Average velocity, me understand

[christian0710]

Hi, can someone explain to my how average velocity is in any way significant? Could it not just be avoided or ignored in physics? Here is why:

If you travel 10 meters forward and 10 meters backwards the average velocity is zero so it tells you nothing about how fast you went, only in one direction, and it tells you nothing about the actual average speed of the whole trip back and forth. To me it seems like average velocity in a forward direction is exactly the same as speed in one direction.

Comments are apprectiated :)

 

[Dale]

You are correct that average velocity is not terribly important in physics. It is mostly used to introduce the concept of velocity for students that do not have any background in calculus.

Nevertheless, it is well defined and any student of physics should be familiar with the definition and able to use it.

 

[faiface]

Average velocity is not very useful in theory, but in reality it is. How do you measure the velocity of an object? Well, you can't really measure it's actual velocity at any precise point of time. You can only measure how long it took for it to travel some short distance. Then you calculate the average velocity that the object had during traveling that short distance. By making this distance shorter and shorter you get more and more precise results but you can never get to the zero distance, which means that you always have to calculate the average velocity.

 

[christian0710]

Average velocity is not very useful in theory, but in reality it is. How do you measure the velocity of an object? Well, you can't really measure it's actual velocity at any precise point of time. You can only measure how long it took for it to travel some short distance. Then you calculate the average velocity that the object had during traveling that short distance. By making this distance shorter and shorter you get more and more precise results but you can never get to the zero distance, which means that you always have to calculate the average velocity.

Why not just calculate the speed in that short distance? if it's a forward direction, it's the same as average velocity.

 

[faiface]

Why not just calculate the speed in that short distance? if it's a forward direction, it's the same as average velocity.

I'm not sure if I understand you, but what I was basically trying to say was that the average velocity has it's use in reality when you want to calculate (or more precisely: measure) the velocity of a moving object.

Example:
Assume you have a video record of a moving car. Now you want to know the speed of the car. So you take the position of the car at the time 0s, you take the position at the time 0.2s, you subtract them and divide by the time it took (0.2s). This way, you get the average velocity over the time of 0.2s which might be enough for you.

 

[christian0710]

However i can't understand that if the acceleration is constant, then the average velocity can be written as 1/2(Vo+ V).
Is it the same as saying The average velocity between 3 points is (V0 + v1+v3)/3 ?? and between 4 points it's (V0 + v1+v3 + v4)/4
And is it a requirement that the acceleration is constant for this to be true? I tried drawing a distance traveled vs. time graph but i can't really get it to make sense.,

 

[faiface]

However i can't understand that if the acceleration is constant, then the average velocity can be written as 1/2(Vo+ V).
Is it the same as saying The average velocity between 3 points is (V0 + v1+v3)/3 ?? and between 4 points it's (V0 + v1+v3 + v4)/4
And is it a requirement that the acceleration is constant for this to be true? I tried drawing a distance traveled vs. time graph but i can't really get it to make sense.,

Rather draw a velocity vs. time graph and I'm sure you'll understand.

 

[Dale]

If you draw a graph of any quantity over time, then you can calculate the area under that curve. The average value is then the height of a rectangle that has the same area. If the acceleration is constant then the area under the curve is a trapezoid and you can easily calculate its area. It matches the area of a rectangle with the height you gave above.

 

[christian0710]

If you draw a graph of any quantity over time, then you can calculate the area under that curve. The average value is then the height of a rectangle that has the same area. If the acceleration is constant then the area under the curve is a trapezoid and you can easily calculate its area. It matches the area of a rectangle with the height you gave above.

Thank you, so you can only express the average velocity as (Vo + V1)/2 or (Vo + V1 + V2)/3 when acceleration is constant right? If accelration was constant for 2 minutes then stopped for 10 minutes (was zero*), and increased by 7m/s^2 constantly, then you would get a straight horizontal line on the V(t) graph for 10 minutes, so the average velocity between point (Vo,t) (before acceleration was zero) and point (V1,t1) (when acceleration was 7m/2^2). Would velocity then not depend on the time diference?

 

[christian0710]

If you choose to do this (V0 + V1)/2, you would not account for the fulcuations and changes in velocity during the two time intervals right? So would it be smarter to integrate it?

 

[Dale]

Would velocity then not depend on the time diference?

Yes, sorry I was not clear. The width of the rectangle I was describing above is the same as the time difference.

 

[christian0710]

Yes, sorry I was not clear. The width of the rectangle I was describing above is the same as the time difference.

Good :) So is the argument this: When acceleration is constant you can write the average velocity between 2 points as (v0+ V1)/2 without the timeinterval involved, because no matter how long or how short the time interval, T1-To is, the function increases linearly (with the same rate of velocity per time for every value of V within that interval of time) but if the rate of velocity changed per time, then we would have to use the formula Average velocity = V1-V0/(t1-t0) ?

 

[christian0710]

I guess I'm asking how to find average values of a function - If you have a function, y=x^2 can you then say the average value of Y between x=1 and x =5 is (y(1) +y(5))/2

Becaue Y_average here represents v_average

 

[christian0710]

Here is my problem (It takes some time to figure out how to express it)
When you say the average velocity =(V0 + V1)/2 you are treating velocity as numbers, but it's a function which depends on x, You are just taking the average between two function values and completely neglecting the x-component, I don't understand why this is okay to do? The function has a direction in another dimension (the x-axis) so how can we just say that the mean value between two points is the mean of the two function values?

