Answer
$\frac{a_y}{a_x} = 1.50$
Work Step by Step
We can find an expression for $a_x$:
$x = x_0+\frac{1}{2}a_x~(\Delta t_1)^2$
$12.0~m = 4.00~m+\frac{1}{2}a_x~(\Delta t_1)^2$
$8.00~m = \frac{1}{2}a_x~(\Delta t_1)^2$
$a_x = \frac{(8.00~m)(2)}{(\Delta t_1)^2}$
$a_x = \frac{16.0~m}{(\Delta t_1)^2}$
We can find an expression for $a_y$:
$y = y_0+\frac{1}{2}a_y~(\Delta t_1)^2$
$18.0~m = 6.00~m+\frac{1}{2}a_y~(\Delta t_1)^2$
$12.0~m = \frac{1}{2}a_y~(\Delta t_1)^2$
$a_y = \frac{(12.0~m)(2)}{(\Delta t_1)^2}$
$a_y = \frac{24.0~m}{(\Delta t_1)^2}$
We can find $\frac{a_y}{a_x}$:
$\frac{a_y}{a_x} = \frac{\frac{24.0~m}{(\Delta t_1)^2}}{\frac{16.0~m}{(\Delta t_1)^2}} = \frac{24.0~m}{16.0~m} = 1.50$
