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Chapter 9: Problem 71

A river flows from east to west. A swimmer on the south bank wants to swim to a point on the opposite shore directly north of her starting point. She can swim at \(2.8 \mathrm{mph},\) and there is a 1 -mph current in the river. In what direction should she head so as to travel directly north (that is, what angle should her path make with the south bank of the river)?

Short Answer

Expert verified
Answer: The swimmer should swim at an angle of approximately 21.8 degrees relative to the south bank.

Step by step solution

01

Define the variables

Let's define the speed of the swimmer relative to water as \(v_S = 2.8 \mathrm{mph}\), the speed of current as \(v_C = 1 \mathrm{mph},\) and the angle between the swimmer's direction and south bank will be \(\theta.\)
02

Represent swimmer's velocity as a vector

We can define the swimmer's velocity vector (relative to water) as \( \vec{v_S} = (v_{Sx}, v_{Sy})\), where \(v_{Sx}\) represents the east-west component, and \(v_{Sy}\) represents the north-south component. Using trigonometry, we have \(v_{Sx} = v_S \sin(\theta)\) \(v_{Sy} = v_S \cos(\theta)\)
03

Represent the current's velocity as a vector

The current's velocity vector will be \(\vec{v_C} = (v_{Cx}, v_{Cy})\). Since the current is moving towards the west, we define its speed as moving towards the negative x-direction (west) and no motion in the north-south direction. Thus, \(v_{Cx} = -v_C = -1 \,\mathrm{mph}\) \(v_{Cy} = 0\)
04

Find the net velocity vector

The swimmer's net velocity vector will be the vector sum of the swimmer's velocity and the current's velocity: \(\vec{v_N} = \vec{v_S} + \vec{v_C}\). Adding the respective components, we have: \(v_{Nx} = v_{Sx} + v_{Cx} = v_S \sin(\theta) - 1\) \(v_{Ny} = v_{Sy} + v_{Cy} = v_S \cos(\theta)\)
05

Equate the net velocity's components to find the angle \(胃\)

Since the swimmer is going directly north, there is no east-west velocity component. So, \(v_{Nx} = 0\). Equating \(v_{Nx}\) to zero and solving for \(\theta\), we have: \(v_S \sin(\theta) - 1 = 0\) \(\sin(\theta) = \frac{1}{v_S} = \frac{1}{2.8}\) \(\theta = \arcsin\left(\frac{1}{2.8}\right)\)
06

Calculate the value of \(\theta\)

Calculate the value of \(\theta\) using a calculator: \(\theta = \arcsin\left(\frac{1}{2.8}\right) \approx 21.8^\circ\) The swimmer should swim at an angle of \(\approx 21.8^\circ\) relative to the south bank to move directly towards the opposite shore in the north direction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometry
Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. In physics, especially in problems involving vectors, trigonometry helps us break down and analyze the components of force, velocity, and more. When thinking about vectors, it's important to realize that they have both magnitude and direction. Trigonometry allows us to decompose a vector into its components using sine, cosine, and tangent functions. For instance, if you have a vector with a magnitude in a specific direction, you can find how much of that vector points along each axis.In the swimmer problem, we used trigonometric functions to express the swimmer鈥檚 velocity components. Here,
  • The sine function (\(\sin(\theta)\)) was used to determine the east-west component.
  • The cosine function (\(\cos(\theta)\)) helped us determine the north-south component.
This method of using trigonometry to express vector components is crucial in vector analysis, as it allows us to isolate and work with individual effects in direction.
Relative Velocity
Relative velocity is the velocity of an object as observed from a particular frame of reference. In our swimmer scenario, the relative velocity considers the speed of the swimmer with respect to the water. Why is this concept so important?
  • In scenarios like the one in the exercise, the swimmer鈥檚 speed is influenced by water currents, hence the observed velocity relative to the shore (the frame of reference) is different from the swimmer's speed in stationary water.
  • The concept of relative velocity helps us determine how the swimmer should compensate for the river's movement to reach the desired destination directly north.
To solve the problem, we formed a net velocity vector by adding the swimmer's velocity vector and the current's velocity. By balancing these vectors such that the east-west component becomes zero, we ensure the swimmer goes directly north, despite the east-to-west flow of the current.
Angle Calculation
Finding the correct angle is crucial for the swimmer to head directly north. This involves understanding the interaction between the different velocities. Since the river flows from east to west, the swimmer needs to adjust her angle to counteract this current.We are asked to calculate this angle (\( \theta \)), which guides her to maintain a course directly north.
  • By setting the east-west component of the net velocity to zero (\( v_{Nx} = 0 \)), we derive the equation (\(v_S \sin(\theta) - 1 = 0 \)). This equation expresses what her angle must achieve: equalizing her eastward motion with the westward current.
  • Solving for (\( \theta \)) using the inverse sine function gives us the specific angle needed relative to the south bank.
The calculation (\( \theta = \arcsin\left( \frac{1}{2.8} \right) \)) results in approximately 21.8掳, which is the angle she should aim her swim to succeed in traveling directly north.

