Answer
a) 11 s
b) 30 m
Work Step by Step
$x=ut+\frac{1}{2}at^{2}$
For both players, u=0
$x=\frac{1}{2}at^{2}$
Distance traveled by first player = $x_{1}$
Distance traveled by second player = $48-x_{1}$
First player:
$x_{1}=\frac{1}{2}(0.50)t^{2}=0.25t^{2}$
Second player:
$48-x_{1}=\frac{1}{2}(0.30)t^{2}$
$x_{1}=48-0.15t^{2}$
We now use a system of linear equations:
a)
$0.25t^{2}=48-0.15t^{2}$
$0.40t^{2}=48$
$t^{2}=120$
$t=\sqrt (120)$ since $t\gt0$
$t\approx 11\; s$
b)
$x_{1}=0.25t^{2}$
$x_{1}=0.25(\sqrt 120)^{2}=30\; m$
