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V=S/T. As per my knowledge i think ratio as division and it don't give any meaning like this much displacement in this much time. So i think physicists only used division as notion for velocity. But why didn't they used addition multiplication instead like V=S+T or V=S*T or V=T/S.

BUT after thinking about it I get to know that multiplication or addition is not a good way to notify velocity as if we use multiplication we have to 1 out of displacement or time to interpret meaning form that velocity and also we r not going to get unique numbers as velocity for totally different conditions like if we say V=10 m/sec then we can interpret it in many ways as 2 meters in 5 sec or 10 m in 1 sec which seems totally wrong

But when I take V=T/S then i didn't seen any impropernece in this. So can u tell me why we preferred division as the notation?

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closed as unclear what you're asking by knzhou, John Rennie, Jon Custer, PM 2Ring, Norbert Schuch Jul 13 at 15:05

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Probably useful reading as to why meters + seconds doesn't make sense: physics.stackexchange.com/q/98241/25301 $\endgroup$ – Kyle Kanos Jul 12 at 11:17
  • $\begingroup$ Related: physics.stackexchange.com/q/378495/109928 $\endgroup$ – Stéphane Rollandin Jul 12 at 12:19
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    $\begingroup$ You certainly cannot interpret 10 m/s as 2 meters in 5 seconds. It would be 2 meters in 1/5 of a second. $\endgroup$ – Stéphane Rollandin Jul 12 at 12:24
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    $\begingroup$ Because $v = \frac{\rm d}{{\rm d}t} x$ can be used to predict the outcome of experiments and all other pseudo-kinematics can't. $\endgroup$ – ja72 Jul 12 at 12:56
  • $\begingroup$ In the second paragraph you realise that multiplication is not useful so why continue to ask the question? The last paragraph just gives the inverse of velocity ("slowness" if you like) and is infinite when you are stopped and so not normally useful either. $\endgroup$ – rghome Jul 12 at 14:17
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This is actually in interesting question, and definitely not worth all of the down votes in my opinion. You're actually asking a different question than you realize. It's not like people were sitting around saying "We need a suitable definition for velocity. What should it be?" This doesn't make sense. Without a definition how would anyone know to even conceive of "velocity". Instead what happened is that the ratio between displacement and the time it takes to achieve that displacement turned out to be a very useful quantity. Because of this it received it's own name.

And so this leads to your actual question. Why has the value $xt$ not gotten any sort of important definition or name? Note that I picked this one because you can't add or subtract values with different units, and $t/x$ is just the inverse of speed so it isn't really anything different from what we already use (although it wouldn't work for a vector displacement, so it's not as important in that regard). It just isn't a useful value. It does not show up often (at all?) when describing the world around us. If we make some new discovery where $xt$ plays a huge role, then I'm sure we could make up a name for it. Displaced time maybe?

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  • $\begingroup$ Thanks sir your answer is very useful for me $\endgroup$ – Nikhil Pant Jul 12 at 16:38
  • $\begingroup$ @NikhilPant Please upvote any useful answers and also select an answer as the accepted answer if it sufficiently answers your question. $\endgroup$ – Aaron Stevens Jul 12 at 16:48
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Well, speed is the rate of change of distance. Which basically means distance per unit time. Distance is a scalar unit and dividing it by another scalar (time) makes speed a scalar unit.

Velocity, however, is the rate of change of Displacement and not distance. Displacement is a vector and unlike distance, when a vector is divided by a scalar, the result is still a vector.

Just as displacement is the vector counterpart to distance, velocity is the vector counterpart to speed. And through velocity, we can derive the vector value for acceleration which is still a vector when derived from speed.

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Note: velocity is the rate of change of displacement with respect to time. It's defined by $v = ds/dt$, not $v = s/t$. $v = s/t$ only if v is a constant, otherwise you're just taking the average velocity.

Technically people can define all sorts of words for every operation between two things but in the end it doesn't matter as long as we're consistent. To most of us, it's more intuitive to define something as "the rate of change of displacement w.r.t time" than "the rate of change of time w.r.t displacement", so that's what we went with. The math gets a lot more confusing if you try to define it the other way. For example, both velocity and displacement are vectors. You can't divide by a vector, so if you wanted to create a function $f = t/s$, then you'd have to find the multiplicative inverse of the vector displacement and define it as $f = ts^{-1}$. This isn't always the case: for example, in current electricity, we have both resistance, = V/i, and conductance = i/V. Both are used when they're convenient because both make intuitive sense to us.

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  • $\begingroup$ Thanks brother for your answer $\endgroup$ – Nikhil Pant Jul 12 at 16:38

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