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A swimmer, capable of swimming at a speed of \(1.4 \mathrm{~m} / \mathrm{s}\) in still water (i.e., the swimmer can swim with a speed of \(1.4 \mathrm{~m} / \mathrm{s}\) relative to the water), starts to swim directly across a 2.8-km- wide river. However, the current is \(0.91 \mathrm{~m} / \mathrm{s}\), and it carries the swimmer downstream, (a) How long does it take the swimmer to cross the river? (b) How far downstream will the swimmer be upon reaching the other side of the river?

Short Answer

Expert verified
(a) It takes 2000 seconds to cross the river. (b) The swimmer is carried 1820 meters downstream.

Step by step solution

01

Identify Known Values

First, identify the given information in the problem. The swimmer's speed in still water is \(1.4 \text{ m/s}\). The width of the river, which is the distance the swimmer needs to cross, is \(2.8 \text{ km}\) or \(2800 \text{ m}\). The current speed is \(0.91 \text{ m/s}\).
02

Calculate Time to Cross the River

To find the time \(t\) it takes to swim directly across the river, use the formula \(t = \frac{d}{v_{s}}\), where \(d\) is the width of the river and \(v_{s}\) is the swimmer's speed across the river.\[t = \frac{2800 \text{ m}}{1.4 \text{ m/s}}\]Calculate the result to find \(t\).
03

Calculate Swimmer's Time to Cross

Perform the calculation: \[t = \frac{2800}{1.4} = 2000 \text{ seconds}\]
04

Calculate Distance Carried Downstream

To find out how far downstream the swimmer is carried, use the formula \(d_{d} = v_{c} \cdot t\), where \(v_{c}\) is the current speed and \(t\) is the time calculated previously.\[d_{d} = 0.91 \text{ m/s} \times 2000 \text{ s}\]Calculate the result to find \(d_{d}\).
05

Calculate Distance Downstream

Perform the calculation:\[d_{d} = 0.91 \times 2000 = 1820 \text{ meters}\]

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Key Concepts

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

Understanding Relative Velocity
In kinematics, **relative velocity** is essential for analyzing motion from different reference frames. It refers to the velocity of one object as observed from another object. For river crossing problems, this concept is critical. Consider a swimmer in a river. The swimmer has a natural speed, but the river's current affects their overall velocity. The swimmer's speed relative to the riverbank is not just their swimming speed, but rather a combination of their swimming speed and the current.
If the swimmer's speed is given as 1.4 m/s in still water, and the river flows at 0.91 m/s, we calculate their actual path by considering these two vectors. Using vector addition, the overall velocity becomes:
  • Swimmer's speed (relative to water): 1.4 m/s
  • River current speed: 0.91 m/s

Vectors are added together to determine the resultant direction and speed of the swimmer relative to a stationary point on the riverbank.
River Crossing Problems Unraveled
**River crossing problems** in kinematics are classic scenarios used to demonstrate the effects of relative velocity. The challenge lies in navigating from one point to another across a moving medium, such as a river.
When crossing a river, the objective is to reach directly across from the start point. However, due to the current, achieving a straight path requires more than straightforward swimming. The swimmer must account for the river's flow.
In our sample problem, even though the swimmer aims to directly cross a 2.8 km-wide river, the downstream current pushes them off course. Therefore, the path they actually take is diagonal. This situation requires these steps for solution:
  • Determine the time to cross using the perpendicular speed across the river.
  • Calculate the downstream distance carried by the current during this time.
By breaking down the motion into perpendicular components—across the river and downstream effects—we simplify the complexity of the problem.
The Dynamics of Swimming in a Current
**Swimming dynamics** in a current involve understanding how a swimmer’s motion changes due to the external water flow. Knowing how to account for these changes is essential.

Firstly, when a swimmer enters the water, their speed gets vectorially combined with the current. Thus, the swimmer's direction and speed alter, as if they were on a moving walkway. The swimmer in our given exercise needs to adjust for this affect, even without changing their speed relative to the water.
  • To determine travel time: Focus on the perpendicular component, using the formula \( t = \frac{d}{v_{s}} \) where \( d \) is the river width and \( v_{s} \) the swimming speed in still water.
  • To ascertain drift: Calculate using \( d_{d} = v_{c} \cdot t \), where \( v_{c} \) is the current speed.

These calculations give insight into real-world swimming where natural environments pose unique challenges, emphasizing the importance of comprehending vector components and how they relate to overall motion.

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Most popular questions from this chapter

A skateboarder, starting from rest, rolls down a 12.0 -m ramp. When she arrives at the bottom of the ramp her speed is \(7.70 \mathrm{~m} / \mathrm{s}\). (a) Determine the magnitude of her acceleration, assumed to be constant, (b) If the ramp is inclined at \(25.0^{\circ}\) with respect to the ground, what is the component of her acceleration that is parallel to the ground?

A rifle is used to shoot twice at a target, using identical cartridges. The first time, the rifle is aimed parallel to the ground and directly at the center of the bull's-eye. The bullet strikes the target at a distance of \(H_{\mathrm{A}}\) below the center, however. The second time, the rifle is similarly aimed, but from twice the distance from the target. This time the bullet strikes the target at a distance of \(H_{\mathrm{B}}\) below the center. Find the ratio \(H_{\mathrm{B}} / H_{\mathrm{A}}\).

The ball is \(26.9 \mathrm{~m}\) from the goalpost. The ball is kicked with an initial velocity of \(19.8 \mathrm{~m} / \mathrm{s}\) at an angle \(\theta\) above the ground. Between what two angles, \(\theta_{1}\) and \(\theta_{2}\), will the ball clear the \(2.74\) -m-high crossbar? (Hint: The following trigonometric identities may be useful: \(\sec \theta=1 /(\cos \theta)\) and \(\sec ^{2} \theta=1+\tan ^{2} \theta\)

Two cars, \(A\) and \(B,\) are traveling in the same direction, although car \(A\) is \(186 \mathrm{~m}\) behind car \(\mathrm{B}\). The speed of \(\mathrm{A}\) is \(24.4 \mathrm{~m} / \mathrm{s},\) and the speed of \(\mathrm{B}\) is \(18.6 \mathrm{~m} / \mathrm{s}\). How much time does it take for A to catch B?

Relative to the ground, a car has a velocity of \(18.0 \mathrm{~m} / \mathrm{s},\) directed due north. Relative to this car, a truck has a velocity of \(22.8 \mathrm{~m} / \mathrm{s}\), directed \(52.1^{\circ}\) south of east. Find the magnitude and direction of the truck's velocity relative to the ground.

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