# Uniform Acceleration problem dealing with time and x variables?

## Homework Statement

An object moving with uniform acceleration has a velocity of 14.0 cm/s in the positive x direction when its x coordinate is 3.00 cm. If its x coordinate 2.00 s later is -5.00 cm, what is its acceleration?

## Homework Equations

deltaX=V naught t + (1/2)at^2)

## The Attempt at a Solution

2(X-Vt)/t^2 = a

I rewrote the equation and plugged in the variables... Didn't seem to get the right answer. I'm all out of ideas... I really don't know where to go from there... I've spent 30mins on this problem so far.

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Delphi51
Homework Helper
Your formula is correct. What answer did you get? If you show the details of the substitution it should be easy to pinpoint the error (if there is one).

= 2((-5.00 cm-3.00 cm) - (14.0 cm/s)(2.00 s)) / (14.0 cm/s)^2

= 2(-8 cm - 28 cm) / 196 cm^2/s^2

= -0.367 cm/s^2

Delphi51
Homework Helper
= 2((-5.00 cm-3.00 cm) - (14.0 cm/s)(2.00 s)) / (14.0 cm/s)^2
In place of your 14.0, I had 2. The time is 2, not 14.

In place of your 14.0, I had 2. The time is 2, not 14.
ahhhhhh thank you!!

I am brilliant haha.

Delphi51
Homework Helper
Everyone has made their share of little mistakes!