Modulo della forza F:
F = √(2^2 + (-6)^2)---> F = 2·√10 N
F = m·a---> a = f/m
m = 2 kg
a = 2·√10/2 = √10 m/s^2
s = 1/2·a·t^2
s = 1/2·√10·1.2^2 = 2.277 m
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Risultante $\small F= \sqrt{2^2+|-6|^2}= \sqrt{2^2+6^2} = 2\sqrt{10}\,N\;(\approx{6,3246}\,N);$
accelerazione $\small a= \dfrac{F}{m}= \dfrac{\cancel2\sqrt{10}}{\cancel2}=\sqrt{10}\,m/s^2\;(\approx{3,162}\,m/s^2);$
distanza percorsa $\small S= \dfrac{a·t^2}{2} = \dfrac{3,162×1,2^2}{2} \approx{2,277}\,m.$