# Why is the rate of change of velocity taken using time and not distance? [closed]

Why is the change in velocity taken with respect to time, why not distance? I am confused, because it makes sense - as the velocity increases, the distance increases or as the velocity decreases, distance covered is decreased. Please don't use calculus to prove, I'd much rather like an explanation.

## closed as unclear what you're asking by lemon, Jon Custer, honeste_vivere, Wolpertinger, M. EnnsAug 18 '17 at 18:22

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• "as the velocity increases, the distance increases" - only if the force applied is constant and co-linear with velocity, which in most cases it isn't. – OrangeDog Aug 18 '17 at 13:24
• And the "as the velocity decreases, distance covered is decreased" is not true. When you slow down you are still moving forward and the distance traveled increases as well. – nasu Aug 18 '17 at 13:34

Simply because, when you're considering $F=ma$ with a constant force $F$, then the rate of change is >>constant<< when taken with respect to time, but not with repsect to distance. If you've got another kind of law in mind, then maybe with respect to distance might make more sense. But most people are pretty happy with Newton:)

Time is not necessarily proportional to distance. When an object is accelerating more distance is travelled per unit of time after it has accelerated for a while compared to before it has accelerated.

Measuring acceleration per distance does not make sense -> we want to find out the rate of change in velocity, independent of current velocity (distance travelled is dependent on current velocity).

• I've got a feeling that I understood what you are trying to say here....................but plz also explain why my idea of - the more I cover distance the more my velocity increases; is wrong ? If you know what I mean. – Geeta Aug 18 '17 at 11:45
• It is true in general that the more distance you travel the more your velocity increases, e.g if you start at 0 velocity v^2 = 2aS where v is ending velocity, a is acceleration and S is displacement. However, since velocity is not directly proportional to displacement but only velocity squared, change in velocity per distance is still dependent on initial velocity making for inconvenient calculations. – Copper Aug 18 '17 at 11:51

This is a definition so ultimately it's a matter of convenience and uniformity. Obviously you define velocity as a rate of change over time: $v=\Delta x/\Delta t$ and so it is natural to define acceleration also as a rate of change over time: $$a=\frac{\Delta v}{\Delta t}=\frac{\Delta^2x}{\Delta t^2}$$ This way, all the information about the motion can be recovered from knowledge of a single position vs time graph.

This way rates are measured using the same quantity: the time interval. Finally, if you use $\Delta v/\Delta x$, then your unit of acceleration would be $(m/s)/m$, i.e. your acceleration would have units of "per second" without reference to any other basic quantity.

• But the derivative of $v$ wrt $x$ can also be recovered from the knowledge of a single position vs time graph. And this answer doesn't really explain why having a unit of one per second for acceleration would be a problem. – JiK Aug 18 '17 at 12:47
• $dv/dx = (dv/dt) / (dx/dt)$. If you can find $dv/dt$, you can find $dv/dx$. – JiK Aug 18 '17 at 13:37
• But I thought this answer claims you can get $dv/dt$ "from knowledge of a single position vs time graph"... What am I missing? – JiK Aug 18 '17 at 13:55
• @JiK Maybe I'm the one missing your point. It is easy enough to get $\Delta v/\Delta t$ from $(x(t+\Delta t)-2x(t)+x(t-\Delta t))/(\Delta t)^2$. – ZeroTheHero Aug 18 '17 at 14:02
• Why isn't it then easy enough to get $\Delta v / \Delta x$ just by dividing your formula for $\Delta v/\Delta t$ by your preferred formula for $\Delta x/ \Delta t$? – JiK Aug 19 '17 at 18:16

It is actually done, for example, in velocity vector fields in a fluid.

However, it is not useful outside those subjects. When you talk about velocity, it is usually referred to one single particle, which is placed anywhere. You compute the rate of change of the velocity following the particle. It's understood that it happens along its movement, but either if it's at rest or not, time is the only quantity that unavoidably runs along.

So, basically, if you consider that velocity "belongs" to the particle, it's kind of "useless" to get $\nabla v$. However, if you consider that the particle "gets" the velocity associated to that point (velocity is a vector field in space, like in fluid mechs.) then it is useful.

Velocity is the change in distance with respect to time. Acceleration is the change in velocity with respect to time. By definition, it doesn't make sense to say 'change in velocity with respect to distance.'

For example, if I increase my velocity from 5ms^-1 to 10ms^-1, my acceleration is 2.5ms^-2. If I increase my velocity from 10ms^-1 to 15ms^-1 in 2 seconds, this acceleration is also 2.5ms^-2. However in the latter case, i will travel a greater distance because the velocity is greater.

• This is just repeating the definition. – JiK Aug 18 '17 at 12:49