# The tourist left the city at 6 in the morning with a speed of five kilometers per hour at 10 in the morning from the city in the same direction the cyclist with a speed of 15 kilometers per hour in what time and in what distance from the city the cyclist will catch up with the tourist left?

Vukol

289

The SOLUTION of 1) t = 10 - 6 = 4 h - the delay time of the cyclist. 2) d = V1 * t = 5 km/h * 4 h = 20 km - a distance at start of the cyclist 3) Vc = V2 - V1 = 15 - 5 = 10 km/h - the speed of rapprochement (pursuit). 4) T = d: Vc = 20 km: 10 km/h = 2 h - time of the movement of the cyclist 5) 10:00 + 02:00 = 12:00 - meeting time (on hours).-The ANSWER 6) S = V2 * T = 15 km/h * 2 h = 30 km - from the city - the ANSWER

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