# Averaging of velocities!

1. Oct 29, 2008

### 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.

2. Oct 29, 2008

### LowlyPion

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?

3. Oct 29, 2008

### ritwik06

v=kt

and v^2=2kx

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

4. Oct 29, 2008

### LowlyPion

5. Oct 29, 2008

### ritwik06

6. Oct 29, 2008

### LowlyPion

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?

7. Oct 29, 2008

### ritwik06

Time Average velocity is total displacement/ total time

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

8. Oct 29, 2008

### robphy

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.

9. Oct 29, 2008

### ritwik06

Yup, I have done that. All I need to know is what is distance averaged velocity???

10. Oct 29, 2008

### 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.

11. Oct 29, 2008

### LowlyPion

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 ...

12. Oct 29, 2008

### ritwik06

$$<v>=\frac{1}{X}\int_{0}^{X} V(x) dx$$
right?
But the thing is that the X is not eliminated!!!

13. Oct 29, 2008

### LowlyPion

T wasn't eliminated either.

14. Oct 30, 2008

### robphy

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

15. Oct 30, 2008

solved