Instantaneous (Midpoint)/Average Velocity?

by cudah
Tags: instantaneous, midpoint or average, velocity
cudah is offline
Nov8-08, 01:00 AM
P: 1
1. The problem statement, all variables and given/known data

Prove the following theorems:
Theorem 1:

Vavg= (Vfinal+Vinitial)/2

TO Theorem 2:

Vavg=Vmp= V(t/2) <--Where V(t/2) represents a function read as V of t/2.
(t/2) is the midpoint in time for a given interval and V(t/2) is the instantaneous velocity at that time

2. Relevant equations

I need to prove these theorems esp theorem 2 (need to prove theorem 1 first) to relate average velocity (measured in lab) to instantaneous velocity (needed to calculate kinetic energy)

3. The attempt at a solution
I was able to prove theorem 1 using kinematic equation:
X-Xo = Vot + 1/2at^2 and V=Vo + at
X-Xo = (2Vot)/2 + (at^2)/2
X-Xo = (2Vot + at^2)/2
(X-Xo)/t = (2Vo + at)/2
(X-Xo)/t = [(Vo + at) + Vo]/2, substitute V=Vo + at
to get,
(X-Xo)/(t-to)= (V + Vo)/2= Vavg

For theorem 2, I have to prove it using a kinematic equation too. I tried but I'm not sure if I'm doing it right.
I tried using X-Xo = 1/2 (V+Vo)t
to get,
X-Xo= [(V+Vo)t]/2
(X-Xo)/(V+Vo) = t/2

Or using V(t/2) = Vo + a(t/2) ----> V of t/2
and plug V(t/2) in X-Xo = 1/2 (V+Vo)t
X-Xo = 1/2 [{(Vo + a(t/2)} +Vo)]t
but my answer didn't make sense.
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heth is offline
Nov8-08, 05:29 AM
P: 109
This is with constant acceleration, right?

Try drawing a graph of displacement vs time, and mark on all the variables that are mentioned in your (correct) solution to theorem 1. That should help you to think your way around the second part.

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