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

Math  Reiny, Friday, July 20, 2012 at 4:32pm
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

Math  MathMate, Friday, July 20, 2012 at 4:35pm
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)
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