From point A to point B, the distance between which is 40 km, a motorist and a cyclist left at the same time.
From point A to point B, the distance between which is 40 km, a motorist and a cyclist left at the same time. It is known that a motorist travels 50 km more per hour than a cyclist. Determine the speed of the cyclist if it is known that he arrived at point B 3 hours 20 minutes later than the motorist.
1. Distance between points: S = 40 km;
2. Speed of the motorist: Vа km / h;
3. Cyclist speed: V in km / h;
4. According to the problem: Va = (Vв + 50) km / h;
5. Driving time of the motorist: That hour;
6. Cyclist trip time: TV hour;
7. By the condition of the problem:
TV – Ta = 3 hours 20 minutes (10/3 hours);
TV – Ta = S / Vv – S / Va = S / Vv – S / (Vv + 50) =
40 / Vv – 40 / (Vv + 50) = 10/3;
200 / (Vв * (Vв + 50)) = 1/3;
Vv ^ 2 + 50 * Vv – 600 = 0;
Vв1,2 = -25 + – sqrt (25 ^ 2 + 600) = -25 + – 35 $
A negative root is meaningless;
Vv = -25 + 35 = 10 km / h.
Answer: the speed of the cyclist is 10 km / h.