Problem 40 - College Physics

Question

\(\bullet\) Crossing the river, I. A river flows due south with a speed of 2.0 \(\mathrm{m} / \mathrm{s} .\) A man steers a motorboat across the river; his velocity relative to the water is 4.2 \(\mathrm{m} / \mathrm{s}\) due east. The river is 800 \(\mathrm{m}\) wide. (a) What is his velocity (magnitude and direction) relative to the earth? (b) How much time is required for the man to cross the river? (c) How far south of his starting point will he reach the opposite bank?

Step-by-Step Solution

Answer

Velocity relative to earth: 4.65 m/s at 25.5° south of east, time to cross: 190.5 s, and 381 m south of the start.

Step 1: Understand the Problem

The problem involves a man in a motorboat crossing a river with a flow speed. We need to determine his resultant velocity with respect to the earth, the time it takes to cross the river, and how far south he ends up.

Step 2: Determine Velocity Relative to Earth

To find the man's velocity relative to the earth, we create a right triangle where the river's velocity (2.0 m/s south) and the boat's velocity relative to water (4.2 m/s east) are the legs of the triangle. The resultant velocity is the hypotenuse.We use the Pythagorean theorem:\[v_{earth} = \sqrt{(2.0)^2 + (4.2)^2}\]Calculating gives:\[v_{earth} = \sqrt{4 + 17.64} = \sqrt{21.64} \approx 4.65\, \text{m/s}\]

Step 3: Determine Direction of Resultant Velocity

The direction (angle \( \theta \)) can be found using the tangent function:\[\tan(\theta) = \frac{2.0}{4.2}\]\[\theta = \arctan\left(\frac{2.0}{4.2}\right) \approx 25.5^\circ\]This angle is measured south of east.

Step 4: Calculate Time to Cross the River

To find the time taken to cross the river, use the width of the river (800 m) and only the eastward velocity component (4.2 m/s), because that is the component perpendicular to the riverbanks:\[ t = \frac{800}{4.2} \approx 190.5 \, \text{seconds}\]

Step 5: Calculate Southward Displacement

The southward component of velocity is 2.0 m/s. Use this to find the southward displacement:\[d = 2.0 \times 190.5 \approx 381 \, \text{m}\]This is how far south of the starting point the man will reach the opposite bank.

Key Concepts

Vector Addition

In physics, vector addition helps us find a resultant vector when multiple vectors are acting together. Imagine vectors as arrows. They have both direction and magnitude. To find the overall effect or resultant vector, we need to "add" these arrows. In our river crossing problem, the boat is subject to two main vectors:

• The current's velocity of 2.0 m/s flowing south.
• The boat's own velocity of 4.2 m/s heading east.

Instead of just adding them like numbers, we treat them geometrically. We form a right triangle where these two velocities are the legs and the resultant velocity is the hypotenuse. This tribunal approach gives us a simple yet effective way to visualize how different forces—or vectors—interact in the real world.

Pythagorean Theorem

The Pythagorean theorem is crucial when dealing with perpendicular vectors. It allows us to calculate the magnitude of the resultant vector. This theorem states that in a right triangle, the square of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b). Mathematically, it's written as: \[ c = \sqrt{a^2 + b^2} \]In our scenario, the boat's eastward speed is 4.2 m/s, and the river's southward speed is 2.0 m/s. These values serve as a and b, respectively. Plugging into the formula, we find:\[ v_{earth} = \sqrt{(2.0)^2 + (4.2)^2} \approx 4.65 \, \text{m/s} \]This resultant speed is the boat's velocity relative to the earth, considering both the river's flow and the boat's motor.

Angle of Direction

Calculating the angle of direction helps us understand where the resultant velocity is pointing. When two vectors are acting, they won't always point straight in one direction. They create an angle, showing the mixed direction of forces. To find this angle, we use tangent, a trigonometric function. Knowing the opposite (southward) and adjacent (eastward) sides of our triangle:\[ \theta = \arctan\left(\frac{\text{opposite}}{\text{adjacent}}\right) \]In this river crossing, it looks like:\[ \theta = \arctan\left(\frac{2.0}{4.2}\right) \approx 25.5^\circ \]So, our angle indicates a path that is about 25.5 degrees south of east. This means the boat moves eastward, but also slightly towards the south due to the river's current.

River Crossing

When crossing a river with a current, understanding both the vector addition and the time it takes is fundamental. Here, even if the motorboat travels directly across the river (eastward), the flow will carry it downstream (southward). Hence, it doesn't end up directly across but displaced somewhat south.To find how long it takes to cross, we consider the distance perpendicular to the current—here, the river's width of 800 meters—and divide by the component of the velocity pointing in that direction (4.2 m/s east). Thus:\[ t = \frac{800}{4.2} \approx 190.5 \, \text{seconds} \]Additionally, the boat is pushed southward during this time by the current. The southward displacement, calculated using the velocity along the current and time, is:\[ d = 2.0 \times 190.5 \approx 381 \, \text{m} \]This value tells us how far downstream the boat ends up from its initial point once it reaches the opposite bank.