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A straight river flows east at a speed of 12 mi/h. A boater starts at the south shore of the river and wants to arrive at a point on the north shore of the river directly opposite the starting point. The motorboat has a speed of 24 mi/h should the boat be headed? N XOW Need Help? Read it Submit Answer

User TC Arlen
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Answer:

The boat should be headed at an angle of 60 degrees west of north (NW) to arrive at the point on the north shore directly opposite the starting point while compensating for the river's flow to the east.

Step-by-step explanation:

To arrive at a point on the north shore of the river directly opposite the starting point, you need to account for the river's eastward flow. You want to navigate the boat at an angle that balances the eastward drift caused by the river's current.

To do this, you can use vector addition. Let's call the angle between the boat's heading and north as θ.

The boat's speed with respect to the ground is 24 mi/h, and the river's flow speed is 12 mi/h to the east. So, you can think of the boat's velocity as a vector with a magnitude of 24 mi/h at an angle θ with respect to the north direction.

To counteract the eastward flow of the river, the boat's velocity vector should have a component in the westward direction equal to the river's velocity, which is 12 mi/h.

So, you have the following equation:

24 * cos(θ) = 12

Now, solve for θ:

cos(θ) = 12 / 24

cos(θ) = 0.5

To find θ, take the arccosine (inverse cosine) of 0.5:

θ = arccos(0.5)

θ ≈ 60 degrees

So, the boat should be headed at an angle of approximately 60 degrees west of north (NW) to arrive at the point on the north shore directly opposite the starting point while compensating for the river's flow to the east.

User Paleo
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