As a log falls in a waterfall, its velocity is given by $v\left(t\right) = –32t – 18$ 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, $s\left(t\right)$.

   Step 1(a):

   Antidifferentiate $v\left(t\right)$.

   Step 2(a):

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

   Answer (a):

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

2. Where is the log after $t$ seconds? This is the position function. What is its relationship to the velocity function?

   Hint (b):

   Refer to the hint in part (a).