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?

Question

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?

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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 · Answer

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

anonymous · Answer

Ah, I see.
Nice job

No problem!

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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 · Answer

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 · Answer

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 · Answer

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

anonymous · Answer

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 · Answer

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 · Answer

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

anonymous · Answer

Which formula? 
Mines or yours?

anonymous · Answer

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

anonymous · Answer

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

sohailiftikhar · Answer

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