Answer
The cart rolled 6.9 meters.
Work Step by Step
The cart's acceleration is $g~sin(3^{\circ})$. We can compare the distance covered by the cart and Julie to find the time $t$.
$\frac{1}{2}at^2 = \frac{1}{2}[g~sin(3^{\circ})]~t^2+20~m$
$\frac{1}{2}at^2 - \frac{1}{2}[g~sin(3^{\circ})]~t^2=20~m$
$t^2 = \frac{40~m}{a-g~sin(3^{\circ})}$
$t = \sqrt{\frac{40~m}{2.0~m/s^2-(9.80~m/s^2)~sin(3^{\circ})}}$
$t = 5.186~s$
We can find the distance the cart has rolled.
$x = \frac{1}{2}(g~sin(3^{\circ}))~t^2$
$x = \frac{1}{2}(9.80~m/s^2)~sin(3^{\circ})~(5.186~s)^2$
$x = 6.9~m$
The cart rolled 6.9 meters.
