# Curious about average velocity (v bar)

1. Oct 29, 2009

### ChristPuncher

1. The problem statement, all variables and given/known data

My question uses the following problem. I am confused as to why s=v(bar) * t works out the same as s=(v0-vfinal)/2 * t Enlighten me please!

Problem: How long does it take to fall 1000ft with no drag? What is the final velocity?

displacement: s=1000ft
acceleration: a=9.8 (gravity)
time: t=?

2. Relevant equations

s=v(bar) * t (distance = average rate * time)
s=(1/2)at2 (distance = one half * acceleration * time squared)
v=at

3. The attempt at a solution

Step 1) Convert 1000ft to 304.8 m
Step 2) Rearrange s=(1/2)at2 to t=sqrt[2a/d]
Step 3) Plugin known variables and solve for t giving 7.887 s
Step 4) Plugin known variables into v=at giving v=77.29 m/s
Step 5) Check answer with s=v(bar) * t
304.8 m = (384.8-0)/(7.887-0) * 7.887 s

I see that step 5 also works if you use this equation s = [(v0-vfinal)/2] * t

Why is this the case?

2. Oct 29, 2009

### DavidDishere

(Start - Final) /2 is just a way to compute the average. Take any 2 numbers and divide them by 2 and you get the average of those 2 numbers. V bar is the average velocity.

3. Oct 30, 2009

### Redbelly98

Staff Emeritus
The expression in paranetheses should be

(v0 + vfinal)​

So your question becomes, why is

vavg = (v0 + vfinal)/2 ?​

If the acceleration is constant, then the velocity is a linear function of time. For linear functions, the average over some interval equals the average at the two endpoints.