How Does a 45 Degree Angle Affect Boat Navigation in a River Current?

[sw3etazngyrl]

1. A boat, whose speed in still water is 2.20m/s, must cross a 260m wide river and arrive at a point 110m upstream from where it starts. To do so, the pilot must head the boat at a 45 degree upstream angle. What's the speed of the river's current?

2. How do you incoporate 45 degrees into the work?

3. So far, I have

a^2+b^2=c^2
c= sqrt(110^2+260^2)=282m

 

[Astronuc]

The boats speed relative to the water is 2.2 m/s and it is pointing 45° with respect to the shoreline. So determine the speed in the direction of the other bank, or normal to the stream. Find the time it takes to go 260 m.

With that time, determine the effective speed to up stream 110 m.

 

[m2003]

1. The relative velocity of a boat is the combination of its own speed in still water and the speed of the current it is traveling against. In this scenario, the boat has a speed of 2.20m/s in still water and needs to cross a 260m wide river to reach a point 110m upstream. To do so, the boat must travel at a 45 degree angle upstream. This means that the boat's velocity will have a horizontal component of 2.20m/s and a vertical component of 2.20m/s. The resultant velocity can be found using the Pythagorean theorem, where the hypotenuse represents the boat's velocity and the sides represent the horizontal and vertical components. Solving for the hypotenuse, we get a velocity of 3.12m/s. This is the relative velocity of the boat.

2. To incorporate the 45 degree angle into the work, we use trigonometric functions. In this case, we can use the cosine function to find the horizontal component of the boat's velocity. The cosine of 45 degrees is equal to the adjacent side (horizontal component) divided by the hypotenuse (boat's velocity). So, the horizontal component can be found by multiplying the boat's velocity (3.12m/s) by the cosine of 45 degrees, which is equal to 2.20m/s. This confirms that the boat will have a horizontal velocity of 2.20m/s.

3. To find the speed of the river's current, we can use the Pythagorean theorem again. This time, the hypotenuse represents the speed of the river's current and the sides represent the horizontal and vertical components. We already know the horizontal component (2.20m/s) and the vertical component (2.20m/s) from the boat's velocity. Solving for the hypotenuse, we get a speed of 3.12m/s for the river's current. This means that the river is flowing at a speed of 3.12m/s in the opposite direction of the boat's travel.