Vận tốc trung bình của người đi xe đạp trên đoạn đường AB là :

(14+16+8) : 3 = 12,6666..... (km/giờ) \(\approx\)12,67 (km/giờ)

Vậy vận tốc trung bình của người đi xe đạp trên đoạn đường AB là 12,67 km/giờ

t₁ = \(\dfrac{S_1}{v_1}\)= \(\dfrac{S}{\dfrac{3}{v_1}}\)
t = \(\dfrac{S_2}{v_2}\)=\(\dfrac{S}{\dfrac{3}{v_2}}\)
t =\(\dfrac{S_3}{v_3}\)=\(\dfrac{S}{\dfrac{3}{v_3}}\)
​⇒vtb = \(\dfrac{S}{\dfrac{S}{\dfrac{3}{v_1}}+\dfrac{S}{\dfrac{3}{v_2}}+\dfrac{S}{\dfrac{3}{v_3}}}\)=\(\dfrac{S}{\dfrac{S}{3v_1}+\dfrac{S}{3v_2}+\dfrac{S}{3v_3}}\)
=\(\dfrac{S}{\text{S( }\dfrac{1}{v_1}+\dfrac{1}{v_2}+\dfrac{1}{v_3}}\) = \(\dfrac{3v_1v_2v_3}{v_2v_3+v_1v_3+v_1v_2}\)
= \(\dfrac{3\cdot12\cdot15\cdot19}{15\cdot19+12\cdot19+12\cdot15}\)=\(\dfrac{10260}{84825}\)= ... km/h

Đáp án B

Thời gian đi trên 1/3 đoạn đường đầu là:

\(t_1=\dfrac{AB}{3v_1}=\dfrac{AB}{3.14}=\dfrac{AB}{42}\left(h\right)\)

Thời gian đi trên 1/3 đoạn đường tiếp theo là:

\(t_2=\dfrac{AB}{3v_2}=\dfrac{AB}{3.16}=\dfrac{AB}{48}\left(h\right)\)

Thời gian đi trên 1/3 đoạn đường cuối cùng là:

\(t_3=\dfrac{AB}{3v_3}=\dfrac{AB}{3.8}=\dfrac{AB}{24}\left(h\right)\)

Vận tốc trung bình trên cả quãng đường là:

\(v_{tb}=\dfrac{AB}{t_1+t_2+t_3}=\dfrac{AB}{\dfrac{AB}{42}+\dfrac{AB}{48}+\dfrac{AB}{24}}=\dfrac{AB}{AB\left(\dfrac{1}{42}+\dfrac{1}{48}+\dfrac{1}{24}\right)}=\dfrac{1}{\dfrac{1}{42}+\dfrac{1}{48}+\dfrac{1}{24}}=\dfrac{336}{29}\left(km/h\right)\)

\(=>Vtb=\dfrac{S}{t1+t2+t3}=\dfrac{S}{\dfrac{\dfrac{1}{3}S}{v1}+\dfrac{\dfrac{1}{3}S}{v2}+\dfrac{\dfrac{1}{3}S}{v3}}\)

\(=>vtb=\dfrac{S}{\dfrac{S}{42}+\dfrac{S}{48}+\dfrac{S}{24}}=\dfrac{S}{\dfrac{S\left(48.24+42.24+48.42\right)}{48384}}=\dfrac{48384}{4176}=11,6km/h\)

Ta có: \(v_{tb}=\dfrac{S_1+S_2+S_3}{t_1+t_2+t_3}=\dfrac{S}{\dfrac{S}{36}+\dfrac{S}{42}+\dfrac{S}{30}}=\dfrac{1260}{107}\left(km/h\right)\) 

\(v_{tb}=\dfrac{\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}}{\dfrac{\dfrac{1}{3}}{12}+\dfrac{\dfrac{1}{3}}{14}+\dfrac{\dfrac{1}{3}}{10}}=\dfrac{1260}{107}\approx11,776\left(\dfrac{km}{h}\right)\)

Vận tốc trung bình của người đi xe đạp trên cả đoạn đường AB là:         v_tb=\(\dfrac{s_1+s_2+s_3}{t_1+t_2+t_3}=\dfrac{s}{\dfrac{s_1}{v_1}+\dfrac{s_2}{v_2}+\dfrac{s_3}{v_3}}=\dfrac{s}{\dfrac{\dfrac{s}{3}}{v_1}+\dfrac{\dfrac{s}{3}}{v_2}+\dfrac{\dfrac{s}{3}}{v_3}}=\dfrac{s}{\dfrac{s}{3v_1}+\dfrac{s}{3v_2}+\dfrac{s}{3v_3}}=\dfrac{1}{\dfrac{1}{3v_1}+\dfrac{1}{3v_2}+\dfrac{1}{3v_3}}=\dfrac{3}{\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{10}}=11.78\)(km/h) 

\(=>t1=\dfrac{\dfrac{1}{3}S}{12}=\dfrac{S}{36}\left(h\right)\)

\(=>t2=\dfrac{\dfrac{1}{3}S}{8}=\dfrac{S}{24}\left(h\right)\)

\(=>t3=\dfrac{\dfrac{1}{3}S}{6}=\dfrac{S}{18}\left(h\right)\)

\(=>vtb=\dfrac{S}{t1+t2+t3}=\dfrac{S}{\dfrac{S}{36}+\dfrac{S}{24}+\dfrac{S}{18}}=\dfrac{S}{\dfrac{432S+648S+864S}{15552}}\)

\(=\dfrac{S}{\dfrac{1944S}{15552}}=\dfrac{15552}{1944}=8km/h\)

Đáp án D

Tham khảo: