# Math Help - Can the average velocity and delta velocity ever be the same?

1. ## Can the average velocity and delta velocity ever be the same?

I think they can be but I'm not sure.. but I also need to know why. I was thinking if acceleration was the same then delta velocity and average velocity could be the same? Help please!

2. Originally Posted by CrazziGirl23
I think they can be but I'm not sure.. but I also need to know why. I was thinking if acceleration was the same then delta velocity and average velocity could be the same? Help please!
The "acceleration was the same" as what?

Of course $\Delta v$ and $v$ can be the same. Given the appropriate acceleration and time interval this could certainly happen. Of course, it isn't likely to happen in an arbitrary situation.

-Dan

3. Originally Posted by CrazziGirl23
I think they can be but I'm not sure.. but I also need to know why. I was thinking if acceleration was the same then delta velocity and average velocity could be the same? Help please!
Even if the acceleration does not remain the same or constant, the delta velocity can never be the same as the average velocity.

That is if by "delta velocity" you mean the difference between two velocities.
Say you have Vo and V1.
delta V = V1 -Vo
average V = (Vo +V1)/2
They can never be the same.

4. Originally Posted by ticbol
Even if the acceleration does not remain the same or constant, the delta velocity can never be the same as the average velocity.

That is if by "delta velocity" you mean the difference between two velocities.
Say you have Vo and V1.
delta V = V1 -Vo
average V = (Vo +V1)/2
They can never be the same.
That's not true. You are talking about arithmetic mean of velocities which is different than average velocity...

The average velocity in the interval [t1, t2] is defined as:
$\frac{1}{t_2 - t_1} \int_{t_1}^{t_2} v(t) dt$

I can easily think of an example where delta v and <v> are the same...

Let's consider motion in interval [0, 2s] on the straight line, with constant accelelation $1 \frac{m}{s^2}$. Let's assume that initial velocity is equal $1 \frac{m}{s}$. One can easily check that: $\Delta v = = 2 \frac{m}{s}$

5. Originally Posted by albi
That's not true. You are talking about arithmetic mean of velocities which is different than average velocity...

The average velocity in the interval [t1, t2] is defined as:
$\frac{1}{t_2 - t_1} \int_{t_1}^{t_2} v(t) dt$

I can easily think of an example where delta v and <v> are the same...

Let's consider motion in interval [0, 2s] on the straight line, with constant accelelation $1 \frac{m}{s^2}$. Let's assume that initial velocity is equal $1 \frac{m}{s}$. One can easily check that: $\Delta v = = 2 \frac{m}{s}$
I see.

So I was wrong. Okay.

In the average velocity as shown in your integration, [which is correct, by the way], I don't know.
Can you also show me an example there?

6. Originally Posted by ticbol
Even if the acceleration does not remain the same or constant, the delta velocity can never be the same as the average velocity.

That is if by "delta velocity" you mean the difference between two velocities.
Say you have Vo and V1.
delta V = V1 -Vo
average V = (Vo +V1)/2
They can never be the same.
Let's use your example instead, since I suspect you will find this more persuasive.

$\Delta v = v_1 - v_0$
$\bar{v} = \frac{v_0 + v_1}{2}$

Set $\Delta v = \bar{v}$. We may solve the resulting equation for $v_1$ in terms of $v_0$:

$v_1 - v_0 = \frac{v_0 + v_1}{2}$

Thus
$v_1 = 3v_0$

So we have the condition for a general situation such that $\Delta v$ and $\bar{v}$ are the same.

-Dan

7. Originally Posted by topsquark
Let's use your example instead, since I suspect you will find this more persuasive.

$\Delta v = v_1 - v_0$
$\bar{v} = \frac{v_0 + v_1}{2}$

Set $\Delta v = \bar{v}$. We may solve the resulting equation for $v_1$ in terms of $v_0$:

$v_1 - v_0 = \frac{v_0 + v_1}{2}$

Thus
$v_1 = 3v_0$

So we have the condition for a general situation such that $\Delta v$ and $\bar{v}$ are the same.

-Dan
Yes. That's true. Nevertheless we usually define average velocity (or speed?) using the integral i wrote previously... If we observe that the distanse made by moving body is equal to:

$s = \int_{t_1}^{t_2} v(t) dt$

We can get a simple relation between distanse and average velocity:

$s = \frac{}{\Delta t}$

If you like arithmetic mean you should observe that "average velocity" is equal to arithmetic mean if accelelation is constant.

8. Originally Posted by albi
Nevertheless we usually define average velocity (or speed?) using the integral i wrote previously...
Actually that isn't true. The integral is indeed the Mathematical average of the velocity function, but the general definition of an average velocity in Physics is
$\bar{v} = \frac{x_f - x_i}{t_f - t_i}$
where the x's are displacements.

I also agree that Ticbol's average velocity is not at all general, as it presupposes a constant acceleration in order to have that form.

-Dan

9. Frankly speaking it is the matter of terminology....

Unfortunatelly I'm not very familiar with physics, english terminology...it can cause some problems...

In polish terminology there is the distinction between "velocity" (prędkość) and "speed" (szybkość)...

As "speed" I consider modulus of "velocity" - scalar quantity. "Velocity" is of a course vector...

The average "speed" is defined with the above integral...

What's with the average "velocity", it is analogous, so it is defined with the integral:

$\frac{1}{t_2-t_1} \int_{t_1}^{t_2} \vec{v} dt$

Obviously this definition is equivalent to yours because:

$\int_{t_1}^{t_2} \vec{v} dt = \int_{t_1}^{t_2} d \vec{r} = \vec{r_1} - \vec{r_0}$

10. Originally Posted by albi
Frankly speaking it is the matter of terminology....

Unfortunatelly I'm not very familiar with physics, english terminology...it can cause some problems...

In polish terminology there is the distinction between "velocity" (prędkość) and "speed" (szybkość)...

As "speed" I consider modulus of "velocity" - scalar quantity. "Velocity" is of a course vector...

The average "speed" is defined with the above integral...

What's with the average "velocity", it is analogous, so it is defined with the integral:

$\frac{1}{t_2-t_1} \int_{t_1}^{t_2} \vec{v} dt$

Obviously this definition is equivalent to yours because:

$\int_{t_1}^{t_2} \vec{v} dt = \int_{t_1}^{t_2} d \vec{r} = \vec{r_1} - \vec{r_0}$
Heh. My bad. I guess I had a brain fart for a moment there...

-Dan