Chapter 4 - Integration - 4.4 Exercises - Page 289: 58

$\approx 150$ ft

$v(t)=\displaystyle \frac{d}{dt}[s(t)]$ so the distance traveled is $\displaystyle \int_{0}^{5}v(t)dt$. We estimate the area under the curve from $0\leq t \leq 5$ to be around $30$ squares, each square representing ($0.5$s)$\times$($10$ ft/s)=$5$ ft $30$($5$) $\approx 150$ ft.