Time and Mean Velocity

Date: 01/16/2003 at 23:03:21
From: Peter
Subject: Why is "s = s0 + v0 t + 1/2 a t^2" true?

I read books about how they come out with the formula and it's 
convincing but when I try to trace it manually, the answer is 
different.

For example,
init position s0 = 0
init velocity v0 = 0
acceleration   a = 1
time           t = 5
s = 0 + 0 + 1/2 (5^2) = 12.5

BUT ...

'v' is change of position in time and 'a' is change of velocity in 
time. So ...

At t0, v0 = 0, then s0 = 0
At t1, v1 = 1 (i.e. v1 = v0 + a), then s1 = 1 (i.e. s1 = s0 + v1)

Also, for other t's ...

 T    S    V
---  ---  ---
 0    0    0
 1    1    1
 2    2    3
 3    3    6
 4    4   10
 5    5   15

so at t = 5, the position 's' should be 15 and not 12.5
What am I understanding wrong on this?

Thanks for your time.

Date: 01/17/2003 at 11:04:19
From: Doctor Rick
Subject: Re: Why is "s = s0 + v0 t + 1/2 a t^2" true?

Hi, Peter.

Did you switch the S and V column labels? The second column correctly 
represents the velocity at time T seconds under constant acceleration 
of 1 meter/second per second. If I understand you correctly, you 
built the third column as follows: Take the position S at time T and 
add 1 second times the velocity at time T+1 to get the position at 
time T+1.

This would be the correct thing to do IF the velocity of, say, 3 
meters/second at time T=3 held constant for the entire one-second 
interval from T=2 to T=3. This is not the case. For instance, in the 
first interval (T=0 to 1) the velocity does not instantaneously jump 
from 0 at T=0 to 1 at T=0.0000000001 second and stay constant for a 
second. It gradually increases; it is 0.25 m/s at T=0.25, 0.5 m/s at 
T=0.5, etc. 

The MEAN velocity during this interval is (0+1)/2 = 0.5 m/s. If you 
use the mean velocity instead of the final velocity for each interval, 
your table will look like this:

T   V   Vmean   S
0   0           0
1   1    0.5    0.5
2   2    1.5    2.0
3   3    2.5    4.5
4   4    3.5    8
5   5    4.5   12.5

Well, what do you know! It comes out correctly this time. Does this 
help answer your question? To give a thorough answer I'd need to get 
into calculus, but this may be enough to show you that it makes sense.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 01/18/2003 at 04:45:23
From: Peter
Subject: Thank you (Why is "s = s0 + v0 t + 1/2 a t^2" true?)

Thanks Doctor Rick for clearing up my confusion.

Peter