/*! 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 44 The drivers of two cars having e... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The drivers of two cars having equal speeds hit their brakes at the same time, but car \(A\) has three times the acceleration of car \(B\). (a) If car \(A\) travels a distance \(D\) before stopping, how far (in terms of \(D)\) will car \(B\) go before stopping? (b) If car \(B\) stops in time \(T\), how long (in terms of \(T\) ) will it take for car \(A\) to stop?

Short Answer

Expert verified
Car B travels 3D, and car A stops in \(\frac{T}{3}\) time.

Step by step solution

01

Understanding the Problem

Car A and car B initially have the same speed and brake at the same time. The acceleration of car A is three times that of car B. We need to find out how far car B travels in terms of D (the distance car A travels before stopping) and how long it takes for car A to stop in terms of T (the time it takes for car B to stop).
02

Establishing the Relationship of Distances

The stopping distance is given by the equation \(d = \frac{v^2}{2a}\) where \(v\) is the initial speed and \(a\) is the acceleration. For car A, this is \(D = \frac{v^2}{2(3a)}\). For car B, this is \(D_B = \frac{v^2}{2a}\). By comparing these, we have \(D_B = 3D\).
03

Establishing the Relationship of Times

The stopping time is given by the equation \(t = \frac{v}{a}\). For car A, this time is \(T_A = \frac{v}{3a}\). For car B, this time is \(T_B = \frac{v}{a}\). Thus, \(T_A = \frac{T}{3}\).
04

Formulating the Final Answers

The distance car B travels is \(3D\), and the time it takes car A to stop is \(\frac{T}{3}\).

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 91Ó°ÊÓ!

Key Concepts

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

Kinematics in Stopping Distance Problems
Kinematics is one of the foundational aspects of physics, dealing with the motion of objects without considering the causes of motion. In stopping distance problems, we mainly focus on how the initial speed, acceleration, and time interact to determine the stopping distance of an object.
When a car is moving at a certain speed and needs to stop, kinematic equations help us to calculate how far it will travel before coming to a complete halt. The key variables that describe this motion include:
  • Initial speed (v): The speed at which the car starts before braking.
  • Acceleration (a): The rate at which the car's velocity changes. In braking, this is a deceleration (a negative acceleration).
  • Stopping distance (d): The complete distance the car travels from the point brakes are applied to when it stops.
Understanding these concepts is crucial in solving physics problems related to motion.
Concept of Stopping Distance
Stopping distance is a critical concept in vehicle dynamics and safety. It refers to the total distance a vehicle travels while stopping from a given speed. The stopping distance is influenced by various factors such as the initial velocity, braking force, and road conditions.
The basic formula for calculating stopping distance in terms of kinematics is derived from the equation: \[ d = \frac{v^2}{2a} \]Where:
  • v: Initial velocity.
  • a: Acceleration (or deceleration in braking scenarios).
This equation tells us how the speed and deceleration rate affect the total stopping distance. Doubling the initial speed, for instance, quadruples the stopping distance, illustrating the critical role of speed in safe driving.
Applying Newton's Laws of Motion
Newton's Laws of Motion play a pivotal role in understanding the principles behind stopping distances. Specifically, these problems often involve Newton's second law, which states that the force on an object is equal to its mass times its acceleration (\(F = ma\)).
When a car is braking, the force applied by the brakes results in a negative acceleration (or deceleration), which slows the vehicle down. This deceleration is the car's response to the braking force applied over time. In the given exercise, understanding the relationship between force, mass, and acceleration helps us determine how car A, with higher deceleration, stops sooner than car B.
This concept highlights the significance of effective braking systems and the impact of force application in reducing stopping distances and ensuring safety.
Understanding Acceleration and Deceleration
Acceleration is all about changes in velocity over time. However, when we talk about braking, the term deceleration is often used to describe the decrease in speed.
In the exercise, car A's acceleration is three times that of car B. This higher deceleration means that car A loses its speed much faster than car B when brakes are applied. The time to stop and the distance covered during slowing down both correlate directly concerning the acceleration magnitude.
Deceleration can be harnessed to understand vehicle performance and improve safety protocols by ensuring that vehicles come to a stop efficiently within minimal distances.
Calculating Braking Distance: A Practical Approach
Braking distance calculations involve using kinematic equations to determine how far a vehicle travels during the braking process. These calculations rely on a few critical parameters, notably the vehicle's initial speed and the deceleration rate.
To calculate braking distance, you need:
  • Initial speed (v): Measure the speed before applying the brakes.
  • Braking deceleration (a): Identify the rate at which the vehicle slows down.
Using the formula:\[ d = \frac{v^2}{2a} \]You can predict how long it will take for a vehicle to stop completely. This is key in scenarios like the given problem, where car B, with lower deceleration, travels a further distance compared to car A.
Understanding and performing these calculations is vital for traffic safety, ensuring that vehicles have adequate stopping space on roads.

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

Astronauts on our moon must function with an acceleration due to gravity of \(1.62 g .\) (a) If an astronaut can throw a certain wrench \(12.0 \mathrm{~m}\) vertically upward on earth, how high could he throw it on our moon if he gives it the same starting speed in both places? (b) How much longer would it be in motion (going up and coming down) on the moon than on earth?

During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, the bags produce a maximum acceleration of \(60 g,\) but lasting only \(36 \mathrm{~ms}\) (or less). How far (in meters) does a person travel in coming to a complete stop in \(36 \mathrm{~ms}\) at a constant acceleration of \(60 \mathrm{~g} ?\)

A healthy heart pumping at a rate of 72 beats per minute increases the speed of blood flow from 0 to \(425 \mathrm{~cm} / \mathrm{s}\) with each beat. Calculate the acceleration of the blood during this process.

A car is traveling in the negative \(x\) direction at \(25 \mathrm{~m} / \mathrm{s}\). After \(5 \mathrm{~s}\) passes, it is traveling at \(50 \mathrm{~m} / \mathrm{s},\) again in the negative \(x\) direction. (a) What is the magnitude of its average acceleration? (b) What is the direction of its average acceleration? (c) How far does the car travel during this period?

According to recent typical test data, a Ford Focus travels \(0.250 \mathrm{mi}\) in \(19.9 \mathrm{~s},\) starting from rest. The same car, when braking from \(60.0 \mathrm{mi} / \mathrm{h}\) on dry pavement, stops in \(146 \mathrm{ft}\). Assume constant acceleration in each part of its motion, but not necessarily the same acceleration when slowing down as when speeding up. (a) Find this car's acceleration while braking and while speeding up. (b) If its acceleration is constant while speeding up, how fast (in \(\mathrm{mi} / \mathrm{h}\) ) will the car be traveling after \(0.250 \mathrm{mi}\) of acceleration? (c) How long does it take the car to stop while braking from \(60.0 \mathrm{mi} / \mathrm{h} ?\)

See all solutions

Recommended explanations on Physics Textbooks

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.