Vận tốc trung bình:

\(v_{tb}=\dfrac{s_1+s_2}{t_1+t_2}=\dfrac{s}{\dfrac{\dfrac{1}{3}s}{25}+\dfrac{\dfrac{2}{3}s}{30}}=\dfrac{s}{\dfrac{1}{75}s+\dfrac{1}{45}s}=28,125\left(\dfrac{km}{h}\right)\)

12,5m/s=45km/h

Quãng đường còn lại : \(75.\left(1-\dfrac{1}{3}\right)=50\left(km\right)\)

\(\dfrac{1}{3}\) quãng đường : \(75\cdot\dfrac{1}{3}=25\left(km\right)\)

a, \(t_1=\dfrac{s_1}{v_1}=\dfrac{25}{50}=0,5\left(h\right)\)

\(t_2=\dfrac{s_2}{v_2}=\dfrac{50}{45}\approx1,1\left(h\right)\)

b,\(v_{tb}=\dfrac{s_1+s_2}{t_1+t_2}=\dfrac{25+50}{50+45}\approx0,79\left(kmh\right)\)

Bài 1:

a. \(\left\{{}\begin{matrix}v_{tb1}=\dfrac{s_1+s_2}{t_1+t_2}=\dfrac{s}{\dfrac{\dfrac{1}{2}s}{20}+\dfrac{\dfrac{1}{2}s}{60}}=\dfrac{s}{\dfrac{1}{40}s+\dfrac{1}{120}s}=30\left(\dfrac{km}{h}\right)\\v_{tb2}=\dfrac{s_1+s_2}{t_1+t_2}=\dfrac{20\cdot\dfrac{1}{2}t+60\cdot\dfrac{1}{2}t}{t}=\dfrac{10t+30t}{t}=40\left(\dfrac{km}{h}\right)\end{matrix}\right.\)

b. Theo đề bài, ta có: \(t_1-t_2=0,5\)

\(\Leftrightarrow\dfrac{s}{30}-\dfrac{s}{40}=0,5\)

\(\Leftrightarrow s=60km\)

Ta có: \(\left\{{}\begin{matrix}v_1=\dfrac{s_1}{t_1}=\dfrac{6}{0,5}=12\left(\dfrac{km}{h}\right)\\v_2=\dfrac{s_2}{t_2}=\dfrac{14}{1}=14\left(\dfrac{km}{h}\right)\end{matrix}\right.\)

Ta thấy: \(14>12\Rightarrow\) xe 2 đi nhanh hơn

Quãng đường 2 xe đi được sau 30p: \(\left\{{}\begin{matrix}s_1=v_1t_1=12\cdot0,5=6km\\s_2=v_2t_2=14\cdot0,5=7\left(km\right)\end{matrix}\right.\)

Khoảng cách 2 xe: \(\Delta s=s_2-s_1=7-6=1km\)

Tốc độ trung bình:

\(v_{tb}=\dfrac{s_1+s_2}{t_1+t_2}=\dfrac{90+25}{\dfrac{90}{30}+0,5}=\dfrac{230}{7}\left(\dfrac{km}{h}\right)\)

Thời gian đi:

\(t=t_1+t_2=\dfrac{90}{30}+0,5=3,5h=3h30p\)

Vậy đến nơi lúc: \(8h+3h30p=11h30p\)

dùng máy tính =))