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Let's see here, our givens are:
a = 2.8m/s^2 initial velocity = 0m/s final velocity = 37m/s Remember that change in velocity = a * t; and we need to know time before we can go further. So, solve for t by doing delta v / a = t and get 13.214... seconds. So now that we have our time, we can solve for our delta x(change in position). We do this by using the equation: delta x = (v initial)t + .5(at^2), but since initial velocity is 0, it simplifies to: delta x = .5(at^2), which equals 243.9m and then do your sig figs. That AP Physics class I am taking is pretty insane...(just wait until you get to friction) There are a few other ways to get this, but they all involve basically the same thing. Above was done using the v^2=2a(delta x) method |
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Let's see here, our givens are: |
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Just use v^2 = u^2 +2as like IS said? With the correct numbers v^2/2a is 244.26m. In the past, when I've been stumped or out of time, I've actually got marks for just writing down the varables, like that! |
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