Answer
$13.6\;\rm m$
Work Step by Step
At the moment Ann starts to move forward, the cart will move backward to maintain the initial momentum, of this isolated system, of zero. The system here is the girl Ann plus the cart.
$$0=m_{Ann}v_{Ann}+m_{cart}v_{cart} $$
where $v_{Ann}=v_{cart}+5$;
$$0=m_{Ann}(v_{cart}+5)+m_{cart}v_{cart} $$
$$0=m_{Ann}v_{cart} +5 m_{Ann}+m_{cart}v_{cart} $$
Solving for $v_{cart}$;
$$v_{cart} =\dfrac{-5 m_{Ann}}{m_{Ann} +m_{cart}}=\dfrac{-5\times 50}{50+500}$$
$$v_{cart} =\bf -0.45\;\rm m/s$$
Now we need to find the time Ann would take to reach the other end of the cart since she is moving at a speed of 5 m/s relative to the cart.
$$\Delta t=\dfrac{L_{cart}}{v_{Ann|cart}}=\dfrac{15}{5}=\bf 3\;\rm s$$
Now we need to find the distance traveled by the cart backward during this time and then we subtract this result from the 15-long of the cart and this will be the distance traveled by the girl relative to the ground.
Thus,
$$\Delta x_{cart}=v_{cart}t=-0.4545\times 3=-1.36\;\rm m$$
Therefore, the distance traveled by the girl relative to the ground is given by
$$\Delta x_{girl}=L_{cart}+\Delta x_{cart}=15-1.36=\color{red}{\bf13.6 }\;\rm m$$
