Chapter 5 - The Integral - 5.6 Net Change as the Integral of a Rate of Change - Exercises - Page 268: 8

$622.9$ m

let $v(t)=100-9.8t$ $\int_0^{15} |100-9.8t| dt$ = $\int_0^{\frac{100}{9.8}} |100-9.8t| dt$ + $\int_{\frac{100}{9.8}}^{15} |100-9.8t| dt$ = $(100t-4.9t^{2} |_0^{\frac{100}{9.8}}$+$(100t-4.9t^{2} |_{\frac{100}{9.8}}^{15}$ = $622.9$ m