Answer
$$5.63 s$$
Work Step by Step
From Equation 2.8: $x_{\text {leader }}+10.0 \mathrm{m}=x_{2 n d}$
$=>v^{0} t+\frac{1}{2} a t^{2}=v_{0} t+\frac{1}{2} a t^{2}+10.0 m$
$(9.50 m / s) t+\left(\frac{1}{2}\right)\left(1.20 m / s^{2}\right) t^{2}=(11.10 m / s) t+\frac{1}{2}\left(0.00 m / s^{2}\right) t^{2}+10.0 m$
Rearranging the terms of this equation:
$=>0=\frac{1}{2}\left(1.20 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}-(1.60 \mathrm{m} / \mathrm{s}) t-10.0 \mathrm{m}$
$=>5.63 \mathrm{s}=t$
