A swimmer can swim in still water at a speed
of 9.31 m/s. He intends to swim directly
across a river that has a downstream current
of 4.35 m/s.
a) How many degrees from straight across
the river should he head? Let upstream be a
positive angle.
Answer in units of degrees
b) What is the magnitude of the swimmer’s
velocity relative to the bank?
Answer in units of m/s
a) Well, the swimmer wants to swim directly across the river, but the river has a downstream current. It's like the river is pulling him in one direction. So, he needs to adjust his heading to compensate for the current. To find the angle, we can use some trigonometry. Let's call the angle he needs to head θ.
The river's downstream current is 4.35 m/s, and the swimmer can swim at a speed of 9.31 m/s in still water.
Using trigonometry, we can find that tan(θ) = 4.35/9.31.
Now, we just need to solve for θ. Taking the arctan of both sides, we get θ ≈ arctan(4.35/9.31).
So, the angle the swimmer should head is approximately θ ≈ arctan(4.35/9.31).
b) To find the magnitude of the swimmer's velocity relative to the bank, we can use vector addition.
The swimmer's velocity in still water is 9.31 m/s, and the river's downstream current is 4.35 m/s.
If we add these two vectors together, we get the swimmer's velocity relative to the riverbank.
Using vector addition, we can find the magnitude of the swimmer's velocity relative to the bank as sqrt((9.31^2) + (4.35^2)).
So, the magnitude of the swimmer's velocity relative to the bank is approximately sqrt((9.31^2) + (4.35^2)).
To solve this problem, we can use trigonometry and vector addition.
a) To determine the angle the swimmer should head to cross the river, we can use the concept of vector addition. The swimmer's velocity relative to the river can be broken down into two components: the downstream component and the cross-river component.
Let θ be the angle the swimmer should head (measured from the upstream direction).
The downstream component of the swimmer's velocity can be expressed as:
V_downstream = swimmer's speed in still water - current speed = 9.31 m/s - 4.35 m/s = 4.96 m/s
The cross-river component of the swimmer's velocity can be expressed as:
V_cross-river = swimmer's speed in still water * sin(θ)
Since the vector addition of the downstream and cross-river components gives the swimmer's velocity relative to the river, we can use the Pythagorean theorem:
(swimmer's velocity relative to the river)^2 = (downstream component)^2 + (cross-river component)^2
V_river^2 = (4.96 m/s)^2 + (9.31 m/s * sin(θ))^2
Solving for θ, we get:
θ = sin^(-1)((V_river/V_swimmer) * sin(θ))
θ = sin^(-1)((4.96/9.31) * sin(θ))
To find the value of θ, we can use numerical methods or calculators. Plugging the values into a calculator, we find θ ≈ 29.7 degrees.
Therefore, the swimmer should head approximately 29.7 degrees from straight across the river.
b) To find the magnitude of the swimmer's velocity relative to the bank, we can use vector addition again. The swimmer's velocity relative to the bank can be found by adding the river's velocity (4.35 m/s) to the swimmer's velocity relative to the river.
V_bank = sqrt((V_river)^2 + (swimmer's speed in still water * cos(θ))^2)
V_bank = sqrt((4.35 m/s)^2 + (9.31 m/s * cos(29.7 degrees))^2)
Plugging the values into a calculator, we find V_bank ≈ 9.994 m/s.
Therefore, the magnitude of the swimmer's velocity relative to the bank is approximately 9.994 m/s.
To find the answers to the questions, we can apply trigonometric concepts and vector addition. Let's break it down step by step:
a) To determine how many degrees from straight across the river the swimmer should head, we can use the concept of vector addition. The swimmer's velocity relative to the ground is a combination of two velocities: the speed at which they can swim in still water and the speed of the river's current.
We can represent the swimmer's velocity relative to the ground as a vector by using the Pythagorean theorem. Let's call the swimmer's velocity in still water Vswimmer and the velocity of the river's current Vcurrent.
The desired resultant velocity (Vresultant) can be obtained by adding these two velocities as vectors:
Vresultant = Vswimmer + Vcurrent
To find the angle at which the swimmer should head, we need to calculate the angle between Vresultant and Vswimmer.
The angle between two vectors can be found using the dot product formula:
Vresultant · Vswimmer = |Vresultant| * |Vswimmer| * cos(theta)
where theta is the angle between the vectors.
Rearranging the formula, we have:
cos(theta) = (Vresultant · Vswimmer) / (|Vresultant| * |Vswimmer|)
Now, we can substitute the magnitudes and dot product values:
cos(theta) = [(Vswimmer + Vcurrent) · Vswimmer] / (|Vswimmer + Vcurrent| * |Vswimmer|)
cos(theta) = [(Vswimmer^2) + (Vswimmer · Vcurrent)] / [sqrt((Vswimmer + Vcurrent)^2) * |Vswimmer|]
cos(theta) = [(9.31^2) + (9.31 * 4.35)] / [sqrt((9.31 + 4.35)^2) * 9.31]
Evaluating this expression will give us the value of cos(theta). Then, we can use the inverse cosine function (arccos) to find the angle theta. Finally, we'll convert the result to degrees.
b) To find the magnitude of the swimmer's velocity relative to the bank, we are interested in the perpendicular component of the swimmer's velocity relative to the ground (Vresultant). This component represents their motion across the river.
Using trigonometry, we know that the magnitude of the swimmer's velocity relative to the bank (Vbank) is given by:
Vbank = |Vresultant| * sin(theta)
where theta is the angle calculated in part a.
Substituting the known values into the equation and evaluating it will give us the magnitude of the swimmer's velocity relative to the bank.
By following these steps and using the given values, we can find the answers to the questions.
(a)tanα =u/v=4.35/9.31=0.467.
α=arctan 0.467=25.04°
(b) v(rel) =sqrt(v²-u²)=sqrt(9.31²-4.35²)=8.23 m/s