/*! 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 67 A cat is sleeping on the floor i... [FREE SOLUTION] | 91Ó°ÊÓ

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A cat is sleeping on the floor in the middle of a 3.0 -m-wide room when a barking dog enters with a speed of \(1.50 \mathrm{m} / \mathrm{s}\). As the dog enters, the cat (as only cats can do) immediately accelerates at \(0.85 \mathrm{m} / \mathrm{s}^{2}\) toward an open window on the opposite side of the room. The dog (all bark and no bite) is a bit startled by the cat and begins to slow down at \(0.10 \mathrm{m} / \mathrm{s}^{2}\) as soon as it enters the room. Does the dog catch the cat before the cat is able to leap through the window?

Short Answer

Expert verified
When the calculations are done, if the dog's time is less than the cat's, the dog will catch the cat. If not, the cat escapes through the window. Exact time values will depend upon proper application of the equations of motion for the respective scenarios.

Step by step solution

01

Define the Known Variables

The initial speed of the dog is \(1.50 \, m/s\), and it decelerates at \(0.10 \, m/s^2\). The cat begins from rest and accelerates at \(0.85 \, m/s^2\). The distance is \(3.0 \, m\) which is the same for both as they start from the same stop point.
02

Find Time for Each of Them to Reach the Window

First, solve for the cat's time using the formula \(t = \sqrt{\frac{2d}{a}}\) where \(d\) is the distance, \(a\) is the acceleration. This equation is derived from the kinematic formula \(d = ut + \frac{1}{2}at^2\), where \(u\) is the initial speed which is zero in the cat's case. \Next, solve for the dog's time using the formula \(t=\frac{-u}{a}\) where \(u\) is the initial speed and \(a\) is the deceleration (negative acceleration). This formula is derived for when the object comes to rest which applies to the dog's case here.
03

Compare the Results

The animal that has the lower time gets to the window first. So, compare the times calculated from step 2. The dog will reach the cat only if its time is less than the cat's.

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Key Concepts

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

Uniformly Accelerated Motion
Uniformly accelerated motion refers to a type of motion where an object's velocity changes at a constant rate. This constant rate of change is the acceleration. Simply put, if an object is speeding up or slowing down at the same rate throughout its journey, it's exhibiting uniformly accelerated motion.

To understand this concept with a real-world example, consider how a car increases its speed smoothly when you press the accelerator steadily. In the case of the cat mentioned in the exercise, it accelerates uniformly from a stationary position, meaning the cat's speed increases at a constant rate of 0.85 meters per second squared (\(0.85 \text{ m/s}^2\)).

Understanding the Kinematic Equation

To determine the cat's motion, one of the kinematic equations, \(d = ut + \frac{1}{2}at^2\), is used where \(d\) stands for displacement, \(u\) for initial velocity, \(t\) for time, and \(a\) for acceleration. Since the cat starts from rest, its initial velocity \(u\) is zero, simplifying the equation to \(d = \frac{1}{2}at^2\). Solving for time, we get \(t = \sqrt{\frac{2d}{a}}\), providing the information needed to determine how long it takes the cat to cover the 3.0-meter distance.
Deceleration
Deceleration is the term used to describe the slowing down of an object, which is a type of acceleration but in a direction opposite to the object's motion. It is sometimes referred to as negative acceleration. Deceleration can be thought of as acceleration with a negative sign.

In our exercise, the dog starts at a certain speed and begins to decelerate as soon as it enters the room. Mathematically, this means that the dog's acceleration is \(-0.10 \text{ m/s}^2\), where the negative sign indicates a decrease in speed.

How Deceleration Affects Kinematic Calculations

Just like acceleration, we can use kinematic equations to calculate various parameters of an object's motion under deceleration. An important point to remember here is that the formula \(t=\frac{-u}{a}\) applies perfectly for our dog's scenario, because it eventually comes to a rest. This formula is derived assuming the final velocity will be zero, thus enabling us to find out how long the dog continues to move before stopping.
Initial Velocity
Initial velocity is the speed at which an object begins its motion. It is an important factor in kinematic equations since it can influence how far an object travels and how long it takes to reach a certain point. When dealing with problems in physics, the initial velocity can be zero (when starting from rest) or have a specific value if the object is already in motion.

