Chapter 12 - Kinematics of a Particle - Section 12.2 - Rectilinear Kinematics: Continuous Motion - Problems - Page 18: 18

$t=5.62$ s

$a ds = v dv$ $\int_0^s(6+0.02s)ds=\int_0^v vdv$ $6s+0.01s^2=\frac{1}{2}v^2$ $v=\sqrt{12s+0.02s^2}$ $ds=vdt$ $\int_0^{100}(\frac{ds}{\sqrt{12s+0.02s^2}})=\int_0^t dt$ $t=\frac{1}{\sqrt{0.02}}\ln [\sqrt{12s+0.02s^2}+s\sqrt{0.02}+\frac{12}{2\sqrt{0.02}}]^{100}_{0}$ $t=5.62$ s