Solving a Physics Problem: Finding Acceleration Without Time | Expert Tips"

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Homework Help Overview

The problem involves finding the acceleration of an object moving with constant acceleration, given its velocities at two different positions without the time variable. The original poster presents their calculations using the kinematic equation but expresses concern over the validity of their result.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the kinematic equation Vf^2 = Vi^2 + 2a(xf - xi) to find acceleration. Some participants question the correctness of the calculations and the assumptions made regarding the problem setup.

Discussion Status

Participants are actively discussing the calculations and potential errors. There is a mix of agreement on the method used, but also uncertainty about the results. Some participants suggest re-evaluating the question and checking for possible mistakes in the setup.

Contextual Notes

The original poster mentions that the problem is part of a three-part question, indicating that the resolution of this first part is necessary for proceeding with the others. There is also mention of submission limits and concerns about significant figures affecting the grading of the answer.

sundrops
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kay well here goes:

An object moving with constant acceleration has a velocity v = 55.2m/s when it is at x = 6.9m and a v = 82.4m/s when it is at x = 12.4m. What is its acceleration?

Now this problem qould be simple if time were given but it isn't so I'm going to need a little bit of help please.

What I did with this one was I plugged the numbers into the following equation:
Vf^2 = Vi^2 +2a(xf - xi)

which gave me,

(82.4)^2 = (55.2)^2 + 2a(12.4 - 6.9)
6789.76 - 3047.04 = 2a(5.5)
a = 340.25m/s^2

whish is needless to say, well...impossible. Is there another way to attack this problem or maybe if someone could please point out my mistake?
 
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That should be right.
 
but it isn't! I submitted my answer (my h/w is done on the net) and it said it was incorrect. too hard!
 
Hmm, let me think for a bit.
 
thanks for taking the time! :)
 
I don't know what's wrong, that answer should be right, maybe I missed something.
 
Last edited by a moderator:
sundrops said:
thanks for taking the time! :)

Can you double check the question? Maybe it was 55.2 m/s at 12.4m and 82.4 m/s at 6.9m?
 
nope its right.

and like its a three part question so I can't attempt the other two parts until i get the first one correct - maybe I'll try submitting it again (we get 3 tries before its marked as wrong)
 
but can something accelerate that quickly? it seems like its accelerating a little too fast... hmmm
 
  • #10
Don't submit it yet! futbol got a different answer! How did you get 134.52? I get the same thing as sundrops: 340.25 m/s^2
 
  • #11
good thing I tried submitting again! I had too many digits in my answer - not the proper amount of sig figs I guess. So the answer is correct! [340.2] (man I hate computers *grumbles*)

thanks to everyone who posted! :D
 
  • #12
wow! you guys are awesome - thanks again!
 
  • #13
Lol, you confused the hell out of me :P Time to get some rest now :P
 
  • #14
sundrops said:
good thing I tried submitting again! I had too many digits in my answer - not the proper amount of sig figs I guess. So the answer is correct! [340.2] (man I hate computers *grumbles*)

thanks to everyone who posted! :D

Cool! Yeesh, that's annoying though.
 
  • #15
g'nite futbol thanks again!
 
  • #16
it sure is! whatever happened to the good ol' days where h/w was done by hand - and marked by people??! lol
 
  • #17
thanks again!
 

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