/*! 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 24 Consult Concept Simulation 2.1 a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 24

Consult Concept Simulation 2.1 at for help in preparing for this problem. A cheetah is hunting. Its prey runs for \(3.0 \mathrm{~s}\) at a constant velocity of \(+9.0 \mathrm{~m} / \mathrm{s}\). Starting from rest, what constant acceleration must the cheetah maintain in order to run the same distance as its prey runs in the same time?

Short Answer

Expert verified
The cheetah must maintain an acceleration of 6.0 m/s虏.

Step by step solution

01

Understanding the Problem

The prey runs at a constant velocity for 3 seconds, covering a certain distance. We need to find the constant acceleration for the cheetah, who starts from rest, to match this distance within the same time frame.
02

Calculate the Distance Covered by the Prey

Since the prey moves at a constant velocity, the distance it covers in the given time is calculated using the formula \(d = v \times t\). Here, \(v = 9.0 \, \text{m/s}\) and \(t = 3.0 \, \text{s}\). So, \(d = 9.0 \, \text{m/s} \times 3.0 \, \text{s} = 27.0 \, \text{m}\).
03

Apply the Equations of Motion for the Cheetah

The cheetah starts from rest, so its initial velocity \(u = 0\). To find the acceleration \(a\), use the equation \(d = ut + \frac{1}{2}at^2\). Plug in \(u = 0\), \(d = 27.0 \, \text{m}\), and \(t = 3.0 \, \text{s}\).
04

Solve for Cheetah's Acceleration

Substitute the known values in the equation: \(27.0 = 0 \times 3.0 + \frac{1}{2}a(3.0)^2\). Simplifying gives \(27.0 = \frac{9}{2}a\). Solving for \(a\) gives \(a = \frac{54}{9} = 6.0 \, \text{m/s}^2\).

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 91影视!

Key Concepts

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

Kinematics
Kinematics is a branch of physics that focuses on the motion of objects. It describes how objects move, but not why they move.
Kinematics involves capturing details like speed, velocity, acceleration, and displacement without considering the underlying forces.
When solving problems in kinematics, you often use equations to find unknown values based on known quantities.
  • Displacement refers to how far an object has moved from its starting position.
  • Velocity is the speed of an object in a particular direction.
  • Acceleration describes how the velocity changes over time.
By understanding these aspects, you can solve many problems related to how things move. In this exercise, we see an application of kinematics to figure out the cheetah's necessary acceleration.
Constant Velocity
Constant velocity means an object moves in a straight line with unchanging speed and direction. This implies that the acceleration is zero.
If an object has constant velocity, the distance it covers over a time period is simply its speed multiplied by the time.
In our example, the prey moves with a constant velocity of 9.0 m/s for 3 seconds.
This behavior can be described by the formula:
  • \[ d = v imes t \]
where
  • \( d \) is the distance,
  • \( v \) is the velocity,
  • \( t \) is the time.
Knowing this, we calculated the prey's total distance as 27 meters.
Understanding constant velocity situations helps in distinguishing them from those involving acceleration.
Acceleration
Acceleration is all about how quickly an object's velocity changes. It can mean speeding up or slowing down.
In the given problem, the cheetah starts from rest, meaning its initial velocity is zero, and it must achieve a particular distance by accelerating.
To determine the acceleration necessary, we rely on the concept of uniform acceleration, which assumes the rate of change of velocity is constant.
For situations starting from rest, the formula used is:
  • \[ d = ut + \frac{1}{2}at^2 \]
where:
  • \( u \) is the initial velocity (0 in this case),
  • \( a \) is the acceleration,
  • \( t \) is the time.
This equation allows us to solve for unknown acceleration when initial velocity is zero, showcasing the critical role acceleration plays in solving such motion problems.
Equations of Motion
The equations of motion provide a way to calculate various aspects of an object's journey, such as its velocity, position, or acceleration, at any given time.
They become especially useful in analyzing objects experiencing uniform acceleration, like our cheetah.
The key equations include:
  • \[ v = u + at \]
  • \[ d = ut + \frac{1}{2}at^2 \]
  • \[ v^2 = u^2 + 2ad \]
These equations relate various parameters:
  • \( v \) is the final velocity,
  • \( u \) is the initial velocity,
  • \( a \) is the acceleration,
  • \( d \) is the distance,
  • \( t \) is the time.
By using these equations, we calculated the exact constant acceleration (6.0 m/s虏) needed for the cheetah to match the prey's 27 meters in 3 seconds. Understanding these equations is essential for solving more complex motion 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

Two runners start one hundred meters apart and run toward each other. Each runs ten meters during the first second. During each second thereafter, each runner runs ninety percent of the distance he ran in the previous second. Thus, the velocity of each person changes from second to second. However, during any one second, the velocity remains constant. Make a position-time graph for one of the runners. From this graph, determine (a) how much time passes before the runners collide and (b) the speed with which each is running at the moment of collision.

(a) What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) when going down a slope for \(5.0 \mathrm{~s} ?\) (b) How far does the skier travel in this time?

For the first \(10.0 \mathrm{~km}\) of a marathon, a runner averages a velocity that has a magnitude of \(15.0 \mathrm{~km} / \mathrm{h}\). For the next \(15.0 \mathrm{~km}\), he averages \(10.0 \mathrm{~km} / \mathrm{h}\), and for the last \(15.0 \mathrm{~km},\) he averages \(5.0 \mathrm{~km} / \mathrm{h}\). Construct, to scale, the position-time graph for the runner.

In a historical movie, two knights on horseback start from rest \(88.0 \mathrm{~m}\) apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of \(0.300 \mathrm{~m} / \mathrm{s}^{2},\) while Sir Alfred's has a magnitude of \(0.200 \mathrm{~m} / \mathrm{s}^{2}\). Relative to Sir George's starting point, where do the knights collide?

Two cars cover the same distance in a straight line. Car \(\mathrm{A}\) covers the distance at a constant velocity. Car B starts from rest, maintains a constant acceleration, and covers the distance in the same time as car A. (a) Is car B's final velocity smaller than, equal to, or greater than car A's constant velocity? Explain. (b) Is car B's average velocity smaller than, equal to, or greater than car A's constant velocity? Why? (c) Given a value for the time to cover the distance and the fact that car B starts from rest, what other variable must be known before car B's acceleration can be determined? Provide a reason for your answer. Problem Both cars cover a distance of \(460 \mathrm{~m}\) in \(210 \mathrm{~s}\). Assume that they are moving in the \(+x\) direction. Determine (a) the constant velocity of car \(\mathrm{A},\) (b) the final velocity of car \(\mathrm{B},\) and \((\mathrm{c})\) the acceleration of car \(\mathrm{B}\). Verify that your answers are consistent with your answers to the Concept Questions.
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Measurements

Read Explanation

Circular Motion and Gravitation

Read Explanation

Wave Optics

Read Explanation

Solid State Physics

Read Explanation

Waves Physics

Read Explanation
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.