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We have the image below displaying the uniform velocity by time-distance graph. At every point velocity is constant but what if distance and time both become zero as at origin in the graph is? The velocity must become undefined as $\frac{0}{0}$ is undefined in mathematics or shall we call it zero? If zero then why?

enter image description here

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    $\begingroup$ i think no body can actually exhibit such a motion if body is at rest at 0 time . we are just beginning to measure the distance(wrto time) after the body reaches uniform motion(a finite speed). so you will see a different graph when thats considered . $\endgroup$ – Gowtham Dec 19 '14 at 10:35
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    $\begingroup$ $0/0$ can be defined as the limit of $\Delta y / \Delta x$ as $\Delta y, \Delta x \rightarrow 0$ if this limit exists, so is not (necessarily) undefined $\endgroup$ – theo Dec 19 '14 at 10:38
  • $\begingroup$ Well, in order to measure average velocity of a particle, you must need at least two points from your graph or else average velocity becomes undefined according to the formula: $$v_{av} = \frac{\Delta x}{\Delta t}.$$ $\endgroup$ – Samama Fahim Dec 20 '14 at 22:18
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what if distance and time both become zero as at origin in the graph is

It appears that you're trying to say that the velocity is equal to the position of the particle divided by the clock time at that position:

$$v= \frac{x(t)}{t}\; ? $$

But this isn't correct.

Average velocity $\bar v$ is defined as displacement $\Delta x = x(t_f) - x(t_i)$ divided by elapsed time $\Delta t = t_f - t_i$:

$$\bar v = \frac{\Delta x}{\Delta t}$$

In this case, we can choose $t_i = 0$ and then $x(t_i) = x(0) = 0$ and then

$$\bar v = \frac{x(t_f)}{t_f}$$

which appears to be the same as the first equation in my answer. However, it isn't since it is in fact

$$\bar v = \frac{x(t_f) - x(0)}{t_f - 0} = \frac{x(t_f)}{t_f}$$

Seen this way, we can't use $t_f = 0$ since, in that case, both the displacement and elapsed time vanish.

However, as $\Delta t$ becomes very small, the average velocity approaches the instantaneous velocity $v$:

$$ \bar v \rightarrow v = \frac{dx(t)}{dt},\; \Delta t \rightarrow 0$$

When the instantaneous velocity is constant, as in this case, the average velocity equals the instantaneous velocity:

$$\bar v = v,\; v =\mathrm{constant}$$

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Velocity is defined as how fast the body changes its position with respect to time. Change of position with respect to a frame is called displacement. Velocity measures how fast the body changes its position. $$ v = \lim_{\Delta t \to 0} \dfrac{\Delta x}{\Delta t}$$.

At $t = 0$ , the body was at a distance of $0$ from the frame. It was not moving and hence was not changing its postition. Since it wasn't displacing, at that point velocity is thus finite and not undefined.

An example:

Suppose, the body travels with uniform velocity $4~\text{m/s}$. At $\Delta t = 1 s$, the body covers $4 m$. At $t = 0.00000000023 s$ , the body is at a distance $9.2 \cdot 10^{-10} m$. Therefore the velocity during the interval $\Delta t = (0.00000000023 - 0)s$ is $$v = \dfrac{9.2 \cdot 10^{-10}}{0.00000000023} = 4$$ . So, at $t = 0$, the velocity is $4~\text{m/s}$. So,one may ask how can there be any velocity at $t = 0$ as the body is at $0 m$ away from the origin. Answer is simple:

Velocity is the measurer of how fast the distance from the frame changes with respect to time. At that moment the body is at $0 m$ away. But what velocity is saying is that at that moment if the body continues its motion like the same way it had at that point ie. $t = 0$, it would travel $4 m$ in $1 s$.

Suppose, the ball that is thrown by a pacer in cricket, has generally a velocity $145 ~\text{km/hr}$ at the time of being hit by a bat. Now, is the pitch $145 km$ ? Nope! It is saying that at that moment, if the ball continued its motion like the motion it had at that same point, it would cover $145 km$ in $1 hr$ . So, velocity just measures the fastness of how the body changes its position.

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  • $\begingroup$ a body can have a finite velocity and still have 0 displacement at "0 time" ? $\endgroup$ – Gowtham Dec 19 '14 at 11:00
  • $\begingroup$ @Gowtham: I have read your above comment. This is indeed answering op's quo. And you are mixing change of position or displacement with distance or position in the later comment. $\endgroup$ – user36790 Dec 19 '14 at 11:28
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    $\begingroup$ i just feel the velocity need not be zero , can be finite , thats all . I agree with everything you say else. $\endgroup$ – Gowtham Dec 19 '14 at 11:32
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Actually you have plotted the graph of displacement $x(t)$, And here the $x(t)=vt$ , $v$ is some constant.

now lets take the slope of the graph i.e $ dx/dt =v $

The slope is constant (equal to $v$ and we physically call it velocity)

now what is the slope at (0,0)?

since the slope is constant it will be still $v$,right? hence at the origin the velocity is $v$ and clearly it is defined.

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At (0,0), before the clock has started, you can say that there is no displacement, and no time measured.

It is only with hindsight, by looking at the rest of the chart after the clock as started, can you see the slope of the line, and hence work out its velocity.

Any single point on the line, without any other data, shows only an average speed, again, we need to see the rest of the line to know that the speed is constant.

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Looking strictly at the graph you provided, it shows that at time=0, distance=0. And at time=6s, distance=60m. That graph shows a uniform velocity of 10m/s. Generally, velocity is not an undefined quantity, it is defined as the rate of change in unit displacement per unit of time, define your units and you have the definition of velocity.

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$$\text{Velocity }= \dfrac{\text{change in postion}}{\text{time taken for that change}} \neq \dfrac{\rm distance}{\rm time}.$$ If we draw a graph for the change in position vs time difference, the case you are talking does not exist. Time difference has to be there when we talking about velocity or speed as they are a measure of rate of change of displacement and distance with respect to time. If there is no time, then velocity and speed do not make sense.

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Let's take an instance of race during Olympics; Say we have a man $A$ at position $a$. Assume, we whistle to start the race but he doesn't start to run and and when the race is finished, after some time intervals, we calculate his average running velocity, we get $0/t$ so the velocity is found to be $zero$ as well. But what happens when again on starting race he don't runs and also we don't start our stopwatch? Then if we calculate velocity according to time of stopwatch, we can make no sense of it and the velocity then w.r.t the time of stopwatch is said to be undefined. But if we calculate velocity according to the time of hand-watch of referee we get velocity to be $zero$ again which makes sense and you can say the man to be steady now. :P

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The velocity is actually determined by the slope of the line:

$ y = mx+b$

Therefore the slope of the line is the velocity:

$dy/dx = m$

Even at x = 0 the slope equals m.

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The simplest answer, that you will easily grasp, would be:

Velocity does not exist at your graph's origin.

Umm... Yes! You can say; Mathematically, it is undefined.

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