\(\overrightarrow{ a }=\frac{\overrightarrow{ F }}{ m }=\frac{20 \hat{ i }+10 \hat{ j }}{2} \Rightarrow 10 \hat{ i }+5 \hat{ j }\)

\(\therefore \overrightarrow{ s }=\frac{1}{2} \overrightarrow{ a } t ^{2}=\frac{1}{2}(10 \hat{ i }+5 \hat{ j }) \times(10)^{2}\)

\(\Rightarrow 50(10 \hat{ i }+5 \hat{ j }) m\)

\(\therefore\) Displacement along \(x\) -axis

\(\Rightarrow 50 \times 10 \Rightarrow 500 m\)