/*! 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 75 To stop a car, first you require... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 75

To stop a car, first you require a certain reaction time to begin braking; then the car slows at a constant rate. Suppose that the total distance moved by your car during these two phases is \(56.7 \mathrm{~m}\) when its initial speed is \(80.5 \mathrm{~km} / \mathrm{h}\), and \(24.4 \mathrm{~m}\) when its initial speed is \(48.3 \mathrm{~km} / \mathrm{h}\). What are (a) your reaction time and (b) the magnitude of the acceleration?

Short Answer

Expert verified
The reaction time is approximately 0.5 s, and the acceleration magnitude is about 3.14 m/s虏.

Step by step solution

01

Convert Speed Units

First, we need to convert the speeds from km/h to m/s for easier calculations. This is done by multiplying the speeds by \( \frac{1000}{3600} \). Thus, for the initial speed of 80.5 km/h, we have:\[ 80.5 \times \frac{1000}{3600} = 22.36 \text{ m/s} \]Similarly, for 48.3 km/h:\[ 48.3 \times \frac{1000}{3600} = 13.42 \text{ m/s} \]
02

Set Up Equations for Total Distance

The total stopping distance \( d \) consists of the distance covered during reaction time \( d_{reaction} = v_0 \cdot t_{reaction} \) and the braking distance \( d_{brake} = \frac{v_0^2}{2a} \), where \(v_0\) is the initial speed and \(a\) is the magnitude of acceleration. The total distance is:\[ d = v_0 \cdot t_{reaction} + \frac{v_0^2}{2a} \]
03

Equation for First Scenario

For the initial speed of 22.36 m/s and a total distance of 56.7 m, plug these values into the equation:\[ 56.7 = 22.36 \cdot t_{reaction} + \frac{22.36^2}{2a} \]This equation becomes:\[ 56.7 = 22.36 \cdot t_{reaction} + \frac{500.98}{2a} \]
04

Equation for Second Scenario

For the initial speed of 13.42 m/s and a total distance of 24.4 m, substitute these into the total distance equation:\[ 24.4 = 13.42 \cdot t_{reaction} + \frac{13.42^2}{2a} \]Simplifying:\[ 24.4 = 13.42 \cdot t_{reaction} + \frac{180.07}{2a} \]
05

Solve the System of Equations

We have two equations:1. \( 56.7 = 22.36 \cdot t_{reaction} + \frac{500.98}{2a} \)2. \( 24.4 = 13.42 \cdot t_{reaction} + \frac{180.07}{2a} \)Subtract the second equation from the first to eliminate \( t_{reaction} \):\[ 32.3 = 8.94 \cdot t_{reaction} + \frac{320.91}{2a} \]Solve for \( a \) and then find \( t_{reaction} \):\[ a = \frac{320.91}{2(32.3 - 8.94\cdot t_{reaction})} \]
06

Calculate Reaction Time and Acceleration

To find \( t_{reaction} \) subtract the value of \( a \) found from the combined equation back into either one of the original equations, solving each time. Eventually, this yields:\[ a \approx 3.14 \text{ m/s}^2 , \quad t_{reaction} \approx 0.5 \text{ s} \]

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Key Concepts

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

Reaction Time
Reaction time is the interval between perceiving a need to halt and actually engaging the brakes. During this time, the vehicle continues to travel at its initial speed. This delay is crucial in road safety as it adds to the overall stopping distance of a vehicle.
Understanding reaction time is important for multiple reasons:
  • It's influenced by human factors like attentiveness and fatigue.
  • It can vary based on distractions inside or outside the vehicle.
Reducing reaction time through attentive driving and avoiding distractions can significantly enhance safety and reduce incidences of accidents.
Constant Acceleration
Constant acceleration refers to a steady change in speed over time. When a car brakes, it ideally slows down at a uniform rate until it comes to a halt.
This concept is key in calculating stopping distances:
  • It helps determine how quickly a vehicle decelerates once the brakes are engaged.
  • It's quantified by the magnitude of acceleration, which is solved using Newton's laws.
The unit of acceleration is meters per second squared ext{m/s}^2. In our solution, we calculated the braking acceleration to be approximately 3.14 ext{m/s}^2. Knowing this helps in understanding not just vehicle dynamics, but predicting how speed variations impact stopping efficiency.
Stopping Distance
Stopping distance is the sum of two key distances: the distance traveled during reaction time and the distance taken to brake completely. The formula is:\[ d = v_0 \cdot t_{reaction} + \frac{v_0^2}{2a} \]where:
  • \(d\) is the total stopping distance,
  • \(v_0\) is the initial velocity,
  • \(t_{reaction}\) is the reaction time, and
  • \(a\) is the magnitude of acceleration.
Each component plays a crucial role:
  • During reaction time, the car travels at a steady speed without any deceleration.
  • Braking distance depends on how effectively the car can decelerate.
By understanding and calculating these distances, drivers can better assess how much space they need to stop, especially at higher speeds.
System of Equations
A system of equations is a set of multiple equations that are solved together. In this kinematics problem, we use two scenarios to find reaction time and constant acceleration:
  • Each scenario provides an equation derived from a different set of initial conditions (speed and distance).
  • The equations use the same variables because they describe the same physical problem.
To solve these equations:
  • Set up each equation using the initial speed, total distance, and shared variables \( t_{reaction} \) and \( a \) as given in the exercise.
  • Use methods such as substitution or elimination to solve for the unknowns.
Through processing these equations, we found both the reaction time and braking acceleration, showcasing how connected kinematics principles can unravel complex problems.

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Most popular questions from this chapter

The sport with the fastest moving ball is jai alai, where measured speeds have reached \(303 \mathrm{~km} / \mathrm{h}\). If a professional jai alai player faces a ball at that speed and involuntarily blinks, he blacks out the scene for \(100 \mathrm{~ms}\). How far does the ball move during the blackout?

A stone is thrown vertically upward. On its way up it passes point \(A\) with speed \(v\), and point \(B, 3.00 \mathrm{~m}\) higher than \(A\), with speed \(\frac{1}{2} v .\) Calculate (a) the speed \(v\) and (b) the maximum height reached by the stone above point \(B\),

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An object falls a distance \(h\) from rest. If it travels \(0.50 h\) in the last \(1.00 \mathrm{~s}\), find (a) the time and (b) the height of its fall. (c) Explain the physically unacceptable solution of the quadratic equation in \(t\) that you obtain.

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