# Homework Help: Simple finding linear acceleration

1. Jun 3, 2014

### mistermill

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

In our lab, we times a cart going 1m. The time was 9.49 s. The cart was pulled by a string, over a pulley, attached to a free-falling mass. The purpose was to calculate the μ, the coefficient of friction.

2. Relevant equations

d = vit + 1/2at$^{2}$

a = (vf-vi)/t

vi = 0 m/s
d = 1.0 m
t = 9.49 s

3. The attempt at a solution

Why does the acceleration come out differently in these two equations?

Is it that the vf is not simply 1m/9.49 seconds?
Is vf = 1/9.49 = 0.11 ?

Because that is the average velocity, and we don't use average velocity because we have a constant acceleration?

In the first equation I get a = 0.02m/s$^{2}$
and in the second equation I get a = 0.01 m/s$^{2}$

I think I know why, but I would like someone to say,"yes that is why you must use the first equation."

Last edited: Jun 3, 2014
2. Jun 3, 2014

### Nathanael

$V_{f}=9.8t$

It's invalid to say $V_{f}=\frac{d}{t}$ because, as you said, that is the average velocity (which is less than the final velocity)

Since initial velocity is zero, the expression $\frac{d}{t}$ will only give you half the final velocity. That is why your two accelerations were different by a factor of 2.