- #1
vysero
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A muon enters a region with a speed of 5.00 x 10^6 m/s and then is slowed at a rate of 1.25 x 10^14 m/s^2. A) How far does the muon take to stop?
After some research on the internet I have found an answer to the problem but that does not help me enough. Here is what I have found:
Im suppose to use this equation: V^2 = Vo^2 + 2a(X-Xo)
As I am sure you can see however I don't know what these symbols in this equation mean. In my book it has the second Vo^2 part with the 2 directly over the o which says to me that it's not saying the square of o. So that confuses me. Secondly here is the answer I found:
x = -1/2(Vo^2/a) = -1/2((5.00 x 10^6)^(2)/(-1.25 x 10^14))
This just confuses me further because they didn't show the algebra steps which I kind of need in order to understand how they got the solution. If someone could please help explain this equation to me and show me the steps in the algebra that would be amazing. I mean knowing the answer isn't going to help me on the test.
After some research on the internet I have found an answer to the problem but that does not help me enough. Here is what I have found:
Im suppose to use this equation: V^2 = Vo^2 + 2a(X-Xo)
As I am sure you can see however I don't know what these symbols in this equation mean. In my book it has the second Vo^2 part with the 2 directly over the o which says to me that it's not saying the square of o. So that confuses me. Secondly here is the answer I found:
x = -1/2(Vo^2/a) = -1/2((5.00 x 10^6)^(2)/(-1.25 x 10^14))
This just confuses me further because they didn't show the algebra steps which I kind of need in order to understand how they got the solution. If someone could please help explain this equation to me and show me the steps in the algebra that would be amazing. I mean knowing the answer isn't going to help me on the test.