8:41
  • $\begingroup$ That would be true if v weren't a function of time as well. If you have $s(v(t),t)$, it can be written just as $s(t)$. Also, it isn't necessary for v(t) to even be in the function of s, which would mean whether or not it changes over time is irrelevant. $\endgroup$
    – Jim
    Commented Apr 25, 2013 at 17:07
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I can't understand why you're asking "Is distance, velocity a function of time?". The question is quite ambiguous because, when we define velocity, acceleration, or jerk in classical mechanics, we're quite sure that we're taking the time derivative of the predecessor. For instance, if you require velocity, then you're taking the time derivative of distance.

$$v(t)=\frac{dx}{dt}=\lim_{\delta t \to 0}\frac{x(t+\delta t)-x(t)}{\delta t}$$

The positions should necessarily be a function of time in order for taking the time derivative. This expression for average velocity simply means that we're putting some digits $\delta t$ to the initial state (position) of the system and determine how the system responds to it (i.e) how it moves (whether it moves or not) along the spatial axis. If it has some finite velocity, its position changes to some other value corresponding to the added time period. Finally, dividing it with the same time period which is to predict how the position is changing over time.

The expression says how the position has changed (numerator) within some period of time (denominator). If $x$ is a function of velocity, then we can say that we multiply it with $t$ and then integrate over a certain limits which you wanna predict. You're somehow arriving to the point that it is a $f(t)$.

What's my point is that units should be conserved when dealing with physical parameters. Whatever you play around (using math) with those expressions, be sure that you arrive at the final conclusion that the velocity is always $m/s$ (in SI)...


then its velocity must remain constant. [...] the distance... ...should remain constant [...] the difference between the velocities should remain constant

There is nothing that the particle should or must follow some trajectory or the laws that we define. We just approximate our current laws accordingly to its activity. So, the answer - It's not necessary..!

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  • $\begingroup$ I've expanded my question.. Please reread it! $\endgroup$ Commented Apr 24, 2013 at 13:09
  • $\begingroup$ So in Newtonian mechanics we assume that position is always a function of time? So we can differentiate and get velocity? $\endgroup$ Commented Jun 9, 2020 at 23:22
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Position is only a function of time. Velocity, acceleration, and jerk, are 1st,2nd, and 3rd order time derivatives of position (this is the number of times you have to take the derivative). Velocity does not have to remain constant, because velocity and position are distinct functions of time, and can be plotted separately.

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