A boat’s speed in still water is vBW = 1.85 m/s. If the boat
is to travel directly across a river whose current has
speed vWS = 1.20 m/s, at what upstream angle must the
boat head?
What is arcsin 1.20/1.85?
35
To determine the angle at which the boat must head, we can use trigonometry.
Let's assume that the boat's velocity relative to the ground is vBG, and the angle between the boat's heading and the direction of the river current is θ.
The boat's velocity relative to the ground can be determined using vector addition:
vBG = √((vBW)^2 + (vWS)^2 + 2(vBW)(vWS)cosθ)
Given:
vBW = 1.85 m/s (boat's speed in still water)
vWS = 1.20 m/s (river current speed)
Substituting these values into the equation, we get:
vBG = √((1.85)^2 + (1.20)^2 + 2(1.85)(1.20)cosθ)
Now, we need to find the value of θ that makes vBG equal to 0 (i.e., the boat is traveling directly across the river).
0 = √(1.85^2 + 1.20^2 + 2(1.85)(1.20)cosθ)
Squaring both sides, we get:
0 = 1.85^2 + 1.20^2 + 2(1.85)(1.20)cosθ
Solving this equation for cosθ, we get:
cosθ = -((1.85^2 + 1.20^2) / (2(1.85)(1.20)))
cosθ = -((3.4225 + 1.44) / (4.44))
cosθ = -4.8625 / 4.44
cosθ = -1.0941
Now, we need to find the angle θ for which cosθ = -1.0941. Since the cosine function has a maximum value of 1 and a minimum value of -1, there is no angle that satisfies this equation. Therefore, the boat cannot travel directly across the river.
In summary, there is no upstream angle at which the boat can head to travel directly across the river.
To find the upstream angle at which the boat must head, we can use the concept of vector addition.
Let's consider two vectors:
1. The velocity of the boat in still water (vBW), which has a magnitude of 1.85 m/s.
2. The velocity of the river current (vWS), which has a magnitude of 1.20 m/s.
The boat's actual velocity will be the vector sum of these two velocities. The angle at which the boat must head upstream will be the angle between the resultant velocity vector and the direction opposite to the current.
To calculate the resultant velocity vector, we can use the Pythagorean theorem.
Let V be the magnitude of the resultant velocity vector, and θ be the angle between the resultant velocity and the current direction (opposite direction).
V^2 = (vBW)^2 + (vWS)^2 - 2(vBW)(vWS)cosθ
In this case, the magnitude of the resultant velocity is equal to the speed of the boat in still water (vBW) since we want the boat to move directly across the river.
1.85^2 = (1.85)^2 + (1.20)^2 - 2(1.85)(1.20)cosθ
Simplifying the equation, we have:
3.4225 = 3.4225 + 1.44 - 4.14cosθ
Rearranging the terms:
4.14cosθ = 1.44
Dividing both sides by 4.14:
cosθ = 1.44 / 4.14
Now, we need to find the angle θ. We can use the inverse cosine function (cos^(-1)).
θ = cos^(-1) (1.44 / 4.14)
Evaluating this expression using a calculator, we find:
θ ≈ 68.81 degrees (rounded to two decimal places)
Therefore, the boat must head at an upstream angle of approximately 68.81 degrees.