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Averaging of velocities!

  • Thread starter ritwik06
  • Start date
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.
 

LowlyPion

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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?
 
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

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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?
 
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

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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.
 
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???
 
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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

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

[tex] V_{Tavg} = \frac{1}{T}\int_{0}^{T} V_{(t)} dt [/tex]

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

[tex] V_{Xavg} = \frac{1}{X}\int_{0}^{X} V_{(x)} dx [/tex]

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

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

robphy

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

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