/*! 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 52 In a marathon race Chad is out i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a marathon race Chad is out in front, running due north at a speed of \(4.00 \mathrm{m} / \mathrm{s} .\) John is \(95 \mathrm{m}\) behind him, running due north at a speed of \(4.50 \mathrm{m} / \mathrm{s} .\) How long does it take for John to pass Chad?

Short Answer

Expert verified
It takes John 190 seconds to pass Chad.

Step by step solution

01

Understand the Given Problem

The problem states that Chad is running north at a speed of \(4.00 \mathrm{m}/\mathrm{s}\) and John is also running north at a speed of \(4.50 \mathrm{m}/\mathrm{s}\), but John starts from a position \(95 \mathrm{m}\) behind Chad. We need to find out how long it takes for John to pass Chad.
02

Set Up the Distance Equation

To find out when John passes Chad, we need to set up an equation where the distance traveled by John equals the distance traveled by Chad plus the initial \(95 \mathrm{m}\) distance gap. Let \(t\) be the time in seconds. Chad's Distance: \(d_C = 4.00t\) John's Distance: \(d_J = 4.50t\)
03

Formulate the Equality

The condition for John passing Chad is that the distance John has traveled equals the distance Chad has traveled plus the initial \(95\) meter gap:\[ 4.50t = 4.00t + 95 \]
04

Solve for Time \(t\)

Rearrange the equation to solve for \(t\):\[ 4.50t - 4.00t = 95 \]Simplify the left side:\[ 0.50t = 95 \]Divide both sides by \(0.50\) to isolate \(t\):\[ t = \frac{95}{0.50} \]\[ t = 190 \]
05

Verify the Solution

We have found \(t = 190\) seconds. To verify, check that John covers the gap correctly:John travels: \(4.50 \times 190 = 855 \mathrm{m}\)Chad travels: \(4.00 \times 190 = 760 \mathrm{m}\)The difference is \(855 - 760 = 95 \mathrm{m}\), which is correct.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Equations of Motion
In the field of physics, equations of motion are key tools for predicting the future position of moving objects. They relate various kinematic variables including velocity, acceleration, time, and displacement. In simple terms, they help us calculate how far an object moves within a given period of time.
For example, if you know the speed of an athlete and the time they have been running, you can use an equation of motion to determine the distance they have traveled. In the exercise, we applied the formula for uniform motion: \(d = vt\), where \(d\) is distance, \(v\) is velocity, and \(t\) is time.
These equations are essential for solving problems involving straight line motion, like running (or marathons). They simplify the complexities of motion, translating them into manageable math tasks.
Distance-Time Relationship
The relationship between distance and time is crucial when solving motion problems. It helps us understand how an object's position changes with time. In the given exercise, this concept was applied by comparing the distances covered by both runners, John and Chad, over time.
The core idea is that distance increases with time when an object moves at a constant speed. By establishing the formula \(d = vt\) for each runner, we could express their distances as a function of time. This relationship allowed us to set up the equation \(4.50t = 4.00t + 95\) and solve for the time \(t\) it takes John to catch up with Chad.
This clear connection between distance and time is essential in systematically approaching and solving relative motion problems.
Speed and Velocity
Speed and velocity are fundamental concepts in the study of motion. Although often used interchangeably in everyday conversation, they have distinct meanings in physics.
Speed is a scalar quantity, which means it only considers the magnitude, or how fast an object is moving. For example, both John and Chad's speeds are given as \(4.50\,\text{m/s}\) and \(4.00\,\text{m/s}\) respectively.
Velocity, on the other hand, is a vector quantity. This means it considers both the speed and the direction of movement. In our exercise, both runners are moving north, so their velocity includes a directional component. Understanding these differences is vital for accurately solving motion problems, especially when direction plays a role.
Problem-Solving in Physics
Problem-solving in physics often requires a structured approach to tackle various scenarios in motion-related exercises. The key steps typically include: understanding the problem, setting up the necessary equations, and solving for the unknowns.
In the provided solution, for example, the approach was methodical. We started by understanding the situation, noted the respective speeds, and distances involved. Then, we framed the problem with an appropriate equation using the distance-time relationship, i.e., aligning the distances of John and Chad.
Such a step-by-step approach ensures clarity and accuracy, allowing even complex problems to be broken down and solved effectively.
  • Begin by thoroughly understanding the problem statement.
  • Sketch or visualize the situation, if possible.
  • Identify and define the known and unknown variables.
  • Formulate the relevant equations.
  • Solve for the unknown variables step-by-step.
  • Finally, verify the solution to ensure correctness.
With practice, this organized method enhances one's ability to tackle physics problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A spacecraft is traveling with a velocity of \(v_{0 x}=5480 \mathrm{m} / \mathrm{s}\) along the \(+x\) direction. Two engines are turned on for a time of 842 s. One engine gives the spacecraft an acceleration in the \(+x\) direction of \(a_{x}=1.20 \mathrm{m} / \mathrm{s}^{2}\), while the other gives it an acceleration in the \(+y\) direction of \(a_{y}=8.40 \mathrm{m} / \mathrm{s}^{2}\) At the end of the firing, find (a) \(v_{x}\) and (b) \(v_{y^{*}}\)

Two friends, Barbara and Neil, are out rollerblading. With respect to the ground, Barbara is skating due south at a speed of \(4.0 \mathrm{m} / \mathrm{s}\). Neil is in front of her. With respect to the ground, Neil is skating due west at a speed of \(3.2 \mathrm{m} / \mathrm{s}\). Find Neil's velocity (magnitude and direction relative to due west), as seen by Barbara.

Two passenger trains are passing each other on adjacent tracks. Train A is moving east with a speed of \(13 \mathrm{m} / \mathrm{s}\), and train \(\mathrm{B}\) is traveling west with a speed of \(28 \mathrm{m} / \mathrm{s}\). (a) What is the velocity (magnitude and direction) of train A as seen by the passengers in train B? (b) What is the velocity (magnitude and direction) of train \(\mathrm{B}\) as seen by the passengers in train A?

A golfer imparts a speed of \(30.3 \mathrm{m} / \mathrm{s}\) to a ball, and it travels the maximum possible distance before landing on the green. The tee and the green are at the same elevation. (a) How much time does the ball spend in the air? (b) What is the longest hole in one that the golfer can make, if the ball does not roll when it hits the green?

A ball is thrown upward at a speed \(v_{0}\) at an angle of \(52^{\circ}\) above the horizontal. It reaches a maximum height of \(7.5 \mathrm{m}\). How high would this ball go if it were thrown straight upward at speed \(v_{0} ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.