Ratio of Space Average Velocity to Time Average Velocity?

[ritwik06]

Homework Statement

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.

 

[LowlyPion]

Homework Statement

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?

 

[LowlyPion]

v=kt

and v^2=2kx

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

How do you take an average?

http://en.wikipedia.org/wiki/Time-average_velocity

 

[ritwik06]

How do you take an average?

http://en.wikipedia.org/wiki/Time-average_velocity

I know about time average velocity. But what is Space average velocity?

 

[LowlyPion]

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]

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]

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]

But the thing is that the X is not eliminated!

T wasn't eliminated either.

 

[robphy]

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

 

[ritwik06]

t wasn't eliminated either.

solved[/color]