A river is 2000 ft wide and flowing at 6 mph from north to south. A woman in a canoe starts on the eastern shore and heads west at her normal paddling speed of 2 mph. In what direction (measured clockwise from north) must she aim her canoe? How long will it take her to go directly across the river? Justify and explain your reasoning
To determine the direction the woman must aim her canoe, we can analyze the vectors involved. Let's break down the velocities involved:
1. River velocity: The river is flowing from north to south at a speed of 6 mph. We can represent this velocity as a vector pointing in the south direction.
2. Canoe velocity: The woman in the canoe is paddling westward at her normal speed of 2 mph. We can represent this velocity as a vector pointing in the west direction.
To determine the direction the woman must aim her canoe, we need to find the resultant velocity, which is the vector sum of the river velocity and the canoe velocity.
Since the canoe is headed west and the river is flowing south, the two velocities are perpendicular to each other. Hence, we can use the Pythagorean theorem to find the magnitude of the resultant velocity:
Resultant velocity = √((canoe velocity)^2 + (river velocity)^2)
= √((2 mph)^2 + (6 mph)^2)
= √(4 mph^2 + 36 mph^2)
= √40 mph^2
= 2√10 mph (approximately)
Now, to find the direction, we can use trigonometry. The direction (measured clockwise from north) can be obtained by finding the angle between the river velocity and the resultant velocity.
Let θ be the angle between the river velocity and the resultant velocity.
sin(θ) = (canoe velocity)/(resultant velocity)
sin(θ) = (2 mph)/(2√10 mph)
θ = sin^(-1)((2 mph)/(2√10 mph))
θ ≈ 18.2 degrees
Therefore, the woman must aim her canoe approximately 18.2 degrees clockwise from north.
Next, to find the time it will take for her to go directly across the river, we can use the concept of relative velocity.
The relative velocity is the velocity of the canoe with respect to the river. In this case, since the canoe is traveling west at 2 mph and the river is flowing south at 6 mph, the relative velocity is the vector difference between the canoe velocity and the river velocity.
Relative Velocity = Canoe Velocity - River Velocity
= 2 mph west - 6 mph south
= -4 mph south
The magnitude of the relative velocity is 4 mph, and the direction is south. Hence, the woman will be directly across the river when the magnitude of her relative velocity is equal to the river's width.
Time taken to cross the river = (River width)/(Relative velocity)
= 2000 ft/(4 mph)
= 500 hours (approximately)
Therefore, it will take her approximately 500 hours to go directly across the river.
To determine the direction the woman must aim her canoe and the time it will take her to go directly across the river, we can break down the given information and use a few principles of vector addition.
1. River Width: The river is 2000 ft wide.
2. River Flow: The river is flowing at 6 mph from north to south. This means that the river's velocity vector points directly southward at a speed of 6 mph.
3. Canoe Speed: The woman's paddling speed is 2 mph westward. Therefore, her canoe's velocity vector points directly westward at a speed of 2 mph.
To find the direction the woman must aim her canoe, we need to determine the resultant velocity vector, representing the combined effect of the river flow and the canoe paddling.
First, we convert the speeds of the river flow and the canoe into vectors. We know that velocity is defined as the rate of change of displacement over time, and in our case, the displacement for both the river and the canoe is constant.
The river's velocity vector can be represented as v_river = 6 mph southward.
The canoe's velocity vector can be represented as v_canoe = 2 mph westward.
Next, we apply vector addition to find the resultant velocity vector.
Since the vectors for the river flow and the canoe paddling are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the resultant velocity vector:
|v_resultant|^2 = |v_river|^2 + |v_canoe|^2
|v_resultant|^2 = (6 mph)^2 + (2 mph)^2
|v_resultant|^2 = 36 mph^2 + 4 mph^2
|v_resultant|^2 = 40 mph^2
Taking the square root of both sides, we get:
|v_resultant| = √(40 mph^2)
|v_resultant| ≈ 6.32 mph
Therefore, the magnitude of the resultant velocity vector (|v_resultant|) is approximately 6.32 mph.
The direction of the resultant velocity vector can be found using the arctan function:
θ = arctan(y-component / x-component)
θ = arctan((v_river)_y / (v_river)_x)
θ = arctan(-6 mph / 0 mph)
(Note: Since in our case, both the x and y components of the river's velocity vector are zero, the arctan function gives an undefined value. This means that the woman should not aim her canoe directly across the river.)
Considering this, the woman should aim her canoe slightly upstream to compensate for the river flow.
To find the time it will take her to cross the river, we can use the formula:
Time = Distance / Speed
Since the woman wants to go directly across the river, the distance she needs to cover is the 2000 ft width of the river.
Time = 2000 ft / 2 mph
Time = 1000 hours
Therefore, it will take her 1000 hours to go directly across the river if she aims her canoe upstream.