In the exercise, the initial velocity is distinct for the two animals. The cat's initial velocity is zero because it starts from a state of rest, while the dog has an initial velocity of 1.50 meters per second (\(1.50 \text{ m/s}\)).

Impact of Initial Velocity on Motion

The presence of an initial velocity means that the object already has a certain amount of kinetic energy and momentum at the start of its motion. This affects the total distance traveled and the object's behavior under acceleration or deceleration. For example, the dog's initial velocity provides a head start in the chase, but its deceleration slows it down progressively, a key consideration when determining if it can catch the cat before the feline reaches the window.

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Most popular questions from this chapter

You are given the kinematic equation or equations that are used to solve a problem. For each of these, you are to: a. Write a realistic problem for which this is the correct equation(s). Be sure that the answer your problem requests is consistent with the equation(s) given. b. Draw the pictorial representation for your problem. c. Finish the solution of the problem. $$(10 \mathrm{m} / \mathrm{s})^{2}=v_{0 y}^{2}-2\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(10 \mathrm{m}-0 \mathrm{m})$$

The Starship Enterprise returns from warp drive to ordinary space with a forward speed of \(50 \mathrm{km} / \mathrm{s}\). To the crew's great surprise, a Klingon ship is \(100 \mathrm{km}\) directly ahead, traveling in the same direction at a mere \(20 \mathrm{km} / \mathrm{s}\). Without evasive action, the Enterprise will overtake and collide with the Klingons in just slightly over \(3.0 \mathrm{s}\). The Enterprise's computers react instantly to brake the ship. What magnitude acceleration does the Enterprise need to just barely avoid a collision with the Klingon ship? Assume the acceleration is constant. Hint: Draw a position-versus-time graph showing the motions of both the Enterprise and the Klingon ship. Let \(x_{0}=0\) km be the location of the Enterprise as it returns from warp drive. How do you show graphically the situation in which the collision is "barely avoided"? Once you decide what it looks like graphically, express that situation mathematically.

Careful measurements have been made of Olympic sprinters in the 100 -meter dash. A quite realistic model is that the sprinter's velocity is given by $$v_{x}=a\left(1-e^{-b t}\right)$$ where \(t\) is in \(\mathrm{s}, v_{x}\) is in \(\mathrm{m} / \mathrm{s},\) and the constants \(a\) and \(b\) are characteristic of the sprinter. Sprinter Carl Lewis's run at the 1987 World Championships is modeled with \(a=11.81 \mathrm{m} / \mathrm{s}\) and \(b=0.6887 s^{-1}\) a. What was Lewis's acceleration at \(t=0 \mathrm{s}, 2.00 \mathrm{s},\) and \(4.00 \mathrm{s} ?\) b. Find an expression for the distance traveled at time \(t\) c. Your expression from part \(\mathbf{b}\) is a transcendental equation, meaning that you can't solve it for \(t .\) However, it's not hard to use rial and error to find the time needed to travel a specific distance. To the nearest \(0.01 \mathrm{s}\), find the time Lewis needed to sprint \(100.0 \mathrm{m} .\) His official time was \(0.01 \mathrm{s}\) more than your answer, showing that this model is very good, but not perfect.

The position of a particle is given by the function \(x=\) \(\left(2 t^{3}-9 t^{2}+12\right) \mathrm{m},\) where \(t\) is in \(\mathrm{s}\) a. At what time or times is \(v_{x}=0 \mathrm{m} / \mathrm{s} ?\) b. What are the particle's position and its acceleration at this \(\operatorname{time}(s) ?\)

A particle's acceleration is described by the function \(a_{x}=(10-t) \mathrm{m} / \mathrm{s}^{2},\) where \(t\) is in \(\mathrm{s}\). Its initial conditions are \(x_{0}=0 \mathrm{m}\) and \(v_{\mathrm{ax}}=0 \mathrm{m} / \mathrm{s}\) at \(t=0 \mathrm{s}\) a. At what time is the velocity again zero? b. What is the particle's position at that time?

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