Đáp án:
1D
....
Giải thích các bước giải:
\(\begin{align}
& 1D \\
& 2A \\
& 3D \\
& 4D \\
& 5D \\
& 6A:a=\frac{2{{S}_{1}}}{t_{1}^{2}}\Rightarrow {{t}_{2}}=\sqrt{2{{S}_{2}}.\frac{t_{1}^{2}}{2{{S}_{1}}}}=\sqrt{\frac{{{S}_{2}}}{{{S}_{1}}}}.{{t}_{1}} \\
& 7D:{{v}_{tb}}=\frac{S}{\frac{S}{2.20}+\frac{S}{2.5}}=15m/s \\
& 8B:1.t=1,2(t-(5.60+30))\Rightarrow t=1980s\Rightarrow AB=1980m \\
& 9C:v={{v}_{0}}+a.t \\
& 10C \\
& 11B:{{v}^{2}}-v_{0}^{2}=2.a.S\Rightarrow a=\frac{{{15}^{2}}-{{10}^{2}}}{2.625}=0,1m/{{s}^{2}} \\
& 12C:a.v<0;x={{x}_{0}}+{{v}_{0}}t+\frac{1}{2}.a.{{t}^{2}} \\
& 13C:a=\frac{{{v}^{2}}-v_{0}^{2}}{2.S}=\frac{-{{5}^{2}}}{2.12,5}=1m/{{s}^{2}} \\
& 14A:v={{v}_{0}}+a.t\Rightarrow a=-\frac{10}{4}=-2,5m/{{s}^{2}} \\
& {{S}_{C}}={{S}_{t}}-{{S}_{(t-1)}}=10.4-\frac{1}{2}.2,{{5.4}^{2}}-(10.3-\frac{1}{2}.2,{{5.3}^{2}})=1,25m \\
\end{align}\)
