Posted by Todd on Friday, July 20, 2012 at 1:52pm.
if 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
x"(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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