Question

Answer 1

Bài 1:

Ta có: \(\left\{{}\begin{matrix}x_1=36t\\x_2=96-28t\end{matrix}\right.\)

Khi 2 xe gặp nhau thì \(x_1=x_2\)

\(\Leftrightarrow36t=96-28t\)

\(\Leftrightarrow t=1,5h\)

Vậy 2 xe gặp nhau cách A \(36\cdot1,5=54\left(km\right)\) lúc \(10h+1,5h=11,5h=11h30p\)

Khi 2 xe cách nhau 32km thì: \(s-s_1-s_2=32\)

\(\Leftrightarrow96-36t-28t=32\)

\(\Leftrightarrow t=1h\)

Bài 2:

Ta có: \(\left\{{}\begin{matrix}x_1=8t\\x_2=120-v_2t\end{matrix}\right.\)

2 xe gặp nhau sau 10s nên: \(x_1=x_2\)

\(\Leftrightarrow8t=120-v_2t\)

\(\Leftrightarrow8\cdot10=120-10v_2\)

\(\Leftrightarrow v_2=4\left(\dfrac{m}{s}\right)\)