/*! 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 17 Two runners start simultaneously... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two runners start simultaneously at opposite ends of a \(200.0 \mathrm{~m}\) track and run toward each other. Runner \(A\) runs at a steady \(8.0 \mathrm{~m} / \mathrm{s}\) and runner \(B\) runs at a constant \(7.0 \mathrm{~m} / \mathrm{s}\). When and where will these runners meet?

Short Answer

Expert verified
They meet at 13.33 seconds, 106.64 meters from runner A's start.

Step by step solution

01

Define Variables and Initial Conditions

Let runner A start from position 0 m and runner B start from position 200 m. The position of runner A at time \( t \) is \( 8t \) meters, and the position of runner B at time \( t \) is \( 200 - 7t \) meters.
02

Set the Equations for Meeting Point

To find when and where the runners meet, set the position equations equal to each other: \( 8t = 200 - 7t \).
03

Solve for Time \( t \)

Rearrange the equation: \( 8t + 7t = 200 \), then combine like terms to get \( 15t = 200 \). Solve for \( t \) by dividing both sides by 15, resulting in \( t = \frac{200}{15} = 13.33 \text{ seconds} \).
04

Calculate the Meeting Point Position

Substitute \( t = 13.33 \) seconds into the position equation for runner A: \( 8t = 8(13.33) = 106.64 \text{ meters}\). Hence, they meet at 106.64 meters from runner A's starting point.

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.

Physics Problems
Physics problems can often seem daunting, but they become manageable with proper understanding and approach. In physics, especially in mechanics, problems usually involve identifying the forces and motions involved. In this exercise, the two runners on a track create a classic relative motion problem, which is a common scenario in physics.
Understanding the problem requires identifying key elements:
  • The constant speeds of both runners: Runner A is at 8 m/s, and Runner B is at 7 m/s.
  • The initial conditions: They start from opposite ends of the 200-meter track.
  • The concept of relative motion: How the motion of one object is perceived with respect to another.
By breaking down the problem, you can see how each runner's position changes over time, ultimately finding the point where they meet.
Kinematics
Kinematics is a branch of mechanics that describes the motion of objects without considering the causes of motion (such as forces). In this problem, kinematics is used to describe the positions of Runner A and Runner B as functions of time.
The kinematic equations depend on the straightforward relationship between velocity, time, and position. Here, the position for each runner is noted as:
  • Runner A's position: \(8t\) meters.
  • Runner B's position: \(200 - 7t\) meters.
In this scenario, the goal is to determine the time and position where these two runners meet.
By setting the runners' position equations equal to each other (\(8t = 200 - 7t\)), you can solve for time \(t\). This approach is a straightforward application of kinematic principles, where constant velocity leads to linear position equations.
Problem Solving
Solving any physics problem requires a structured approach. Begin by understanding the problem description clearly and identifying the variables involved.
This exercise is a relative motion problem where the key steps include:
  • Defining initial conditions, such as the starting positions and velocities of the runners.
  • Setting up equations that represent the position of each runner as a function of time.
  • Solving the equations for time \(t\) when the runners meet.
Then, evaluate this time in one of the position equations to find the meeting point. Practice makes perfect, so try solving similar problems to solidify your understanding. Each problem you solve builds your intuition and problem-solving skills.

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

Suppose that you drop a marble from the top of the Burj Khalifa building in Dubai, which is about \(830 \mathrm{~m}\) tall. If you ignore air resistance, (a) how long will it take for the marble to hit the ground? (b) How fast will it be moving just before it hits?

If a pilot accelerates at more than \(4 g\), he begins to "gray out" but does not completely lose consciousness. (a) What is the shortest time that a jet pilot starting from rest can take to reach Mach 4 (four times the speed of sound) without graying out? (b) How far would the plane travel during this period of acceleration? (Use \(331 \mathrm{~m} / \mathrm{s}\) for the speed of sound.)

A car in the northbound lane is sitting at a red light. At the moment the light turns green, the car accelerates from rest at \(2 \mathrm{~m} / \mathrm{s}^{2}\). At this moment, there is also a car in the southbound lane that is \(200 \mathrm{~m}\) away and traveling at a constant \(25 \mathrm{~m} / \mathrm{s}\). The northbound car maintains its acceleration until the two cars pass each other. (a) How long after the light turns green do the cars pass each other? (b) How far from the red light are they when they pass each other?

A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of \(15.0 \mathrm{~m} / \mathrm{s}\). (a) What is its speed after falling freely for \(2.00 \mathrm{~s}\) ? (b) How far does it fall in \(2.00 \mathrm{~s} ?\) (c) What is the magnitude of its velocity after falling \(10.0 \mathrm{~m} ?\)

A brick is released with no initial speed from the roof of a building and strikes the ground in \(2.50 \mathrm{~s},\) encountering no appreciable air drag. (a) How tall, in meters, is the building? (b) How fast is the brick moving just before it reaches the ground? (c) Sketch graphs of this falling brick's acceleration, velocity, and vertical position as functions of time.

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.