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anonymous
 one year ago
A particle confined to motion along the xaxis moves with constant acceleration from
x = 2.0 m
to
x = 8.0 m
during a 2.5s 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
 one year ago
A particle confined to motion along the xaxis moves with constant acceleration from x = 2.0 m to x = 8.0 m during a 2.5s 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?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x _{f}=x _{0}+vt+\frac{ 1 }{ 2 }at^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What do I use as acceleration? I know that the answer is 1.0 m, I just do not see it yet

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Acceleration unit is m/s^2. What makes you think you know the answer? I solved it and it was not 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nevermind, I got it. .32 m/s^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Um, 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
 one year ago
Best ResponseYou've already chosen the best response.0I was looking at the other problem for some reason, this one i did get with the formula :) Thanks!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, I see. Nice job No problem!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I thought I had copied the other problem here that I cant figure out haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh no wonder. Actually the answer is 0.33 not positive 0.32. Forgot to tell you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0Nope, the first formula is 0.33 m/s^2 The second formula is 0.33m/s^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0Could 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
 one year ago
Best ResponseYou've already chosen the best response.0Yes it is better and way faster. I did it both ways and I always got .32 m/s^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Which formula? Mines or yours?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yours... here is what I did with yours,

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea, it is a shortcut. That is why they made it separate formula.

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.2as he said the acc is constant you it means increase in velocity in equal interval of time is same ..
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