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A motor boat, moving upstream passess a drifting raft. 1 hour later, the boat's motor breaks. it takes 45 min to fix it. while the boat is drifting downstream. after that, the boat begins to chase the raft and passes it. \(S=11 \text{ km}\) away from the place where they met first. What is the current \(v\) of the river.

Note by Nosa Wahyu
10 months ago

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We have that the raft travels 11 km. Lets call his traveled distance \( S_1 = v_r * t_1= 11 km \). His speed is \( v_1= v \) of the river ( \(v_r\) ), so \( v_1 = v_r \). The distance that the boat travels for that 1 hour after it meets the raft is \( S_2 = v_2 * t_2\) . \(t_2 = 1\) and \(v_2 = v_b - v_r\) {\(v_b\) is the speed of the boat} . \(S_2 = v_b - v_r\) . \(S_3\) is the distance traveled during the engine was broken. \(v_3 = v_r\) and \(t_3 = 3/4\) h. \(S_3 = 3/4 * v_r\) . \(S_4\) is the distance from when he fixes the engine to where he passes the raft. \(v_4 = v_b + v_r\). We have \(t_4\) and \(S_4 = v_4 * t_4 . t_1 = t_2 + t_3 + t_4\) ; \(t_1 = 7/4 + t_4\) ; \(t_4 = t_1 - 7/4\)

\(S_1+ S_2 = S_3 + S_4\) \(11 + v_b - v_r = 3/4 * v_r + (v_b + v_r) * t_4\) \(v_r * (7/4 + t_4) - v_m * (1 - t_4) = 11\) \(v_r * t_1 - v_m * (1 - t_4) = 11\) \(11 - v_m * (1 - t_4) = 11\) \(v_m * (1 - t_4) = 0\) /(v_m = 0) or \(1- t_4 = 0\) \(v_m\) is bigger than zero because it is speed so \(1 - t_4 = 0\); \(t_4 = 1\) \(t_1 = 7/4 + t_4 = 11/4\); \(S_1 = t_1 * v_r\); \(v_r = 4 km/h\) Jack Nikolov · 9 months, 3 weeks ago

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