Chapter 4 - Integration - 4.4 Exercises - Page 289: 57

$\approx 540$ ft.

$v(t)=\displaystyle \frac{d}{dt}[s(t)]$ so the distance traveled is $\displaystyle \int_{0}^{8}v(t)dt$. We estimate the area under the curve from $0\leq t\leq 8$ to be around 18 squares, each square representing (2s)$\times$(15 ft/s)=30 ft 18(30) $\approx 540$ ft.