/*! 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 42 You are arguing over a cell phon... [FREE SOLUTION] | 91Ó°ÊÓ

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You are arguing over a cell phone while trailing an unmarked police car by \(25 \mathrm{~m}\); both your car and the police car are traveling at \(110 \mathrm{~km} / \mathrm{h}\). Your argument diverts your attention from the police car for \(2.0 \mathrm{~s}\) (long enough for you to look at the phone and yell, "I won't do that!"). At the beginning of that \(2.0 \mathrm{~s}\), the police officer begins braking suddenly at \(5.0 \mathrm{~m} / \mathrm{s}^{2} .(\mathrm{a})\) What is the separation between the two cars when your attention finally returns? Suppose that you take another \(0.40 \mathrm{~s}\) to realize your danger and begin braking. (b) If you too brake at \(5.0 \mathrm{~m} / \mathrm{s}^{2}\), what is your speed when you hit the police car?

Short Answer

Expert verified
(a) 35.4 m is the separation after 2.4 seconds. (b) Your speed is zero when you hit the police car, as both are braking at the same rate.

Step by step solution

01

Understanding the Problem

This problem involves two cars: your car and an unmarked police car. Initially, both are moving at the same speed of \(110 \text{ km/h}\). As your attention is diverted, the police car starts decelerating at \(5.0 \text{ m/s}^2\) for \(2.0 \text{ s}\). You need to find the separation between the two cars after this period and the speed of your car when it hits the police car.
02

Convert Speeds to Meters per Second

Convert the speed of both cars from \(110 \text{ km/h}\) to \(\text{m/s}\). Use the conversion: \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\).\[110 \text{ km/h} = \frac{110}{3.6} \approx 30.56 \text{ m/s}\]
03

Calculate Police Car's Distance During Initial 2 Seconds

The police car decelerates at \(5.0 \text{ m/s}^2\) for \(2.0 \text{ s}\). Use the equation of motion:\[s = ut + \frac{1}{2} a t^2\]where \(u = 30.56\text{ m/s}\), \(a = -5.0 \text{ m/s}^2\), \(t = 2 \text{ s}\):\[s = (30.56 \times 2) + \frac{1}{2} \times (-5.0) \times (2^2) = 61.12 - 10 = 51.12 \text{ m}\]
04

Calculate Your Car's Distance During Initial 2 Seconds

Since your car continues at a constant speed of \(30.56 \text{ m/s}\) (no braking), calculate the distance traveled in \(2\) seconds:\[s = u \times t = 30.56 \times 2 = 61.12 \text{ m}\]
05

Calculate the Initial Separation after 2 Seconds

Initial separation was \(25\text{ m}\). Your car travels \(61.12 \text{ m}\) while the police car travels \(51.12 \text{ m}\):\[\text{Separation} = 25 + (61.12 - 51.12) = 35.0 \text{ m}\]
06

Calculate Additional Distance for 0.4 Seconds Reaction Time

During the reaction time, your car continues at \(30.56 \text{ m/s}\):\[s = 30.56 \times 0.4 = 12.224 \text{ m}\]The police car continues to decelerate:\[s = (30.56 \times 0.4) + \frac{1}{2} \times (-5.0) \times (0.4^2)\]\[s \approx 12.224 - 0.4 = 11.824 \text{ m}\]
07

Calculate Final Separation before Braking

Add the distances from the reaction time to the separation:\[\text{Separation} = 35.0 + (12.224 - 11.824) = 35.4 \text{ m}\]
08

Calculate Your Speed at Collision

Use equations of motion to find your speed when you hit the police car, given you and the police car brake at \(5.0 \text{ m/s}^2\):Both cars will continue to decelerate until they collide. We can calculate the time of collision by:Suppose, at collision police final speed = \(v_p\) and your final speed = \(v_y\). Then use \(v^2 = u^2 + 2as\).However, since collisions usually mean zero final relative separation single v = u + at.Given initial relative velocity \(0\) + \(5\) relative acceleration = \(-5 m/s^2\).Collision: relative velocity \(= 0 + 0 \times t\) must quickly check when stopped.
09

Calculate Final Speed before Collision

Set up the equation considering both decelerating equally but when collision.Since we arrange for both cars breaking simultaneously with similar rates means time using speed exit zero in reliant observation condition.Calculate based on time relations:\[v_{your} = 5t\]Calculates any specific errors in surpassing phase calculation matching worst-case spaces variability.

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Key Concepts

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

Kinematics
Kinematics is the study of motion without considering the forces that cause the motion. In our example, both the police car and your car are initially moving at the same velocity. To solve problems involving kinematics, we often use equations of motion that connect an object's initial velocity, final velocity, acceleration, time, and displacement.

For this exercise, we needed to convert the speed from \(110 \text{ km/h}\) to \(\text{m/s}\) using the conversion factor \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\). This gives us a common unit to work with throughout the problem. Remember, consistent units help avoid confusion and ensure accuracy.

Calculations then proceed with the formula \[ s = ut + \frac{1}{2} a t^2 \] where \(u\) is the initial speed, \(a\) is the acceleration (or deceleration, if negative), and \(t\) is the time. Such formulas allow us to find out how far a car travels in a particular time interval when its speed changes due to acceleration or deceleration.
Deceleration
Deceleration refers to a decrease in the speed of an object over time. It is essentially negative acceleration, meaning the object is slowing down rather than speeding up. In this scenario, the police car begins to decelerate at \(5.0 \text{ m/s}^2\), indicating it is reducing its speed over each second of travel.

When calculating how deceleration affects movement, we again use modified kinematics equations. Here, it’s crucial to distinguish deceleration by taking acceleration values as negative. This is reflected in the formula \[ s = ut + \frac{1}{2} a t^2 \] where the negative \(a\) impacts total distance traveled head-on, allowing us to determine the real separation between two moving objects, i.e., how much one has moved in context to decelerating pace.

Through steps, you first find out how far the police car moves while decelerating for \(2.0 \text{ s}\), taking it through reduced speed measures, while your vehicle travels at consistent speed. Understanding deceleration helps capture how promptly one should act to avoid collisions when observing too long or immersing in other tasks.
Collision Mechanics
Collision mechanics involves understanding the behavior of objects (in this case, vehicles) when they come in contact with each other. It is essential to know how their speeds and distances change over time, especially under deceleration when preventing or analyzing collisions.

In our problem, the core concept is to calculate if and when your car will collide with the police car as both decelerate at \(5.0 \text{ m/s}^2\). Initially, we determine how close the cars come based on their differing actions during a shared path. The moment of collision is derived from balancing the relative speeds and variable deceleration to forecast the endpoint speed of your car during the collision.

Analyzing interlaced distance traveled by each within various timeframes leads to assessing relative velocity; it defines at what speed your car will strike the police vehicle based on remaining space and speed moderation. Understanding these aspects of collision mechanics can help predict how changes in driver response and braking can prevent accidents, serving as learning nodes for safer travel practices.

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Most popular questions from this chapter

A shuffleboard disk is accelerated at a constant rate from rest to a speed of \(6.0 \mathrm{~m} / \mathrm{s}\) over a \(1.8 \mathrm{~m}\) distance by a player using a cue. At this point the disk loses contact with the cue and slows at a constant rate of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) until it stops. (a) How much time elapses from when the disk begins to accelerate until it stops? (b) What total distance does the disk travel?

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