At \(t=0\) a particle starts at rest and moves along a line in such a way that at time \(t\) its acceleration is \(36t^2\) feet per second per second. Through how many feet does the particle move during the first 2 seconds?

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The particle's acceleration is given:

$$ a(t)=36t^2$$

Integrating with respect to time to obtain the velocity equation:

$$ \int a(t)   dt = \int 36t^2   dt $$ $$ v(t)= 12t^3+C $$

Since the velocity starts at rest, we are given \(v(0)=0\),

$$ v(0)=12(0)^3+C=0 $$ $$ C=0$$ $$ v(t)=12t^3 $$

Integrating again to obtain the position equation:

$$ \int v(t)   dt= \int 12t^3   dt $$ $$ s(t) = 3t^4 + C$$

To find the distance traveled from \(t=0\) to \(t=2\),

$$ s(2)-s(0)=3(2)^4-3(0)^4 $$ $$ = \boxed{48} $$