# Homework Help: Average velocity and acceleration

1. Sep 2, 2008

### Kajayacht

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

What is the average acceleration between time t = 2.2 s and time t = 9.6 s?
http://img98.imageshack.us/my.php?image=prob20av2yo4.gif
I've already found the average velocity between the two points to be 1.09×10^-1 m/s

2. Relevant equations

a= "delta"V / "delta" T
V= d/t

3. The attempt at a solution

v1= 9/2.2 = 4.09
v2= 10/9.6 = 1.04

a = (1.04-4.09) / (9.6-2.2) = -3.05 / 7.4 = -.412 m/s^2

2. Sep 2, 2008

3. Sep 2, 2008

### Redbelly98

Staff Emeritus
Actually, v is the slope of the position-vs.-time graph.

4. Sep 2, 2008

### Kajayacht

Ok, so I tried with the slope between 2.2 and 9.6

V= (10-9)/(9.6-2.2) = 1/7.4 = .135

a= .135/7.4 = .0183m/s^2

still says I'm wrong.

5. Sep 2, 2008

### Kajayacht

Got it, the main problem here was it involved calculus and my calc professor was sick the day she would've taught us that (that's a big thing)

Anyway, I had to take the slopes as it approached 2.2 and 9.6 which I found to be 5 and -10 plug and chug, shoop dah woop and it came out to -2.03 m/s^2 which the lon-capa thing told me yes.

and I only used up 18 of my 30 tries on it.