Problem 138 - Fundamentals of Physics

Question

A woman can row a boat at \(6.40 \mathrm{~km} / \mathrm{h}\) in still water. (a) If she is crossing a river where the current is \(3.20 \mathrm{~km} / \mathrm{h}\), in what direction must her boat be headed if she wants to reach a point directly opposite her starting point? (b) If the river is \(6.40 \mathrm{~km}\) wide, how long will she take to cross the river? (c) Suppose that instead of crossing the river she rows \(3.20 \mathrm{~km}\) down the river and then back to her starting point. How long will she take? (d) How long will she take to row \(3.20 \mathrm{~km} u p\) the river and then back to her starting point? (e) In what direction should she head the boat if she wants to cross in the shortest possible time, and what is that time?

Step-by-Step Solution

Answer

(a) Head 30° upstream. (b) 1.16 hours. (c) 1.33 hours. (d) 1.33 hours. (e) Head straight; 1 hour.

Step 1: Calculating Angle for Direct Cross (a)

To go directly across the river, the woman must head upstream at an angle that compensates for the current. Her speed relative to the ground is the resultant vector of her speed in still water and the river's current. Therefore, we have:\[ v_{b} = 6.40 \text{ km/h} \] (boat speed in still water)\[ v_{r} = 3.20 \text{ km/h} \] (river current speed)She must row at an angle \( \theta \) such that the component of her speed that cancels out the river's current is:\[ v_{b} \sin{\theta} = v_{r} \]Solving for \( \theta \):\[ \sin{\theta} = \frac{v_{r}}{v_{b}} = \frac{3.20}{6.40} = 0.5 \]\[ \theta = \arcsin(0.5) = 30^\circ \]She must head at an angle of \(30^\circ\) upstream from the direct line across the river.

Step 2: Calculating Time to Cross River (b)

To find the time taken to cross the river, we need to consider the width of the river and the effective speed perpendicular to the current:\[ \text{Width of the river} = 6.40 \text{ km} \]The effective speed across the river is the component of her velocity perpendicular to the current:\[ v_{\perp} = v_{b} \cos{\theta} = 6.40 \cos{30^\circ} = 6.40 \times \frac{\sqrt{3}}{2} = 5.54 \text{ km/h} \]Now, calculate the time:\[ t = \frac{\text{Distance}}{\text{Speed}} = \frac{6.40}{5.54} = 1.16 \text{ hours} \]It takes approximately 1.16 hours to cross the river.

Step 3: Time for Round Trip Down the River (c)

On her downstream trip, she travels with the current:\[ v_{d} = v_{b} + v_{r} = 6.40 + 3.20 = 9.60 \text{ km/h} \]Time to row \(3.20\) km downstream:\[ t_{d} = \frac{3.20}{9.60} = 0.33 \text{ hours} \]Returning upstream against the current, her speed is:\[ v_{u} = v_{b} - v_{r} = 6.40 - 3.20 = 3.20 \text{ km/h} \]Time to row \(3.20\) km upstream:\[ t_{u} = \frac{3.20}{3.20} = 1.00 \text{ hour} \]Total time for round trip:\[ t_{total} = t_{d} + t_{u} = 0.33 + 1.00 = 1.33 \text{ hours} \]The woman takes 1.33 hours to row downstream and back.

Step 4: Time for Round Trip Up the River (d)

The tasks here are essentially reversed from part (c). For the upstream part:\[ v_{u} = 3.20 \text{ km/h} \]Time for \(3.20\) km upstream:\[ t_{u} = \frac{3.20}{3.20} = 1 \text{ hour} \]For the downstream return:\[ v_{d} = 9.60 \text{ km/h} \]Time for \(3.20\) km downstream:\[ t_{d} = \frac{3.20}{9.60} = 0.33 \text{ hours} \]Total time for round trip:\[ t_{total} = t_{u} + t_{d} = 1.00 + 0.33 = 1.33 \text{ hours} \]It takes the same 1.33 hours to row upstream and back.

Step 5: Shortest Time Across the River (e)

The shortest time occurs when she aims straight across the river. In this case, her effective speed is her full speed in still water, because no part of her speed counters the current:\[ \text{Shortest time speed} = v_{b} = 6.40 \text{ km/h} \]The direction is straight across, perpendicular to the bank. The time is:\[ t = \frac{\text{Distance}}{\text{Speed}} = \frac{6.40}{6.40} = 1 \text{ hour} \]She should head straight across the river, and it will take her 1 hour in this case.

Key Concepts

Vector Components

When solving physics problems involving vectors, it's essential to understand the concept of vector components. Vectors are quantities that have both magnitude and direction, such as velocity or force. To make calculations easier, we often break a vector down into its components using trigonometry. These components lie along the axes of a coordinate system, typically the x-axis and y-axis in a two-dimensional setup. For example, if a vector represents a woman's rowing velocity, one component aligns with the direction across the river, while the other counters the river's current.

In the exercise given, the rowing speed across the river can be considered as a vector. The horizontal component, which helps the woman reach directly across, is calculated using the cosine function. This component ignores the influence of the river current. The other component, which negates the current, is found using the sine function.

Understanding vector components allows us to assess different real-world situations, enabling precise control and navigation in tasks like rowing a boat or piloting an aircraft.

Relative Velocity

Relative velocity refers to the measurement of one object's velocity in relation to another object. It's a crucial concept when analyzing situations involving motion in environments such as rivers or air currents. In the exercise, we calculate the woman's velocity relative to the ground by taking into account both her rowing speed in still water and the current's speed.

To solve such problems, identify the velocities of both the moving object and the medium it moves through. Then, determine the resultant velocity vector, which considers both speeds. For instance, a downstream trip is faster because the current adds to the boat's speed, while an upstream trip is slower as the current opposes the motion.

Relative velocity helps us predict the actual trajectory and speed of the object in motion. By understanding this concept, we can effectively calculate travel time and direction, which is vital for planning in navigation scenarios.

Trigonometry in Physics

Trigonometry plays a key role in physics, especially in problems involving angles and vectors. It helps resolve the magnitude and direction of vectors into parallel and perpendicular components. In this exercise, trigonometry helps calculate the correct angle for the woman to row to counter the current and reach her destination directly across the river.

Key trigonometric functions like sine, cosine, and tangent help break down a vector into its components. For example, sine is used to find the component that counters the river current, and cosine finds the component perpendicular to it. Knowing the angle, we can use relations like \( \sin{\theta} = \text{Opposite}/\text{Hypotenuse} \) and \( \cos{\theta} = \text{Adjacent}/\text{Hypotenuse} \) to find these components.

Understanding trigonometry's applications in physics enables us to solve complex vector problems and calculate angles and distances more precisely. This understanding is vital for accurately navigating scenarios involving multiple direction changes, like crossing a river.

Kinematics in Two Dimensions

Kinematics is the study of motion without considering the forces that cause it. It becomes more complex when motion occurs in two dimensions, as with the woman's journey in this exercise. Here, her movement involves crossing a river flowing in one direction while she rows in another.

Analyzing kinematics in two dimensions often requires breaking the motion into independent components along two perpendicular axes. In our scenario, these are across and along the river. We calculate the effective velocity in each direction and time to determine how long her trips take, considering both the vector components of her speed and the river's current.

• The time taken to cross relies on the velocity component perpendicular to the current.
• The round trip times alter based on the current aiding or opposing the journey.

Kinematics in two dimensions is essential for accurately modeling and predicting motion in contexts involving natural elements or artificial settings, enhancing our understanding and problem-solving abilities in physics.