Question

Un biciclist a plecat intr-o excursie din orasul A in orasul B.Prima jumatate de drum a mers cu viteza V1 = 12 km/h. Jumatate din timpul ramas a mers cu viteza V2 = 8,0 km/h iar restul drumului a mers pe jos cu viteza V3 = 4,0 km/h. Care a fost viteza medie a biciclistului?

Intrebare de 99 de puncte. Doresc seriozitate maxima + rapiditate.

Answer 1

[tex]\displaystyle \text{Se da:}\\ \\ d_1=\frac d2\\ \\ v_1=12\frac{km}{h}\\ \\ t_2=\frac{t-t_1}{2}\\ \\ v_2=8\frac{km}h\\ \\ d_3=d-d_1-d_2\\ \\ v_3=4\frac{km}{h}\\ \\ v=?\frac{km}{h}\\ \\ \\[/tex]


$\displaystyle \text{Formule:}\\ \\ v=\frac dt\\ \\ \\ t=t_1+t_2+t_3\\ \\ \\ t_1=\frac{d_1}{v_1}\\ \\ t_1=\frac{d}{2\times v_1}\\ \\ \\ t_2=\frac{t-t_1}{2}\\ \\ 2\times t_2=t-\frac{d}{2\times v_1}\\ \\ 2\times t_2=\frac{2\times v_1\times t-d}{2\times v_1}\\ \\ t_2=\frac{2\times v_1\times t-d}{4\times v_1}$


[tex]\displaystyle t_3=\frac{d_3}{v_3}\\ \\ t_3\times v_3=d-d_1-d_2\\ \\ t_3\times v_3=d-\frac{d}2-v_2\times t_2\\ \\ t_3\times v_3=\frac{d}2-v_2\times t_2\\ \\ t_3\times v_3=\frac d2-\frac{v_2\times 2\times v_1\times t-v_2\times d}{4\times v_1}\\ \\ t_3\times v_3=\frac{2\times v_1\times d-2\times v_2\times v_1\times t+v_2\times d}{4\times v_1}\\ \\ t_3=\frac{2\times v_1\times d-2\times v_2\times v_1\times t+v_2\times d}{4\times v_1\times v_3}\\ \\ \\[/tex]


[tex]\displaystyle t=\frac{d}{2\times v_1}+\frac{2\times v_1\times t-d}{4\times v_1}+\frac{2\times v_1\times d-2\times v_2\times v_1\times t+v_2\times d}{4\times v_1\times v_3}\\ \\ t=\frac{2\times v_3\times d+2\times v_1\times v_3\times t-v_3\times d+2\times v_1\times d-2\times v_2\times v_1\times t}{4\times v_1\times v_3}\\ \frac{+v_2\times d}{}\\ \\ 4\times v_1\times v_3\times t=\\v_3\times d+2\times v_1\times v_3\times t+2\times v_1\times d-2\times v_2\times v_1\times t+v_2\times d\\ \\[/tex]

[tex]\displaystyle 2\times v_1\times v_3\times t=v_3\times d+2\times v_1\times d-2\times v_2\times v_1\times t+v_2\times d\\ \\ 2\times v_1\times v_3\times t+2\times v_2\times v_1\times t=v_3\times d+2\times v_1\times d+v_2\times d\\ \\ 2\times v_1\times t\times(v_2+v_3)=d\times(v_3+ 2\times v_1+v_2)\\ \\ t=\frac{d\times(2\times v_1+v_2+v_3)}{2\times v_1\times(v_2+v_3)}\\ \\ \\ [/tex]


[tex]\displaystyle \boxed{v=\frac{2\times v_1\times(v_2+v_3)}{2\times v_1+v_2+v_3}}\\ \\ \\ \text{Calcule:}\\ \\ v=\frac{2\times 12\times(8+4)}{2\times 12+8+4}=8\frac {km}h[/tex]