anonymous
  • anonymous
A particle confined to motion along the x-axis moves with constant acceleration from x = 2.0 m to x = 8.0 m during a 2.5-s time interval. The velocity of the particle at x = 8.0 m is 2.8 m/s. What is the acceleration during this time interval?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[x _{f}=x _{0}+vt+\frac{ 1 }{ 2 }at^2\]
anonymous
  • anonymous
What do I use as acceleration? I know that the answer is -1.0 m, I just do not see it yet
anonymous
  • anonymous
Acceleration unit is m/s^2. What makes you think you know the answer? I solved it and it was not 1

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More answers

anonymous
  • anonymous
nevermind, I got it. .32 m/s^2
anonymous
  • anonymous
Um, the answer is -1.0m is the answer? Are you sure? The units should be m/s^2 Either that is the wrong solution or the solution you saw is another one.
anonymous
  • anonymous
I was looking at the other problem for some reason, this one i did get with the formula :) Thanks!
anonymous
  • anonymous
Ah, I see. Nice job No problem!
anonymous
  • anonymous
I thought I had copied the other problem here that I cant figure out haha
anonymous
  • anonymous
Oh no wonder. Actually the answer is -0.33 not positive 0.32. Forgot to tell you
anonymous
  • anonymous
Why do you get a negative answer? I got positive .032... and all the answers that I get to choose from are positive as well
anonymous
  • anonymous
Oh sorry. Because the correct formula was \[x_{f}=x _{0}+vt-\frac{ 1 }{ 2 }at^2\] I wrote the wrong formula and you solved it?
anonymous
  • anonymous
I had done that one in a different way and got the same result :) I can show you hold on. Also, when I copied the formula I looked at my notes from class too
anonymous
  • anonymous
Nope, the first formula is -0.33 m/s^2 The second formula is 0.33m/s^2
anonymous
  • anonymous
I write even the simplest thing on the notes, so that when going back I know where things came from. Even if I can do it in my head..... I know its long sorry
anonymous
  • anonymous
Could have told me you used a different formula. I wrote the first formula wrong, but it is better to use mines, since it just 1 formula. It is faster.
anonymous
  • anonymous
Yes it is better and way faster. I did it both ways and I always got .32 m/s^2
anonymous
  • anonymous
Which formula? Mines or yours?
anonymous
  • anonymous
yours... here is what I did with yours,
anonymous
  • anonymous
Yea, it is a shortcut. That is why they made it separate formula.
sohailiftikhar
  • sohailiftikhar
as he said the acc is constant you it means increase in velocity in equal interval of time is same ..

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