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?

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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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Answer add