From points A and B, the distance between which is 10 km, two cyclists drive towards each other at the same time. After their meeting, the first arrives at point B in 48 minutes, and the second at point A in 27 minutes. How much time (in minutes) elapsed from the start of the cyclists’ movement to their meeting, if the cyclists were moving at a constant speed?
1. The distance between points A and B is equal to: S = 10 km;
2. Time of movement of cyclists until the moment of meeting: T min;
3. Path traveled by the first cyclist to the meeting point: S1 km;
4. Path traveled by the second cyclist before the meeting: S2 km;
5. After the meeting, the first cyclist arrived at point B in: T1 = 48 minutes;
T1 = S2 / V1 = (V2 * T) / V1 = 48 min;
T = (48 * V1) / V2 min;
6. The second cyclist arrived at point A in: T2 = 27 minutes;
T2 = S1 / V2 = (V1 * T) / V2 = 27 min;
T = (27 * V2) / V1 min;
7. Equate the right-hand sides:
T = (48 * V1) / V2 = (27 * V2) / V1;
16 * V1 ^ 2 = 9 * V2 ^ 2;
(4 * V1) ^ 2 = (3 * V2) ^ 2;
4 * V1 = 3 * V2;
V1 / V2 = 3/4;
8. Time T:
T = 48 * (V1 / V2) = 48 * (3/4) = 36 min;
Or T = 27 * (V2 / V1) = 27 * (4/3) = 36 minutes.
Answer: 36 minutes passed from the beginning of the movement of cyclists to their meeting.