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March 1, 2015

March 1, 2015

Posted by **Todd** on Friday, July 20, 2012 at 1:52pm.

- Math -
**Reiny**, Friday, July 20, 2012 at 4:32pmif a = 4 - t^2 , then

v = 4t - (1/3)t^3 + c

given: when t=0, v = 2 mm/s

2 = 0 - 0 + c

c = 2

v = 4t - (1/3)t^2 + 2

s = 2t^2 - (1/12)t^4 + 2t + k

when t=0 , s = 0 , so k = 0

s = 2t^2 - (1/12)t^4 + 2t

check by taking the derivative twice

- Math -
**MathMate**, Friday, July 20, 2012 at 4:35pmx"(t)=d²x/da²=a=4-t^2

Integrate two times and substitute the initial conditions to find the required equation:

x'(t)

=∫a dt

=∫(4-t^2)dt

=4t-(t^3/3)+C1

From x'(t)=2 mm-t^-1, we get

2 mm-t^-1 = 4(0)-(0^3/3)+C1

=>

C1=2mm-t^-1

x(t)=∫x'(t)dt

=∫(4t-(t^3/3)+2) dt

=2t²-(1/12)t^4+2t+C2

x(0)=0 => C2=0

=>

x(t)=2t² -(1/12)t^4 +2t

(in millimetres)

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