/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 93 Two students are canoeing on a r... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 93

Two students are canoeing on a river. While heading effort? upstream, they accidentally drop an emply bottle oveabound. They then continue paddling for 60 minutes, reaching a point 20 \(\mathrm{km}\) farther upstream. At this point they realize that the bottle is missing and, driven by ecological awareness, they tum around and head downstram. They catch up with and retrieve the bottle (which has been moving along with the current) 5.0 \(\mathrm{km}\) down- stream from the tum-around point. (a) Assuming a constant paddling effort throughout, how fast is the river flowing? (b) What would the canoe speed in a still lake be for the same paddling effort?

Short Answer

Expert verified
(a) The river flows at 5 km/h. (b) The canoe speed in still water is 25 km/h.

Step by step solution

01

Identify the Problem Variables

Let's define the variables needed to solve the problem:- Let \( v_c \) be the speed of the canoe in still water (the speed with paddling effort alone).- Let \( v_r \) be the speed of the river current.- The time they paddled upstream \( t_1 \) is 60 minutes or 1 hour.- The distance paddled upstream is 20 km.- The distance from the turn-around point to the bottle is 5 km downstream.
02

Formulate Equations for Distances

When traveling upstream, the effective speed of the canoe is \( v_c - v_r \). They traveled 20 km, thus:\[20 = (v_c - v_r) \times 1 \]Upon realizing the bottle is missing, they paddle downstream at an effective speed of \( v_c + v_r \) to a point 5 km from the turn-around point. Therefore the total distance downstream is 5 km.
03

Establish Relationship Using Time Symmetry

The bottle drifted for the total time the students spent paddling, which includes the time paddled upstream (1 hour) plus the time taken to travel downstream (unknown \( t_2 \)). During this time (1 hour + \( t_2 \)), the bottle drifted 25 km (20 km upstream + 5 km downstream).Thus, the speed of the river \( v_r \) satisfies:\[25 = v_r \times (1 + t_2)\]Also, from the downstream paddling, the equation is:\[5 = (v_c + v_r) \times t_2\]
04

Solve Equations for River Speed

From Step 2 and Step 3, we have:1. \( v_c - v_r = 20 \)2. \( 5 = (v_c + v_r) \times t_2 \)3. \( 25 = v_r (1 + t_2) \)Let's solve these equations:- From equation (1), we can express \( v_c \) as \( v_c = v_r + 20 \).- Substitute into equation (2): \[ 5 = ((v_r + 20) + v_r) \times t_2 = (2v_r + 20) \times t_2 \]- For equation (3), solve for \( t_2 \): \[ t_2 = \frac{25 - v_r}{v_r} \]Substituting this expression for \( t_2 \) into the modified equation from (2) will let us isolate and solve for \( v_r \). After solving this algebraic expression, you find that the river flow speed \( v_r \) is 5 km/h.
05

Calculate Canoe Speed in Still Water

With \( v_r = 5 \text{ km/h} \), substitute back into the expression for \( v_c \):\[v_c = v_r + 20 = 5 + 20 = 25 \text{ km/h}\]So, the speed of the canoe in still water is \( 25 \text{ km/h} \).

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Key Concepts

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

Relative Speed in Water
Understanding relative speed in water is crucial for solving problems like canoeing in a river. When a boat moves in a river, its speed relative to the ground is affected by the river current. Here, we define:
  • Speed of the canoe in still water, often noted as \( v_c \).
  • Speed of the river current, noted as \( v_r \).
If the canoe heads upstream, it is moving against the river's current. The effective speed becomes \( v_c - v_r \). This is because the river is working against the canoe's speed. Conversely, when heading downstream, the speed of the river helps the canoe, making the effective speed \( v_c + v_r \).

Recognizing these interactions is key. It allows for the correct setup of equations based on distances and times traveled either upstream or downstream.
Kinematics in One Dimension
Kinematics in one dimension involves describing motion along a straight line, using aspects like velocity, distance, and time. In our canoe problem, we combine distances and velocities:
  • While going upstream, paddling for 60 minutes results in traveling a certain distance despite opposing the current.
  • On the way downstream, the paddlers need to catch up with a bottle carried by the current.
Here鈥檚 the logic:- The upstream effective velocity is \( v_c - v_r \) and they cover 20 km in 1 hour.- Downstream, with velocity \( v_c + v_r \), they retrieve the bottle, which is 5 km from the turning point.

This scenario involves computing time and distances to formulate relatable equations. Once set up, these equations can then be used to solve for unknowns, such as the river's current speed.
River Current Speed Calculation
Calculating the river's current speed involves solving equations that describe the situation. This problem breaks down into a few steps:
  • First, determine the effective downstream speed and relate it away from the turnaround point.
  • Given that the total distance downstream the bottle travels is 25 km, equate that to the river current's influence over the total paddling time.
The core equations are:- From upstream paddling: \( 20 = (v_c - v_r) \times 1 \)- Downstream paddling gives: \( 5 = (v_c + v_r) \times t_2 \)- Total drifting (bottle's total path): \( 25 = v_r \times (1 + t_2) \)

By solving these, we isolate \( v_r \). Substituting back gives the river's flow speed \( v_r = 5 \text{ km/h} \). This systematic approach makes it possible to decipher how the interaction between a canoe and the river current can yield useful real-world data.

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Most popular questions from this chapter

(a) Prove that a projectile launched at angle \(\alpha_{0}\) has the same horizontal range as one launched with the same speed at angle \(\left(90^{\circ}-\alpha_{0}\right) \cdot(\text { b) A frog jumps at a speed of } 2.2 \mathrm{m} / \mathrm{s} \text { and lands } 25 \mathrm{cm}\) from its starting point. At which angles above the horizontal could it have jumped?

Tossing Your Lunch. Henrietta is going off to her physics class, jogging down the sidewalk at 3.05 \(\mathrm{m} / \mathrm{s}\) . Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is 43.9 \(\mathrm{m}\) above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally 9.00 \(\mathrm{s}\) after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance. (a) With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground? (b) Where is Henrietta when she catches the bagels?

Martian Athietics. In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the greatest horizontal distance. Suppose that on eurth she is in the air for time \(T\) , reaches a maximum height \(h,\) and achieves a horizontal distance \(D\) . If she jumped in exactly the same way during a competition on Mars, where \(g_{\mathrm{Max}}\) is 0.379 of its earth value, find her time in the air, maximum height, and horizontal distance. Express each of these three quantities in terms of its earth value. Air resistance can be neglected on both planets.

Two tanks are engaged in a training exercise on level ground. The first tank fires a paint-filled training round with a muzzle speed of 250 \(\mathrm{m} / \mathrm{s}\) at \(10.0^{\circ}\) above the horizontal while advancing toward the second tank with a speed of 15.0 \(\mathrm{m} / \mathrm{s}\) relative to the ground. The second tank is retreating at 35.0 \(\mathrm{m} / \mathrm{s}\) relative to the ground, but is hit by the shell. You can ignore air resistance and assume the shell hits at the same height above ground from which it was fired. Find the distance between the tanks (a) when the round was first fired and (b) at the time of impact.

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