## anonymous one year ago 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?

1. anonymous

$x _{f}=x _{0}+vt+\frac{ 1 }{ 2 }at^2$

2. anonymous

What do I use as acceleration? I know that the answer is -1.0 m, I just do not see it yet

3. anonymous

Acceleration unit is m/s^2. What makes you think you know the answer? I solved it and it was not 1

4. anonymous

nevermind, I got it. .32 m/s^2

5. 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.

6. anonymous

I was looking at the other problem for some reason, this one i did get with the formula :) Thanks!

7. anonymous

Ah, I see. Nice job No problem!

8. anonymous

I thought I had copied the other problem here that I cant figure out haha

9. anonymous

Oh no wonder. Actually the answer is -0.33 not positive 0.32. Forgot to tell you

10. 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

11. 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?

12. 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

13. anonymous

Nope, the first formula is -0.33 m/s^2 The second formula is 0.33m/s^2

14. 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

15. 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.

16. anonymous

Yes it is better and way faster. I did it both ways and I always got .32 m/s^2

17. anonymous

Which formula? Mines or yours?

18. anonymous

yours... here is what I did with yours,

19. anonymous

Yea, it is a shortcut. That is why they made it separate formula.

20. sohailiftikhar

as he said the acc is constant you it means increase in velocity in equal interval of time is same ..