Đáp án:

a) \(\dfrac{{{d_0}{v_A}}}{{\sqrt {v_A^2 + v_B^2} }}\)

b) \(t = \dfrac{{{d_0}}}{{\sqrt {v_B^2 - v_A^2} }}\)

Giải thích các bước giải:

a) Khoảng cách giữa 2 tàu là:

\(\begin{array}{l}
d = \sqrt {{{\left( {{d_0} - {v_B}t} \right)}^2} + {{\left( {{v_A}t} \right)}^2}} \\
 \Rightarrow d = \sqrt {{t^2}\left( {v_A^2 + v_B^2} \right) - 2{d_0}{v_B}t + d_0^2} 
\end{array}\)

Ta có:

\(\begin{array}{l}
d\min  \Leftrightarrow t = \dfrac{{{d_0}{v_B}}}{{v_A^2 + v_B^2}}\\
 \Rightarrow d = \sqrt {{{\left( {\dfrac{{{d_0}{v_B}}}{{v_A^2 + v_B^2}}} \right)}^2}\left( {v_A^2 + v_B^2} \right) - 2{d_0}{v_B}.\dfrac{{{d_0}{v_B}}}{{v_A^2 + v_B^2}} + d_0^2} \\
 \Rightarrow d = \sqrt {\dfrac{{{{\left( {{d_0}{v_B}} \right)}^2}}}{{v_A^2 + v_B^2}} - 2.\dfrac{{{{\left( {{d_0}{v_B}} \right)}^2}}}{{v_A^2 + v_B^2}} + d_0^2} \\
 \Rightarrow d = \sqrt {\dfrac{{d_0^2\left( {v_A^2 + v_B^2} \right) - {{\left( {{d_0}{v_B}} \right)}^2}}}{{v_A^2 + v_B^2}}} \\
 \Rightarrow d = \dfrac{{{d_0}{v_A}}}{{\sqrt {v_A^2 + v_B^2} }}
\end{array}\)

b) Để tàu B bắt kịp tàu A:

\(\begin{array}{l}
{\left( {{v_A}t} \right)^2} + d_0^2 = {\left( {{v_B}t} \right)^2}\\
 \Rightarrow {t^2}\left( {v_B^2 - v_A^2} \right) = d_0^2\\
 \Rightarrow t = \dfrac{{{d_0}}}{{\sqrt {v_B^2 - v_A^2} }}
\end{array}\)

Phương chạy hợp với AB một góc là: 

\(\cos \alpha  = \dfrac{{{d_0}}}{{{v_B}t}} = \dfrac{{{d_0}}}{{{v_B}.\dfrac{{{d_0}}}{{\sqrt {v_B^2 - v_A^2} }}}} = \dfrac{{\sqrt {v_B^2 - v_A^2} }}{{{v_B}}}\)