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Large cockroaches can run as fast as 1.50 \(\mathrm{m} / \mathrm{s}\) in short bursts. Suppose you turn on the light in a cheap motel and see one scurrying directly away from you at a constant 1.50 \(\mathrm{m} / \mathrm{s}\) . If you start 0.90 \(\mathrm{m}\) behind the cockroach with an initial speed of 0.80 \(\mathrm{m} / \mathrm{s}\) toward it, what minimum constant acceleration would you need to catch up with it when it has traveled \(1.20 \mathrm{m},\) just short of safety under a counter?

Short Answer

Expert verified
Minimum acceleration needed is 1.75 m/s².

Step by step solution

01

Understand the Problem

The problem describes two objects: a cockroach running away at 1.50 m/s and a person (you) trying to catch it starting from 0.80 m/s, initially 0.90 m behind. You need to determine the minimum acceleration required to catch the cockroach when it has traveled 1.20 m.
02

Define Variables and Equations

Let: - Initial distance to the cockroach be \(d_0 = 0.90 \, \mathrm{m}\). - Cockroach speed \(v_c = 1.50 \, \mathrm{m/s}\). - Time \(t\) is the same for both you and the cockroach until they meet.- Your initial speed \(v_0 = 0.80 \, \mathrm{m/s}\).The cockroach's equation of motion: \[ x_c = v_c \cdot t \]The equation of motion for you (person): \[ x_p = d_0 + v_0 \cdot t + \frac{1}{2} a \cdot t^2 \]Where \(a\) is the acceleration you need to find.
03

Set Up the Equation to Catch the Cockroach

We need to catch the cockroach when it has traveled 1.20 m, so:\[ x_c = 1.20 \, \mathrm{m} \]Substituting the cockroach equation into this, we get:\[ 1.50 \cdot t = 1.20 \rightarrow t = \frac{1.20}{1.50} \]Calculate \(t\) from this equation to find the time it takes the cockroach to cover 1.20 m.
04

Calculate Time

Calculate the time \(t\) it takes for the cockroach to travel 1.20 m:\[ t = \frac{1.20}{1.50} = 0.80 \, \mathrm{s} \]
05

Distance You Need to Travel

You start 0.90 m behind the cockroach and need to reach the same position as the cockroach when it has traveled 1.20 m. Therefore, the distance you need to cover is:\[ 1.20 + 0.90 = 2.10 \, \mathrm{m} \]
06

Equation for Your Motion

Using the distance and the equation of motion for you, substitute the time and distance needed:\[ d_f = d_0 + v_0 \cdot t + \frac{1}{2} a \cdot t^2 \]\[ 2.10 = 0.90 + 0.80 \cdot 0.80 + \frac{1}{2} a \cdot (0.80)^2 \]
07

Solve for Acceleration

Simplify the equation:\[ 2.10 = 0.90 + 0.64 + 0.32a \]\[ 0.56 = 0.32a \]\[ a = \frac{0.56}{0.32} \]Calculate \(a\).
08

Calculate the Acceleration

\[ a = \frac{0.56}{0.32} = 1.75 \, \mathrm{m/s^2} \]
09

Conclusion

You need a minimum constant acceleration of 1.75 m/s² to catch the cockroach before it reaches safety.

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Key Concepts

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

Constant Acceleration
In kinematics, constant acceleration means that the rate of change of velocity is uniform over time. It signifies that an object's velocity changes at the same rate in every second it moves. This concept simplifies the analysis of motion because predictable patterns emerge, making it easier to calculate variables like velocity, time, and displacement.

Let's break down why constant acceleration is significant in solving problems, like the one involving the cockroach and you:
  • It allows for the use of specific formulas in kinematics, facilitating calculation by providing known values and relationships.
  • When acceleration is constant, we know that:
    - The velocity changes linearly over time.
    - The displacement can be calculated using standard equations like:
    \[d = v_0 \cdot t + \frac{1}{2} a \cdot t^2\]
In the given scenario, finding the minimum constant acceleration allows us to determine how you can match the cockroach's motion to catch up with it.
Relative Motion
Relative motion refers to the calculation of the movement of an object in relation to another. It involves understanding that the kinematic equations we use should include the positions, velocities, and accelerations of both the objects when they are in motion relative to each other.

In the exercise, the cockroach and you are moving simultaneously: you towards the cockroach, and the cockroach away from you. Key aspects to note include:
  • The cockroach starts moving initially at a given speed, while you start with a different speed but require additional acceleration due to the initial distance between both.
  • The initial distance is crucial in setting up the equation because it represents the gap you need to close in order to "catch" the cockroach.
    When dealing with relative motion, always remember that both speeds and positions are evaluated concerning each other.
This comparative setup lets us establish equations such as:
\[x_c = x_p\] By adjusting time, speed, and acceleration within these equations, we're able to simulate catching up on the cockroach.
Equations of Motion
The equations of motion are the backbone of solving kinematic problems such as the one described with the cockroach. These equations provide the necessary links between velocity, time, displacement, and acceleration. With these, you can understand how one aspect of motion affects another.

Important equations include:
  • The basic velocity equation:
    \[v = v_0 + at\] Which shows how velocity changes over time.
  • The displacement equation, useful in the problem:
    \[d = v_0 \cdot t + \frac{1}{2} a \cdot t^2\]
These equations allowed us to determine at what point you would catch the cockroach. When applied, the equations of motion give us a structured way to isolate and solve for variables like acceleration, highlighting the methodical nature of solving kinematics issues.

By substituting known values like initial speeds and distances traveled, we could isolate unknowns, such as acceleration, concluding that you'd need a constant acceleration of 1.75 m/s² to catch the cockroach just in time.

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Most popular questions from this chapter

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