SOLUTION: Starting from rest, a boulder rolls down a hill with constant acceleration and travels 2.00m during the first second. How far does it travel during the second second? How fast is

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Question 263100: Starting from rest, a boulder rolls down a hill with constant acceleration and travels 2.00m during the first second.
How far does it travel during the second second?
How fast is it moving at the end of the first second?
How fast is it moving at the end of the second second?
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
One formula we can use is
(i) h%28t%29+=+-9.8t%5E2+%2B+V%28sub+0%29+t+%2B+S%28sub+0%29
we know that v(sub0) = 0
and if t= 1 then h(1) = 2
so, w get
2 = -9.8(1)^2 + S(sub0)
S(sub0) = 11.8
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Our equation becomes
(ii) h%28t%29+=+-9.8t%5E2+%2B+11.8
take a derivative to get speed as
(ii) s(t) = -19.6t
take another derivative to get acceleration as
(iii) a(t) = -19.6
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in the second second, we get
h(2) = -9.8*2^2 + 11.8
h(2) = -27.4 or 27.4 feet down hill.
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how fast at end of second second sounds like average speed or
average = (h(2)-h(1)) / (2-1)
or
average = (-27.4-2)/1 = -29.4
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how fast at end of first second sounds like average speed or
average = (h(1)-h(0)) / (1-0)
or
average = (2-11.8)/1 = -8.2
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