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?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.
v=ktTry 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?
How do you take an average?v=kt
and v^2=2kx
where t is time
x is displacement
and k is the constant acceleration
now what shall I do?
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?I know about time average velocity. But what is Space average velocity?
Time Average velocity is total displacement/ total timeYou 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?
Can you derive the statement thatTime Average velocity is total displacement/ total time
Is distance average velocity? total distance / total displacement ??? I really have no idea. Please help!!
Yup, I have done that. All I need to know is what is distance averaged velocity???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.
OK. If Time averaged velocity is given by:Yup, I have done that. All I need to know is what is distance averaged velocity???
[tex]<v>=\frac{1}{X}\int_{0}^{X} V(x) dx[/tex]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.
T wasn't eliminated either.But the thing is that the X is not eliminated!!!
solvedt wasn't eliminated either.