Finding acceleration given initial velocity and displacement

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OUmecheng
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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.
 

Discussion

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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^2NOT the correct answer.
 
= 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.
 
Delphi51 said:
In place of your 14.0, I had 2. The time is 2, not 14.

ahhhhhh thank you!

I am brilliant haha.
 
Everyone has made their share of little mistakes!