# Kinematics Eq'n Problem

1. Sep 17, 2013

### harujina

1. The problem statement, all variables and given/known data
A student riding a bicycle begins to go downhill and accelerates at a rate of 1.8m/s2. If the acceleration lasts for 2.4s, and the final speed of the bicycle is 10.2m/s, at what speed was he initially travelling?

a = 1.8
t = 2.4
vf = 10.2

vi = ?

2. Relevant equations
a = vf - vi / t

3. The attempt at a solution
I first isolated vi to solve for the problem and got : vi = vf-a/t and then plugged in the variables but it didn't work.

2. Sep 17, 2013

### SteamKing

Staff Emeritus
a = (vf - vi) / t is the correct equation. Parentheses make a difference. Also, using correct algebra.

However, this equation is good only over short time intervals.

3. Sep 17, 2013

### collinsmark

Be careful with your parenthesis, they make a difference.

the correct equation for uniform acceleration is

a = ( vf - vi )/t

There's an algebra mistake in there somewhere. It shouldn't contain the term a divided by t. It's something else.

[Edit: SteamKing beat me to the response.]

4. Sep 17, 2013

### harujina

Ok, then is this correct?:

a = (vf - vi)/t
at = vf - vi
at + vi = vf
vi = vf - at

5. Sep 17, 2013

### SteamKing

Staff Emeritus
Looks Good

6. Sep 17, 2013

### harujina

Ohh, I was so confused since I thought I had to divide t from both sides since it was a*t...
Anyways, thank you!!