Averaging of velocities

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In summary: my problem for me. distance averaged velocity is (total distance/total displacement)*100 solved my problem for me. distance averaged velocity is (total distance/total displacement)*100
  • #1
ritwik06
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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.
 
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  • #2
ritwik06 said:

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?
 
  • #3
LowlyPion said:
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?
 
  • #6
ritwik06 said:
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?
 
  • #7
LowlyPion said:
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!
 
  • #8
ritwik06 said:
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.
 
  • #9
robphy said:
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?
 
  • #10
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
ritwik06 said:
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 ...
 
  • #12
borgwal said:
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!
 
  • #13
ritwik06 said:
But the thing is that the X is not eliminated!

T wasn't eliminated either.
 
  • #14
If I understood the problem correctly, it appears that the ratio desired is a dimensionless factor... a simple fraction.
 
  • #15
lowlypion said:
t wasn't eliminated either.

solved
 

1. What is "averaging of velocities"?

"Averaging of velocities" refers to the process of determining the average velocity of an object over a certain period of time. This is done by dividing the total distance traveled by the total time taken.

2. Why is averaging of velocities important in scientific research?

Averaging of velocities is important in scientific research because it allows scientists to accurately measure and analyze the motion of objects. By calculating the average velocity, scientists can make predictions and draw conclusions about the behavior of objects.

3. How is averaging of velocities different from finding the instantaneous velocity?

While averaging of velocities involves calculating the average speed of an object over a period of time, finding the instantaneous velocity involves determining the speed of an object at a specific moment in time. Instantaneous velocity takes into account any changes in speed during the given time interval.

4. Can averaging of velocities be applied to both linear and curved motion?

Yes, averaging of velocities can be applied to both linear and curved motion. In linear motion, the average velocity is simply the ratio of the total displacement to the total time taken. In curved motion, the average velocity is the ratio of the total arc length to the total time taken.

5. How can errors affect the accuracy of the averaged velocity?

Errors, such as measurement errors or human errors, can affect the accuracy of the averaged velocity. The larger the error, the less accurate the average velocity will be. It is important to minimize errors as much as possible in order to obtain reliable results.

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