Đáp án:

 \(t = \dfrac{{23}}{{14}}\left( s \right)\)

Giải thích các bước giải:

Ta có: 

\(\begin{array}{l}
{x_1} = 30 - 20t\\
{x_2} = 50 - 30t\\
{d^2} = x_1^2 + x_2^2 - 2{x_1}{x_2}\cos 60\\
 \Rightarrow {d^2} = {\left( {30 - 20t} \right)^2} + {\left( {50 - 30t} \right)^2} - 2\left( {30 - 20t} \right)\left( {50 - 30t} \right).\dfrac{1}{2}\\
 \Rightarrow {d^2} = 900 - 1200t + 400{t^2} + 2500 - 3000t + 900{t^2} - \left( {1500 - 1900t + 600{t^2}} \right)\\
 \Rightarrow {d^2} = 700{t^2} - 2300t + 1900\\
 \Rightarrow {d^2} = 700\left( {{t^2} - 2.\dfrac{{23}}{{14}}.t + \dfrac{{529}}{{196}}} \right) + 10,71\\
 \Rightarrow {d^2} = 700{\left( {t - \dfrac{{23}}{{14}}} \right)^2} + 10,71\\
{\left( {t - \dfrac{{23}}{{14}}} \right)^2} \ge 0 \Rightarrow {d^2} \ge 10,71 \Rightarrow d \ge 3,27\left( m \right)
\end{array}\)

Dấu = xảy ra khi: \(t - \dfrac{{23}}{{14}} = 0 \Rightarrow t = \dfrac{{23}}{{14}}\left( s \right)\)