As a log falls in a waterfall, its velocity is $v(t) = -32t-18$ in feet per second. The position of the log at time $t = 0$ was at the top of the waterfall, $500$ feet above sea level. 

1. Where is the log after $1$ second? $2$ seconds? $3$ seconds?

   Hint (a):

   You need to find the position function, $s(t)$.

   Step 1(a):

   Antidifferentiate $v(t)$.

   Step 2(a):

   Solve for $C$, by evaluating the given point $s(0)=500$.

   Answer (a):

   Check your work:
   $s(1) = 466$ ft
   $s(2) = 400$ ft
   $s(3) = 302$ ft

2. Where is the log after $t$ seconds? This is the position function $s(t)$. What is its relationship to $v(t)$ ?

   Hint (b):

   Refer to the hint in part (a).