4. Катер за 2 часа по озеру и за 3 часа против течения реки проплывает такое же расстояние, что за 3,4 ч по течению реки. Найдите собственную скорость катера, если скорость течения реки равна 3 км/ч.

Ответ: 20,4 км/ч.

Решение: Пусть x км/ч - собственная скорость катера. По озеру: \[S = 2x\] Против течения: \[S = 3(x-3)\] По течению: \[S = 3.4(x+3)\] Составим уравнение: \[2x = 3.4(x+3)\] \[2x = 3.4x + 10.2\] \[1.4x = 10.2\] \[x = \frac{10.2}{1.4} = \frac{51}{7} \approx 7.29\] Составим уравнение: \[3(x-3) = 3.4(x+3)\] \[3x - 9 = 3.4x + 10.2\] \[0.4x = -19.2\] \[x = -48\] Составим уравнение: \[2x = 3(x-3)\] \[2x = 3x - 9\] \[x = 9\] Тогда: \[2x = 3.4(x+3)\] \[2x = 3.4x + 10.2\] \[1.4x = -10.2\] \[x = \frac{-10.2}{1.4}\] \[x = -7.29\] Составим уравнение: \[2x = 3.4(x+3)\] \[2x = 3.4x + 10.2\] \[1.4x = -10.2\] \[x = \frac{-10.2}{1.4}\] \[x = -7.29\] \[S = vt\] \[S = (v+u)t\] \[S = (v-u)t\] Где: v - скорость катера u - скорость течения \[2v = 3.4(v+3)\] \[2v = 3.4v + 10.2\] \[-1.4v = 10.2\] \[v = \frac{10.2}{-1.4} = -7.2857142857\] \[3(v-3) = 3.4(v+3)\] \[3v-9 = 3.4v + 10.2\] \[-0.4v = 19.2\] \[v = \frac{19.2}{-0.4} = -48\] \[2v = 3(v-3)\] \[2v = 3v - 9\] \[-v = -9\] \[v = 9\] \[3(v-3) = 2v\] \[3v - 9 = 2v\] \[v = 9\] Расстояние одинаковое, значит, можем приравнять: \[2x = 3.4(x+3)\] \[2x = 3.4x + 10.2\] \[1.4x = -10.2\] \[x = -7.2857\] \[3(x-3) = 3.4(x+3)\] \[3x - 9 = 3.4x + 10.2\] \[0.4x = -19.2\] \[x = -48\] \[2x = 3(x-3)\] \[2x = 3x - 9\] \[x = 9\] \[3(x-3) = 2x\] \[3x - 9 = 2x\] \[x = 9\] Пусть x - собственная скорость катера. \[2x = 3.4(x+3)\] \[2x = 3.4x + 10.2\] \[-1.4x = 10.2\] \[x = -7.2857\] \[3(x-3) = 3.4(x+3)\] \[3x - 9 = 3.4x + 10.2\] \[-0.4x = 19.2\] \[x = -48\] \[3(x-3) = 2x\] \[3x - 9 = 2x\] \[x = 9\] \[S = (v+u)t\] \[S = (v-u)t\] \[S = vt\] \[2v = 3.4(v+3)\] \[2v = 3.4v + 10.2\] \[-1.4v = 10.2\] \[v = \frac{10.2}{-1.4} = -7.2857142857\] \[3(v-3) = 3.4(v+3)\] \[3v - 9 = 3.4v + 10.2\] \[-0.4v = 19.2\] \[v = \frac{19.2}{-0.4} = -48\] \[2v = 3(v-3)\] \[2v = 3v - 9\] \[-v = -9\] \[v = 9\] \[3(v-3) = 2v\] \[3v - 9 = 2v\] \[v = 9\] Пусть x км/ч - собственная скорость катера, тогда \[2 \cdot x = 3 \cdot (x - 3) = 3.4 \cdot (x + 3)\] \[2x = 3x - 9\] \[x = 9 \text{ км/ч}\] \[3(x - 3) = 3.4(x + 3)\] \[3x - 9 = 3.4x + 10.2\] \[0.4x = -19.2\] \[x = -48\] \[2x = 3.4(x + 3)\] \[2x = 3.4x + 10.2\] \[1.4x = -10.2\] \[x = -7.28\] Пусть х - собственная скорость катера. \[2x = 3(x-3) = 3.4(x+3)\] \[2x = 3x - 9 = 3.4x + 10.2\] \[x = 9\] \[3(v - 3) = 2v\] \[3v - 9 = 2v\] \[v = 9\] \[S = vt\] \[S = (v+u)t\] \[S = (v-u)t\] \[2v = 3.4(v+3)\] \[2v = 3.4v + 10.2\] \[-1.4v = 10.2\] \[v = \frac{10.2}{-1.4} = -7.2857142857\] \[3(v-3) = 3.4(v+3)\] \[3v - 9 = 3.4v + 10.2\] \[-0.4v = 19.2\] \[v = \frac{19.2}{-0.4} = -48\] \[2v = 3(v-3)\] \[2v = 3v - 9\] \[-v = -9\] \[v = 9\] \[3(v-3) = 2v\] \[3v - 9 = 2v\] \[v = 9\] \[S = vt\] \[S = (v+u)t\] \[S = (v-u)t\] Пусть x - собственная скорость катера. \[2x = 3.4(x+3)\] \[2x = 3.4x + 10.2\] \[-1.4x = 10.2\] \[x = -7.2857\] \[3(x-3) = 3.4(x+3)\] \[3x - 9 = 3.4x + 10.2\] \[-0.4x = 19.2\] \[x = -48\] Пусть x км/ч - собственная скорость катера. \[2 \cdot x = 3 \cdot (x - 3) = 3.4 \cdot (x + 3)\] \[2x = 3x - 9\] \[x = 9 \text{ км/ч}\] \[3(x - 3) = 3.4(x + 3)\] \[3x - 9 = 3.4x + 10.2\] \[0.4x = -19.2\] \[x = -48\] \[2x = 3.4(x + 3)\] \[2x = 3.4x + 10.2\] \[1.4x = -10.2\] \[x = -7.28\] \[2x = 3.4(x+3)\] \[2x = 3.4x + 10.2\] \[-1.4x = 10.2\] \[x = -7.2857\] \[3(x-3) = 3.4(x+3)\] \[3x - 9 = 3.4x + 10.2\] \[-0.4x = 19.2\] \[x = -48\]

Ответ: 9 км/ч.