Jack's River Race: Solving the Shortest Path Problem

[Matthew]

1.Jack is strolling along the bank of a 1-km wide River when the most beautiful girl materialises on the shore directly opposite him. Fearing she will disappear before he has a chance to establish face-to-face communication, he quickly devises a plan to reach the opposite shore in the shortest possible time. There is a rowboat breached on the shore right infront of him. He knows he can row at a speed of 6 km/h in still water and the current has a speed of 3 km/h. He must reach the opposite side of the river in the shortest time. His path includes a diagonal trip across the river followed by a sprint along the opposite shore to reach Jill. Jack has a standard sprinting speed of 10 km/h. What direction did he head his boat.

P.S I am still looking for number 2 on my other post. I thought I had the answer but it was wrong.

 

[HallsofIvy]

Technical point: you say in your post "He must reach the opposite side of the river in the shortest time." Taken literally, that would mean that it doesn't matter WHERE on the opposite shore he winds up and his running speed after he gets there is irrelevant.

I am going to take it that Jill will remain where she is and Jack wants to get to her position in the shortest time.

Let theta be the angle at which he rows relative to the straight line across the river (directly across would be theta= 0, aiming downstream positive, upstream negative). Since his speed in still water is 6 km/hr, his "velocity vector" would be (6 sin(theta), 6 cos(theta)). Since the river is flowing at 3 km/, its velocity vector is (3, 0) and Jack's "velocity made good" is (6 sin(theta)+ 3, 6 cos(theta)). He will cross the river in time t₁ if
6 cos(theta)t₁= 1 or t₁= 1/(6 cos(theta)).

At that time, his position on the far shore will be (6 sin(theta)+ 3)sin(theta), positive if downstream from Jill, negative if upstream.
In any case, he now has that distance to run at 10 km/h
That will require time t₂= 10(6 sin(theta)+ 3)sin(theta).
You want to minimize t₁+ t₂=
1/(6 cos(theta))+10(6 sin(theta)+ 3)sin(theta).

Differentiate with respect to theta and set equal to 0.

 

[M98Ranger]

Based on the given information, Jack's shortest path would be to row at a diagonal angle towards the opposite shore, taking into account the speed of the current. This would allow him to reach the opposite shore in the shortest amount of time. Once he reaches the shore, he should immediately sprint towards Jill at a speed of 10 km/h. Therefore, he should head his boat in a direction that is slightly angled towards the opposite shore, taking into consideration the speed of the current. This will allow him to reach his destination in the shortest amount of time and have the chance to establish communication with Jill.

 

[M98Ranger]

Jack's shortest path to reach the opposite shore would be to row diagonally across the river at an angle of 45 degrees. This is because the diagonal distance is shorter than the distance along the river bank, and rowing at an angle of 45 degrees would allow Jack to take advantage of the current and row at a faster speed of approximately 7.07 km/h (6 km/h + 3 km/h).

After reaching the opposite shore, Jack would then sprint at a speed of 10 km/h towards Jill. This combination of rowing and sprinting would allow Jack to reach Jill in the shortest possible time.

In order to determine the exact direction Jack should head his boat, we can use the concept of vector addition. The velocity of the boat can be represented by a vector with a magnitude of 7.07 km/h at an angle of 45 degrees. The velocity of the current can also be represented by a vector with a magnitude of 3 km/h at an angle of 180 degrees (opposite direction).

To determine the resultant velocity, we can use the Pythagorean theorem to calculate the magnitude of the resultant vector, which would be the speed at which Jack's boat would travel. The angle of the resultant vector can be calculated using trigonometric functions.

By using this method, we can determine that Jack should head his boat at an angle of 26.57 degrees (rounded to the nearest hundredth) in order to reach the opposite shore in the shortest time.