Puzzle of the Day- Boat Challenge: Can you find the missing times?

I'm going to work in the frame of the river. From this perspective, the wooden block is stationary, and the float tube and the camp move upstream at constant speed. Also, measure distance in arbitrary units (a.u.) such that the boat travels 1 a.u./min.

The boat reaches the wooden block after 75 minutes of traveling downstream from its furthest upstream point, so it must have also passed it 75 minutes of traveling upstream before this point. Therefore, YYY is 75 minutes before 11:45, i.e., YYY = 10:30.

Let t be the time, in minutes, between 10:20 and XXX. The boat travels 10 a.u. upstream between its first and second meetings with the float tube, so between these meetings, the float tube travels 10 a.u. in 10 + t minutes. Between XXX and 11:45, the boat travels 85 - t a.u. upstream, and between 12:00 and 12:10, it travels 10 a.u. downstream. So, the third meeting with the float tube, at 12:10, is 85 - t a.u. upstream from the first one, at 12:10, and between these meetings, the float tube traveled 85 - t a.u. in 120 minutes. Since the float tube is traveling at constant speed, 10/(10 + t) = (85 - t)/120, so 0 = t^2 - 75t + 350 = (t - 5)(t - 70). Since XXX must be before YYY, we must take t = 5. Therefore, XXX = 10:25.

Since t = 5, the furthest upstream point of the boat is 10 + 85 - 5 = 90 a.u. upstream from its first meeting with the float tube. As this occurred at 10:10, after 100 minutes of traveling, the first meeting with the float tube was 100 a.u. upstream from the position the camp occupied at 08:30, making the upstream point 100 + 90 = 190 a.u. upstream from the initial position of the camp. The block is 75 a.u. downstream from this, so it is 190 - 75 = 115 a.u. upstream from the initial position of the camp. Since the wooden block meets the camp at 14:15, 345 minutes after 08:30, the camp has travelled 115 a.u. in 345 minutes, so it is moving at 1/3 a.u./minute. Therefore, at 12:00, the camp has moved 210/3 = 70 a.u. upstream from its initial position, so it is 190 - 70 = 120 a.u. from the boat's position. The boat and the camp close at 4/3 a.u./minute, so they rendezvous 120/(4/3) = 90 minutes after 12:00, at ZZZ = 13:30.

The answer then is that the missing times are XXX = 10:25, YYY = 10:30, and ZZZ = 13:30.

the situation may be solved without algebra. i will artwork in the reference physique of the river. We began rowing downstream at 11:34, fifty six minutes till now we handed the shirt at 12:30. So, between the time we handed the shirt for the 1st time and the time we circled the boat, we ought to have spent fifty six finished minutes rowing upstream. 5 minutes of this have been between 10:10 and 10:15, leaving fifty one minutes between the time as quickly as we handed the blue jetty for the 2d time and 11:sixteen. So, we handed the blue jetty for the 2d time at 10:25. this suggests that it took the marina quarter-hour (between 10:10 and 10:25) to conceal the area we coated in 5 minutes (between 10:10 and 10:15). as a result, the marina, and all different factors on the floor, are shifting at a million/3 our velocity. degree distance in arbitrary gadgets (a.u.) such that the area between the shirt and our factor furthest upstream is fifty six*3 = 168 a.u. Then we are rowing at 3 a.u./min and the floor is shifting at a million a.u./min. So, between 10:10 and 11:34, the marina has moved eighty 4 a.u. upstream, and at 11:34, we are 168 - eighty 4 = eighty 4 a.u. upstream from it. After 11:34 we close with the marina at 3 + a million = 4 a.u./min, so we rendezvous with it for the third time at eighty 4/4 = 21 minutes after 11:34, at 11:fifty 5. It took us a hundred minutes (between 08:30 and 10:10) to row from the camp to the shirt, so the region the camp became into in at 08:30 is 3 hundred a.u. downstream from the shirt. 240 minutes later, at 12:30, it has moved to 3 hundred - 240 = 60 a.u. downstream from the shirt. We close with the camp at 4 a.u./min, so we attain camp at 60/4 = quarter-hour after 12:30, at 12:forty 5. The shirt closes with the camp at a million a.u./min, so it reaches camp 60 minutes after 12:30, at 13:30. This became right into a constructive puzzle. i found it efficient to attraction to a spacetime diagram plotting the trajectories of the boat, shirt, jetty, and camp, with time because of the fact the x axis and area because of the fact the y axis.

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