The width of the river is d = 300 m, the velocity of the river is V1 = 1 m / s, the velocity of the ship in still water is V2 = 3 M / s, and the course of the ship forms a 30 degree angle with the upstream bank How long does it take for it to cross the river (3) How can the course reach the opposite shore (4) How to cross the river in the shortest time, (2) 320m upstream （3）cosθ=1/3 (4) The bow is always perpendicular to the other side of the river, t = 100s

The width of the river is d = 300 m, the velocity of the river is V1 = 1 m / s, the velocity of the ship in still water is V2 = 3 M / s, and the course of the ship forms a 30 degree angle with the upstream bank How long does it take for it to cross the river (3) How can the course reach the opposite shore (4) How to cross the river in the shortest time, (2) 320m upstream （3）cosθ=1/3 (4) The bow is always perpendicular to the other side of the river, t = 100s

When the velocity is divided into two directions, the velocity in the vertical direction is v vertical = V2 * cos 30 ° along the river direction and the direction perpendicular to the river
It takes time t = D / V2 * cos30 ° to cross the river
The velocity along the river is v along = V2 * sin30 ° - V1 = 0.5m/s
Then the position on the opposite bank is v * t
The third question is that to reach the opposite bank, V edge must be equal to 0, then we know that V2 * sina-v1 = 0, Sina = 1 / 3, a = arcsin1 / 3
In the fourth question, if the time is the shortest, then the V sag is the maximum, and V sag = V2 * cosa. If the V sag is the maximum, then a = 0 degree, then the ship's heading forward opposite shore time is the shortest