### Home > CALC > Chapter 4 > Lesson 4.4.1 > Problem4-127

4-127.

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?

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

Antidifferentiate $v(t)$.

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

$s(1) = 466$ 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 $v(t)$ ?