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Chapter 4: Problem 71

A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking \(2.50 \mathrm{~s}\). Then security agents appear, and the man runs as fast as he can back along the sidewalk to his starting point, taking \(10.0 \mathrm{~s}\). What is the ratio of the man's running speed to the sidewalk's speed?

Short Answer

Expert verified
The man's running speed is \( \frac{5}{3} \) times the sidewalk's speed.

Step by step solution

01

Define Variables

Let's define the variables involved. Let \( v_m \) be the man's running speed, \( v_s \) be the sidewalk's speed, and \( d \) be the length of the sidewalk. When moving from one end to the other, the man's speed relative to the ground is \( v_m + v_s \), and when returning, it is \( v_m - v_s \).
02

Formulate Equations

The total distance covered in both directions is the length of the sidewalk \( d \). Moving forward, we have: \[ t_1 = \frac{d}{v_m + v_s} \] For returning, we have: \[ t_2 = \frac{d}{v_m - v_s} \] Given, \( t_1 = 2.50 \) s and \( t_2 = 10.0 \) s.
03

Express Time Equations

Using the given time, express the length \( d \) in terms of speeds and time for both forwards and backwards movements: \[ d = (v_m + v_s)t_1 \] \[ d = (v_m - v_s)t_2 \]
04

Set Equations Equal

Set the equations derived for \( d \) equal to each other: \[ (v_m + v_s) \cdot 2.5 = (v_m - v_s) \cdot 10.0 \] Simplifying gives: \[ 2.5v_m + 2.5v_s = 10v_m - 10v_s \]
05

Solve for Speed Ratio

Rearrange the equation to solve for \( \frac{v_m}{v_s} \): \[ 2.5v_m + 2.5v_s = 10v_m - 10v_s \] \[ 12.5v_s = 7.5v_m \] \[ \frac{v_m}{v_s} = \frac{12.5}{7.5} \] Simplify the fraction to find the ratio: \[ \frac{v_m}{v_s} = \frac{5}{3} \]
06

Final Step: Conclusion

The man's running speed is \( \frac{5}{3} \) times the sidewalk's speed.

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Key Concepts

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

Relative Velocity
Relative velocity is the concept of evaluating the speed of one object with respect to another object. This is important in scenarios where two different velocities interact, such as a person running on a moving sidewalk.
In the given exercise, the man's speed relative to the ground changes depending on whether he is running with the sidewalk or against it. When moving in the same direction as the moving sidewalk, his effective speed is the sum of his running speed and the sidewalk's speed.
  • Forward speed: \( v_m + v_s \)
  • Return speed: \( v_m - v_s \)
Understanding relative velocity helps us determine how long it takes for him to traverse the sidewalk in each direction.
Kinematics
Kinematics is a branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. In this problem, we use kinematic equations to understand the motion of the man along the sidewalk and solve the exercise.
The key kinematic equation here connects distance, speed, and time:
  • Distance \( d \) = Speed \( v \) 脳 Time \( t \)
Given the time durations for the forward and return trips, and knowing the relation between distance, speed, and time, we can set up equations for both parts of the journey. The use of kinematics allows us to transform the problem into one that can be approached via algebraic manipulation.
Equation Solving
Solving equations is fundamental to breaking down complex problems in physics, such as this exercise. Here, we derived two expressions for the sidewalk's distance, each based on the time taken to complete the journey in different directions, and set them equal to each other.
This approach has allowed us to simplify the system involving the man's speed \( v_m \) and the sidewalk's speed \( v_s \):
  • \((v_m + v_s) \cdot 2.5 = (v_m - v_s) \cdot 10.0\)
  • Rearrange and simplify to find \(\frac{v_m}{v_s} = \frac{5}{3}\)
By handling these equations carefully, we obtain a solution that accurately describes the speed ratio between the man and the sidewalk.
Motion Analysis
Motion analysis involves breaking down and understanding how movement happens within different contexts or scenarios. It's about understanding the intricacies of distance, speed, and time relationships.
In this exercise, we analyze motion in two distinct parts:
  • Forward trip with the sidewalk's help
  • Return trip against the sidewalk's movement
By examining these scenarios and applying our understanding of speeds, times, and distances, we can gain insights into the dynamics involved. The sidewalk adds an extra layer that adjusts the man's effective speed, providing a challenge typical of motion analysis problems.

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Most popular questions from this chapter

A \(200-\mathrm{m}\) -wide river flows due east at a uniform speed of \(2.0 \mathrm{~m} / \mathrm{s}\). A boat with a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) relative to the water leaves the south bank pointed in a direction \(30^{\circ}\) west of north. What are the (a) magnitude and (b) direction of the boat's velocity relative to the ground? (c) How long does the boat take to cross the river?

A plane flies \(483 \mathrm{~km}\) east from city \(A\) to city \(B\) in \(45.0 \mathrm{~min}\) and then \(966 \mathrm{~km}\) south from city \(B\) to city \(C\) in \(1.50 \mathrm{~h}\). For the total trip, what are the (a) magnitude and (b) direction of the plane's displacement, the (c) magnitude and (d) direction of its average velocity, and (e) its average speed?

A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wall to break apart the wall. The machine was not placed near the wall because then arrows could reach it from the castle wall. Instead, it was positioned so that the stone hit the wall during the second half of its flight. Suppose a stone is launched with a speed of \(v_{0}=28.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(\theta_{0}=40.0^{\circ}\). What is the speed of the stone if it hits the wall (a) just as it reaches the top of its parabolic path and (b) when it has descended to half that height? (c) As a percentage, how much faster is it moving in part (b) than in part (a)?

A moderate wind accelerates a pebble over a horizontal \(x y\) plane with a constant acceleration \(\vec{a}=\left(5.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(7.00 \mathrm{~m} / \mathrm{s}^{2}\right) \mathrm{j}\). At time \(t=0\), the velocity is \((4.00 \mathrm{~m} / \mathrm{s})\) i. What are the (a) magnitude and (b) angle of its velocity when it has been displaced by \(12.0\) \(\mathrm{m}\) parallel to the \(x\) axis?

An ion's position vector is initially \(\vec{r}=5.0 \hat{\mathrm{i}}-6.0 \hat{\mathrm{j}}+2.0 \hat{\mathrm{k}}\), and \(10 \mathrm{~s}\) later it is \(\vec{r}=-2.0 \hat{\mathrm{i}}+8.0 \hat{\mathrm{j}}-2.0 \hat{\mathrm{k}}\), all in meters. In unit- vector notation, what is its \(\vec{v}_{\text {ave }}\) during the \(10 \mathrm{~s}\) ?
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