Answer
52.8 m
Work Step by Step
$x=ut+\frac{1}{2}at^{2}$
For both knights, u=0
$x=\frac{1}{2}at^{2}$
$t=\sqrt \frac{2x}{a}$ since $t\gt0$
Distance traveled by Sir George = $x_{G}$
Distance traveled by Sir Alfred = $88.0-x_{G}$
Sir George:
$t=\sqrt \frac{2(x_{G})}{0.300}=\sqrt \frac{20x_{G}}{3}$
Sir Alfred:
$t=\sqrt \frac{2(88.0-x_{G})}{0.200}=\sqrt (10(88.0-x_{G}))$
We now use a system of linear equations:
$\sqrt \frac{20x_{G}}{3}=\sqrt (10(88.0-x_{G}))$
$ \frac{20x_{G}}{3}=10(88.0-x_{G})$
$20x_{G}=30(88.0-x_{G})$
$20x_{G}=2640-30x_{G}$
$50x_{G}=2640$
$x_{G}=52.8\; m$
