Problem P. 4970. (November 2017)

P. 4970. A vehicle travels from city \(\displaystyle A\) to city \(\displaystyle B\). During the first part of the journey its average speed is \(\displaystyle v_1\) whilst in the rest of the journey it is \(\displaystyle v_2\). By what factor is the length of the second part of the journey longer than that of the first part if the average speed calculated for the whole journey is

\(\displaystyle a)\) the arithmetic mean;

\(\displaystyle b)\) the geometric mean;

\(\displaystyle c)\) the harmonic mean;

\(\displaystyle d)\) the weighted arithmetic mean of ratio \(\displaystyle 1:k\) of the average speeds calculated for the two parts of the journey?

(4 pont)

Deadline expired on December 11, 2017.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Az első, \(\displaystyle s_1\) hosszúságú szakaszt \(\displaystyle t_1=s_1/v_1\) idő alatt teszi meg a jármű, a másodikat \(\displaystyle t_2=s_2/v_2\) idő alatt. A teljes útra vonatkozó átlagsebessége:

\(\displaystyle \bar{v}=\frac{s_1+s_2}{t_1+t_2}=\frac{s_1+s_2}{\frac{s_1}{v_1}+\frac{s_2}{v_2}}.\)

Innen az útszakaszok hosszának aránya:

\(\displaystyle \frac{s_2}{s_1}=\frac{1-\frac{\bar{v}}{v_1}}{\frac{\bar{v}}{v_2}-1}. \)

\(\displaystyle a)\) Ha \(\displaystyle \bar{v}=\frac{v_1+v_2}{2},\) akkor \(\displaystyle \frac{s_2}{s_1}= \frac{v_2}{v_1}\). Ilyenkor a \(\displaystyle t_1\) időtartam megegyezik \(\displaystyle t_2\)-vel.

\(\displaystyle b)\) Ha \(\displaystyle \bar{v}= {\sqrt{v_1 v_2}} ,\) akkor \(\displaystyle \frac{s_2}{s_1}= \sqrt{\frac{v_2}{v_1}} \).

\(\displaystyle c)\) Ha \(\displaystyle \bar{v}=\frac{2v_1 v_2}{v_1+ v_2},\) akkor \(\displaystyle \frac{s_2}{s_1}= 1\).

\(\displaystyle d)\) Ha \(\displaystyle \bar{v}=\frac{v_1+kv_2}{1+k},\) akkor \(\displaystyle \frac{s_2}{s_1}= k \frac{v_2}{v_1} \).

Statistics:

84 students sent a solution.
4 points:	Balaskó Dominik, Bartók Imre, Békési Ábel, Bíró Dániel, Bukor Benedek, Conrád Márk, Czett Mátyás, Debreczeni Tibor, Fajszi Bulcsú, Fekete András Albert, Fekete Balázs Attila, Garamvölgyi István Attila, Geretovszky Anna, Gulácsi Máté, Hervay Bence, Illés Gergely, Jánosdeák Márk, Jánosik Áron, Keltai Dóra, Kolontári Péter, Kondákor Márk, Kovács 111 Bence, Kovács Gergely Balázs, Lipták Gergő, Mamuzsics Gergő Bence, Markó Gábor, Máth Benedek, Merkl Gergely, Merkl Levente, Molnár 957 Barnabás, Morvai Orsolya, Németh Csaba Tibor, Olosz Adél, Pácsonyi Péter, Pszota Máté, Rusvai Miklós, Sal Dávid, Schneider Anna, Schrott Márton, Selmi Bálint, Seres Soma, Stirling András, Tafferner Zoltán, Takács Árpád, Tófalusi Ádám, Turcsányi Máté.
3 points:	18 students.
2 points:	6 students.
1 point:	5 students.
0 point:	9 students.

Problems in Physics of KöMaL, November 2017