Math
posted by Todd .
the acceleration of a particle is defined by the equation a= 4t^2 with an initial velocity of two millimeters per second. Find the exact equation for the position of the particle if its initial position was zero.

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=4t^2
Integrate two times and substitute the initial conditions to find the required equation:
x'(t)
=∫a dt
=∫(4t^2)dt
=4t(t^3/3)+C1
From x'(t)=2 mmt^1, we get
2 mmt^1 = 4(0)(0^3/3)+C1
=>
C1=2mmt^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)