What is the difference between two types of velocity?

Question

What is the difference between $v=\frac{s}{\Delta t}$ and $\bar{v} =\frac{\Delta\bar{x}}{\Delta t}$, are they the same?

Answer 1

I am guessing in your first you you mean $\frac{\Delta S}{\Delta t}$. If this is the case then the difference would be in that you seem to use vector notation in the second equation and not in the first one. This means that in the first one you talk about the speed of something, i.e. that rate at which it moves. In the second one, since there is also a vector-notation involved, this would indicate that here you have a vector with a direction and magnitude, the magnitude representing the value from the first equation, the rate of change in position, and the direction being the direction in which this change is happening.

Think if it as the first one is saying "We are moving at 2m/s", and the second one saying "we are moving at 2m/s to the left" or something like that.

Other than the direction-component to my answers, the equations represent the same thing. They both essentially say $\frac{\text{how much we have moved}}{\text{the time it took us to move this distance}}$ with the second one having added in which direction we moved as well.

Speed is rate of movement, velocity is rate of movement in a certain direction.

Answer 2

The first expression $\frac s {\Delta t}$ is the average speed over the interval $\Delta t$ and the second $\frac {\Delta \vec x}{\Delta t}$ is the average velocity over the same interval.

The difference is that if I travel around a circle with circumference $10$ metres in $5$ seconds then my average speed is $\frac {10} 5=2$ metres per second, but my average velocity is zero because $\Delta \vec x=0$ since I finish where I started.

Answer 3

There are two types of speed - instantaneous and average. Average is defined as, total displacement over time period : $$ \overline v = \frac {\Delta x}{\Delta t} $$

By definition, instant speed is average speed when time period goes to zero :

$$ v = \lim_{\Delta t \to 0} \overline v = \lim_{\Delta t \to 0} \frac {\Delta x}{\Delta t} = \frac {dx}{dt} $$

EDIT

As about your direct question => $$\begin{align} \overline v &= \frac 1N \left( \frac{\Delta x_1}{\Delta t} + \frac{\Delta x_2}{\Delta t} + \ldots + \frac{\Delta x_n}{\Delta t} \right) \\&=\frac {1}{\Delta t} \left( \frac{\Delta x_1}{N} + \frac{\Delta x_2}{N} + \ldots + \frac{\Delta x_n}{N} \right) \\&=\frac {\overline {\Delta x}}{\Delta t} \\&= \frac{\Delta x_1}{N\Delta t} + \frac{\Delta x_2}{N\Delta t} + \ldots + \frac{\Delta x_n}{N\Delta t} \\&=\frac {\Delta x_1 + \Delta x_2 + \ldots + \Delta x_n}{N \Delta t} \\&=\frac {\Delta s}{\tau} \end{align} $$

Thus, your both given definitions are equivalent.