 

[jtbell]

If you have a function, y=x^2 can you then say the average value of Y between x=1 and x =5 is (y(1) +y(5))/2

No. This works only if the function is linear: y = ax + b, where a and b are constants.

For constant acceleration, v is a linear function of t: v = v₀ + at.

 

[Dale]

I guess I'm asking how to find average values of a function - If you have a function, y=x^2 can you then say the average value of Y between x=1 and x =5 is (y(1) +y(5))/2

Becaue Y_average here represents v_average

I explained how you find the average value of a function in post 8. You plot that function, you calculate the area under the plot, and you find the rectangle whose area matches. The height of that rectangle is the average value. (The rectangle has the width of the time difference). Equivalently, you can simply calculate the area under the plot and then divide that by the time difference. Either way, you have to calculate the area under the curve.

If the function is a straight line then you can simply use the formula for the area of a trapezoid to calculate the area. That gives you the easy formula you posted. However, if the function is a curve, like y=x^2, then you need to use calculus to calculate the area.

 

[christian0710]

Thank you, I think i might have misunderstood the book. Can you please confirm that this is corect? I would really appreciate it. :)Here is the specific problem from the book:

A runner runs for from x0=zero meters to z1=100 meters with a constant velocity in 11seconds, this means his average velocity is 100m/11s = 9,1m/s. This is a straight line on the distance vs time plot.If I used the formula Average velocity = (v0 +v1)/2 =( 0+100)/2 = 50m/s, which is NOT 9,1m/s. But I think I might have misunderstood what the book said on page 42, I quote “Since the velocity increases uniformly with time, the average value of the velocity is

simply the average of the initial value and the final values, or Average velocity = (v0 +v1)/2”

So my question is this: Does average velocity and average value of velocity mean two different things? If this is correct then it makes sense what you wrote

The average VALUE of the velocity is like finding the average value of a function, and as jtbell said this can be done like (y0+y1)/2 only for linear functions? And for other functions we integrate 1/(b-a)'integral(f(x),a,b) right?

 

[Dale]

A runner runs for from x0=zero meters to z1=100 meters with a constant velocity in 11seconds, this means his average velocity is 100m/11s = 9,1m/s. This is a straight line on the distance vs time plot.If I used the formula Average velocity = (v0 +v1)/2 =( 0+100)/2 = 50m/s, which is NOT 9,1m/s.

Here is your mistake: (v0 +v1)/2 =( 0+100)/2. 0 is the initial position, x0, not the initial velocity, v0. Similarly 100 is the final position, x1, not the final velocity, v1. As you noted, it is a position time graph, so applying that formula gives you the average position, which is indeed 50 m.

Does average velocity and average value of velocity mean two different things?

No, that is the same thing, but what you calculated above was the average position, not the average velocity.

the average value of a function, and as jtbell said this can be done like (y0+y1)/2 only for linear functions? And for other functions we integrate 1/(b-a)'integral(f(x),a,b) right?

Yes.

 

[christian0710]

Here is your mistake: (v0 +v1)/2 =( 0+100)/2. 0 is the initial position, x0, not the initial velocity, v0. Similarly 100 is the final position, x1, not the final velocity, v1. As you noted, it is a position time graph, so applying that formula gives you the average position, which is indeed 50 m.

Yess but the thing I'm really trying to understand is this: Why does the book say that The average velocity can be calculated as v_average= (position1-position0)/(t1-t0)
AND at the same time it says that the average velocity can be calculated as 1/2(V0 +V) i just prooved that they give two different values, so I don't understand how they both can represent the Average velocity. :( I really want to understand it.

 

[christian0710]

Here is what the book sais: please help me out :)

 

[Dale]

Yess but the thing I'm really trying to understand is this: Why does the book say that The average velocity can be calculated as v_average= (position1-position0)/(t1-t0)

This formula is always true, it is essentially the definition of average velocity.

AND at the same time it says that the average velocity can be calculated as 1/2(V0 +V)

This formula is only true in the case of constant acceleration.

i just prooved that they give two different values, so I don't understand how they both can represent the Average velocity. :( I really want to understand it.

No, you didn't prove that. You simply misused the second formula. In the scenario above the v0 is not 0, it is 9.1. Similarly, v1 is not 100, it is 9.1. Then, if you use the formula correctly, you get 1/2(9.1+9.1)=9.1 which agrees with the other method.

This scenario has a constant velocity, so the acceleration is 0, which is also a constant, so both formulas are applicable. However, you have to plug in the correct values for the formula to work.

 

[Dale]

Here is what the book sais: please help me out :)

The book is correct. You just made a small mistake in your calculation (put x0 and x1 in the formula where it asked for v0 and v1).

 

[christian0710]

Thank you so much for explaining that, somehow I could not see it. I really appreciate your help.

 

[Dale]

Thank you so much for explaining that, somehow I could not see it. I really appreciate your help.

You are welcome. I am glad to help.