# Homework Help: Finding the acceleration of an object w/ kinematic equations

1. Apr 3, 2006

### oooo

This is a problem that appeared quite simple but is for some reason incorrect. I am not quite sure where I made my mistake so I spelled out my thought process below:

Question:
"Suppose that a cart was released from rest at a position of 3.5 cm down the ramp and continued to travel down the ramp to a position of 6.7 cm.The time the cart took to pass through these points is 0.2545 seconds.
What is the acceleration between these two points on the ramp?"

What I did:
I used the kinematic equation of: displacement = initial velocity* time passed + 1/2 * acceleration * (time passed )^2

I knew the initial velocity = 0 because it was released from rest and the time passed was given. I had to solve for the acceleration.
So, I converted my centimeters to meters and plugged in my values into the equation to get:
0.32 = 0*0.2545 + 1/2* a*(0.2545)^2
and then 0.32= 1/2 *a* 0.06477025
and solved for a. I got the answer of 9.88 m/s^2 approx. This, however was marked incorrect.

The question then led me to solve for the velocity at given periods of time. For example, "what is the velocity at t=0.509 seconds?" If I had a correct acceleration, wouldn't I just use the kinematic equation of acceleration= change in velocity/change in time?

2. Apr 3, 2006

### Kurdt

Staff Emeritus
you used 0.32 for displacement when 3.2 cm is 0.032 meters. so the acceleration should be 0.988 m/s^2. v=a*t for the last bit.

3. Apr 3, 2006

### oooo

Now I feel kind of stupid, but that made a huge difference. Thanks

4. Apr 4, 2006

### Kurdt

Staff Emeritus
No worries. We all ocassionally go insane while staring the obvious in the face. I've certainly done it many many times. Just need to take a step back now and again.