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Most popular questions from this chapter

Suppose \(\mathbf{u}=\langle a, b\rangle\) and \(\mathbf{v}=\langle c, d\rangle\) are nonzero parallel vectors. (a) If \(c \neq 0,\) show that \(\mathbf{u}\) and \(\mathbf{v}\) lie on the same nonvertical straight line through the origin. (b) If \(c \neq 0,\) show that \(\mathbf{v}=\frac{a}{c} \mathbf{u}\) (that is, \(\mathbf{v}\) is a scalar multiple of \(\mathbf{u}\) ). [Hint: The equation of the line on which \(\mathbf{u}\) and v lie is \(y=m x\) for some constant \(m\) (why?), which implies that \(b=m a\) and \(d=m c .\) ] (c) If \(c=0,\) show that \(\mathbf{v}\) is a scalar multiple of \(\mathbf{u} .[\) Hint: If \(c=0,\) then \(a=0\) (why?), and hence, \(b \neq 0\) (otherwise, \(\mathbf{u}=\mathbf{0}) .]\)

Let \(\boldsymbol{u}=\langle a, b\rangle, \boldsymbol{v}=\langle c, d\rangle,\) and \(\boldsymbol{w}=\langle r, s\rangle\) Verify that the given property of dot products is valid by calculating the quantities on each side of the equal sign. $$0 \cdot \mathbf{u}=0$$

A plane is flying in the direction \(200^{\circ}\) with an air speed of \(500 \mathrm{mph}\). Its course and ground speed are \(210^{\circ}\) and \(450 \mathrm{mph}\) respectively. What are the direction and speed of the wind?

Let \(\mathbf{v}\) be the vector with initial point \(\left(x_{1}, y_{1}\right)\) and terminal point \(\left(x_{2}, y_{2}\right),\) and let \(k\) be any real number. (a) Find the component form of \(\mathbf{v}\) and \(k \mathbf{v}\). (b) Calculate \(\|\mathbf{v}\|\) and \(\|k \mathbf{v}\|\). (c) Use the fact that \(\sqrt{k^{2}}=|k|\) to verify that \(\|k \mathbf{v}\|=|k| \cdot\|\mathbf{v}\|\). (d) Show that \(\tan \theta=\tan \beta,\) where \(\theta\) is the direction angle of \(\mathbf{v}\) and \(\beta\) is the direction angle of \(k \mathbf{v} .\) Use the fact that \(\tan t=\tan \left(t+180^{\circ}\right)\) to conclude that \(\mathbf{v}\) and \(k \mathbf{v}\) have either the same or opposite directions. (e) Use the fact that \((c, d)\) and \((-c,-d)\) lie on the same straight line on opposite sides of the origin (Exercise 85 in Section 1.3 ) to verify that \(\mathbf{v}\) and \(k \mathbf{v}\) have the same direction if \(k>0\) and opposite directions if \(k<0\).

If \(\mathbf{u}\) and \(\mathbf{v}\) are nonzero vectors such that \(\mathbf{u} \cdot \mathbf{v}=0,\) show that u and \(v\) are orthogonal. \([\text {Hint} \text { : If } \theta \text { is the angle between } \mathbf{u}\) and \(\mathbf{v}, \text { what is } \cos \theta \text { and what does this say about } \theta ?]\)
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