×
Log in to StudySoup
Get Full Access to Physics - Textbook Survival Guide
Join StudySoup for FREE
Get Full Access to Physics - Textbook Survival Guide

Automobile Accident Analysis. You are called as an expert

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 80P Chapter 8

University Physics | 13th Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

4 5 0 341 Reviews
16
2
Problem 80P

Automobile Accident Analysis.? You are called as an expert witness to analyze the following auto accident: Car B, of mass 1900 kg, was stopped at a red light when it was hit from behind by car A, of mass 1500 kg. The cars locked bumpers during the collision and slid to a stop with brakes locked on all wheels. Measurements of the skid marks left by the tires showed them to be 7.15 m long. The coefficient of kinetic friction between the tires and the road was 0.65. (a) What was the speed of car A just before the collision? (b) If the speed limit was 35 mph, was car A speeding, and if so, by how many miles per hour was it exceeding? the speed limit?

Step-by-Step Solution:

Solution 80P Step 1: We know that, the change in kinetic energy of car B is converted as the work done to move the car A. Therefore, we can write, KE finalKE initial That is, ½ mv final½ mv initialW “m” is the sum of the masses of both cars. Mass of car A, m = 15A kg Mass of car B, m = 19B kg m = m +m = 1500 A B vfinal0 m/s since, the cars will come to rest due to frictional force. 2 Therefore, 0 - ½ mv initial W Work done, W = FS Where, F - force applied S - Displacement for the body F = - k Actually, the force is generated from car A. Therefore, F = - m g k A Provided, the coefficient of kinetic friction, = 0.65 k Therefore, W = - m g S k A Where, S - braking distance = 7.15 m 2 W = - 0.65 × 1500 kg × 9.8 m/s × 7.15 m = - 68318 J Step 2: 2 a) We can write, - ½ mv initial - 68318 J mv initial 2 × 68318 J = 136636 J 3400 kg × v 2= 136636 J initial 2 2 2 vinitial 136636 J / 3400 kg = 40.19 m /s Taking square root on both sides, v = 6.34 m/s initial This is the combined velocity after collision.

Step 3 of 4

Chapter 8, Problem 80P is Solved
Step 4 of 4

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

Automobile Accident Analysis. You are called as an expert

×
Log in to StudySoup
Get Full Access to Physics - Textbook Survival Guide
Join StudySoup for FREE
Get Full Access to Physics - Textbook Survival Guide
×
Reset your password