Curious about average velocity (v bar)

[ChristPuncher]

Homework Statement

My question uses the following problem. I am confused as to why s=v(bar) * t works out the same as s=(v₀-v_final)/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=?

Homework Equations

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

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-v_final)/2] * t

Why is this the case?

 

[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.

 

[Redbelly98]

Homework Statement

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

The expression in paranetheses should be

(v₀ + v_final)​

So your question becomes, why is

v_avg = (v₀ + v_final)/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.