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Chapter 3: Problem 41

Crossing the River I. A river flows due south with a specd 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 to cross the river? (c) How far south of his starting point will he reach the opposite bank?

Short Answer

Expert verified
(a) 4.65 m/s at 25.4掳 south of east; (b) 190.48 s; (c) 381 m south.

Step by step solution

01

Understand the Components of Velocity

The man's motion with respect to the earth involves two components: the west-east component due to the boat's velocity of 4.2 m/s, and the north-south component due to the river's current of 2.0 m/s.
02

Calculate Resultant Velocity's Magnitude

Use the Pythagorean theorem to calculate the magnitude of the resultant velocity. The formula for the resultant velocity is defined as \( v = \sqrt{(v_{east})^2 + (v_{south})^2} \), where \( v_{east} = 4.2 \) m/s and \( v_{south} = 2.0 \) m/s.\[ v = \sqrt{(4.2)^2 + (2.0)^2} = \sqrt{17.64 + 4.00} = \sqrt{21.64} \approx 4.65 \text{ m/s} \]
03

Calculate Direction of the Resultant Velocity

Use the tangent function to find the direction. The direction \( \theta \) can be calculated using \( \tan(\theta) = \frac{v_{south}}{v_{east}} = \frac{2.0}{4.2} \). Solving for the angle:\[ \theta = \tan^{-1}\left(\frac{2.0}{4.2}\right) \approx 25.4^\circ \] The angle is measured from the east towards south.
04

Calculate Time to Cross the River

The time required to cross the river only depends on the east-west velocity component, as the river is 800 m wide. Use the formula \( t = \frac{distance}{velocity} \), where the distance is 800 m and the velocity component is 4.2 m/s.\[ t = \frac{800}{4.2} \approx 190.48 \text{ seconds} \]
05

Calculate Southward Displacement

The southward displacement is calculated based on the time it takes to cross the river and the southward velocity component. Use the formula \( d = v_{south} \times t \), where \( v_{south} = 2.0 \) m/s and \( t \approx 190.48 \) s.\[ d = 2.0 \times 190.48 \approx 380.96 \text{ m} \]

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Key Concepts

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

Vector Components
Understanding vector components is essential in analyzing motion in two dimensions. In the context of the river crossing problem, the boat's movement involves two perpendicular vector components: one from the boat trying to move east and the other from the river current moving south. These components can be represented as two separate velocities:
  • Eastward velocity of the boat: 4.2 m/s
  • Southward velocity of the river: 2.0 m/s
These components form a right-angled triangle where the hypotenuse represents the resultant velocity. In physics, breaking down the velocity into components helps us solve for quantities like the resultant velocity and direction without getting overwhelmed by the problem's complexity.
Resultant Velocity
The resultant velocity is a crucial concept when dealing with relative motion problems like the crossing of a river. It represents the actual path and speed an object takes relative to a stationary observer. To find this resultant velocity, we use the Pythagorean theorem since we are dealing with perpendicular components:\[v = \sqrt{(v_{east})^2 + (v_{south})^2}\]For the given problem:\[v = \sqrt{(4.2)^2 + (2.0)^2} = \sqrt{21.64} \approx 4.65 \text{ m/s}\]This velocity tells us how fast the man in the boat is actually moving across the ground. Its direction is also important and can be determined using trigonometric functions like tangent to find the angle of movement:\[\theta = \tan^{-1}\left(\frac{v_{south}}{v_{east}}\right) \approx 25.4^\circ\]This angle is measured from the east towards the south, showing the path the boat takes as it crosses the river.
Crossing the River Problem
The crossing the river problem is a classic scenario in physics that involves understanding relative motion in two dimensions. To find solutions to this problem, it is crucial to know how long it takes to cross the river and where the boat will end up on the opposite bank.For the time required to cross an 800 m wide river, only the eastward velocity component is used. It is because this component is solely responsible for moving the boat across the width of the river:\[t = \frac{distance}{velocity} = \frac{800}{4.2} \approx 190.48 \text{ seconds}\]In terms of final position, the southward current causes a drift from the boat's intended path. This drift can be calculated via the formula for displacement:\[d = v_{south} \times t = 2.0 \times 190.48 \approx 380.96 \text{ m}\]The calculation shows how the man will end up approximately 380.96 meters south of his starting point upon reaching the opposite bank. Understanding these concepts helps one predict the outcomes of dealing with multiple simultaneous motions.

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Most popular questions from this chapter

A faulty model rocket moves in the \(x y\) -plane (the positive \(y\) - direction is vertically upward. The rocket's acceleration has components \(a_{x}(t)=\alpha t^{2}\) and \(a_{y}(t)=\beta-\gamma t,\) where \(\alpha=2.50 \mathrm{m} / \mathrm{s}^{4}\) , \(\beta=9.00 \mathrm{m} / \mathrm{s}^{2},\) and \(\gamma=1.40 \mathrm{m} / \mathrm{s}^{3} .\) At \(t=0\) the rocket is at the origin and has velocity \(\vec{v}_{0}=v_{\mathrm{ux}} \hat{\mathrm{i}}+v_{\mathrm{oy}} \hat{\mathrm{j}}\) with \(v_{\mathrm{ox}}\) \(=1.00 \mathrm{m} / \mathrm{s}\) and \(v_{0 y}=7.00 \mathrm{m} / \mathrm{s}\) , (a) Calculate the velocity and position vectors as functions of time. (b) What is the maximum height reached by the rocket? (c) Sketch the path of the rocket. (d) What is the horizontal displacement of the rocket when it returns to \(y=0 ?\)

Bang! \(\quad\) A student sits atop a platform a distance \(h\) above the ground. He throws a large firecracker horizontally with a speed \(v\) . However, a wind blowing parallel to the ground gives the fire- cracker a constant horizontal acceleration with magnitude \(a\) . This results in the firecracker reaching the ground directly under the student. Determine the height \(h\) in terms of \(v, a,\) and \(g\) . You can ignore the effect of air resistance on the vertical motion.

A pistol that fires a signal flare gives it an initial velocity (muzzle velocity) of 125 \(\mathrm{m} / \mathrm{s}\) at an angle of \(55.0^{\circ}\) above the borizontal. You can ignore air resistance. Find the flare's maximum height and the distance from its firing point to its landing point if it is fired (a) on the level salt flats of Utah, and (b) over the flat Sea of Tranquility on the Moon, where \(g=1.67 \mathrm{m} / \mathrm{s}^{2}\) .

A daring \(510-N\) swimmer dives off a cliff with a running horizontal leap, as shown in Fig. 3.39 . What must her mini- mum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom. which is 1.75 \(\mathrm{m}\) wide and 9.00 \(\mathrm{m}\) below the top of the cliff?

Raindrops. When a train's velocity is 120 \(\mathrm{m} / \mathrm{s}\) eastward, raindrops that are falling vertically with respect to the earth make traces that are inclined \(30.0^{\circ}\) to the vertical on the windows of the train. (a) What is the horizontal component of a drop's velocity with respect to the earth? With respect to the train? (b) What is the magnitude of the velocity of the raindrop with respect to the earth? With respect to the train?
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