X=Y/2
X=Y
X=Y/4
X=Y/3

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Can use basic kinematics equation $\Delta x = v_0 t + \frac{1}{2} a t^2$

From position 0 to position X+Y in 20 seconds:

$\begin{array}{c} (X+Y)-0 = (0)(20) + \frac{1}{2}a(20)^2 = 200a \\ X + Y = 200a \\ \end{array}$

From position 0 to X in 10 seconds:

$\begin{array}{c} X-0 = (0)(10) + \frac{1}{2}a(10)^2 = 50a \\ a = \frac{X}{50} \\ \end{array}$

Therefore:

$\begin{array}{c} X + Y = 200 \cdot \frac{X}{50} \\ X + Y = 4X \\ Y = 3X \\ \end{array}$