# Averaging of velocities!

#### ritwik06

1. The problem statement, all variables and given/known data
A particle starts from rest with constant acceleration. Find the ratio of space average velocity with time average velocity.

Explain to me the underlined words in order that I may proceed to solve this problem.

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#### LowlyPion

Homework Helper
1. The problem statement, all variables and given/known data
A particle starts from rest with constant acceleration. Find the ratio of space average velocity with time average velocity.

Explain to me the underlined words in order that I may proceed to solve this problem.
Try writing out the equations for V as a function of t and V as a function of x. Then perhaps you can arrive at what taking the average over each function might yield?

#### ritwik06

Try writing out the equations for V as a function of t and V as a function of x. Then perhaps you can arrive at what taking the average over each function might yield?
v=kt

and v^2=2kx

where t is time
x is displacement
and k is the constant acceleration
now what shall I do?

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#### LowlyPion

Homework Helper
I know about time average velocity. But what is Space average velocity?
You have V(x) over the distance X in the same way that you have V(t) over T for how you might figure the time average velocity. You don't think you can come up with an expression for the distance averaged V?

#### ritwik06

You have V(x) over the distance X in the same way that you have V(t) over T for how you might figure the time average velocity. You don't think you can come up with an expression for the distance averaged V?
Time Average velocity is total displacement/ total time

Is distance average velocity? total distance / total displacement ??? I really have no idea. Please help!!

#### robphy

Homework Helper
Gold Member
Time Average velocity is total displacement/ total time

Is distance average velocity? total distance / total displacement ??? I really have no idea. Please help!!
Can you derive the statement that
"Time Average velocity is total displacement/ total time"?
You may wish to first consider a simpler situation in which your total trip has just two velocities:
v1 for time-interval t1, and then v2 for time-interval t2.

#### ritwik06

Can you derive the statement that
"Time Average velocity is total displacement/ total time"?
You may wish to first consider a simpler situation in which your total trip has just two velocities:
v1 for time-interval t1, and then v2 for time-interval t2.
Yup, I have done that. All I need to know is what is distance averaged velocity???

#### borgwal

Example: For 2km I go at a velocity 30m/s, then for 1km, I go at a velocity 15m/s. The distance-averaged velocity would be (2x30m/s+1x15m/s)/3=25m/s.

The time-averaged velocity will be smaller than 25m/s, as I go an equal amount of time at the two different speeds: the time-averaged velocity is (30+15)/2 m/s=22.5 m/s.

#### LowlyPion

Homework Helper
Yup, I have done that. All I need to know is what is distance averaged velocity???
OK. If Time averaged velocity is given by:

$$V_{Tavg} = \frac{1}{T}\int_{0}^{T} V_{(t)} dt$$

Then won't the distance averaged velocity be given by:

$$V_{Xavg} = \frac{1}{X}\int_{0}^{X} V_{(x)} dx$$

Now you have already supplied V as a function of x and V as a function of t.

So ...

#### ritwik06

Example: For 2km I go at a velocity 30m/s, then for 1km, I go at a velocity 15m/s. The distance-averaged velocity would be (2x30m/s+1x15m/s)/3=25m/s.

The time-averaged velocity will be smaller than 25m/s, as I go an equal amount of time at the two different speeds: the time-averaged velocity is (30+15)/2 m/s=22.5 m/s.
$$<v>=\frac{1}{X}\int_{0}^{X} V(x) dx$$
right?
But the thing is that the X is not eliminated!!!

#### LowlyPion

Homework Helper
But the thing is that the X is not eliminated!!!
T wasn't eliminated either.

#### robphy

Homework Helper
Gold Member
If I understood the problem correctly, it appears that the ratio desired is a dimensionless factor... a simple fraction.

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