Physics for Scientists and Engineers, Vol. 1
Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 2, Problem 88P

(a)

To determine

Minimum negative acceleration to avoid collision.

(a)

Expert Solution
Check Mark

Answer to Problem 88P

Minimum negative acceleration to avoid collision is 0.75m/s2 .

Explanation of Solution

Given:

Velocity of passenger train is 29m/s .

Separation between passenger and freight train when engineer sees it is 360m .

Velocity of freight train is 6m/s .

Reaction time of engineer is 0.4s

Formula used:

Write the expression for relative velocity.

  vir=(v1v2)  ........(1)

Here, vir is initial relative velocity, v1 is velocity of passenger train and v2 is velocity of freight train

Write the expression for distance covered between them during reaction time .

  s=(v1v2)tr  ........(2)

Here, s is distance covered between them during reaction time and tr is reaction time

Write the expression for remaining distance between trainsafter reaction time

  s1=(360s)  ........(3)

Here, s1 is remaining distance between trains after reaction time.

Write the expression for final relative velocity between the trains

  vrf2=(vri22amins1)  ........(4)

Here, vrf is final relative velocity which will be equal to zero in the case.

Calculation:

Substitute 29m/s for v1 and 6m/s for v2 in equation (1).

  vir=(296)vir=23m/s .

Substitute 29m/s for v1 , 6m/s for v2 and 0.4s for tr in equation (2).

  s=(296)0.4s=9.2m .

Substitute 9.2m for s in equation (3)

  s1=(3609.2)s1=350.8m

Substitute 23m/s for vir and 350.8m for s1 in equation (4)

  0=2322(amin)(350.8)amin=0.75m/s2

Conclusion:

Thus, the minimum negative acceleration to avoid collision is 0.75m/s2

(b)

To determine

Velocity of passenger train at minimum acceleration to avoid collision

(b)

Expert Solution
Check Mark

Answer to Problem 88P

Velocity of passenger train at minimum acceleration to avoid collision is 10.07m/s .

Explanation of Solution

Given:

Reaction time is 0.8s

Formula used:

Write expression for distance remaining between trains after reaction time

  s2=360(v1v2)tr  ........(5)

Here, s2 is distance remaining between trains after reaction time and tr is reaction time.

Write expression for final relative velocity.

  vfr=vir2as2  ........(6)

Here, vfr is final relative velocity.

Write expression for final relative velocity.

  vfr=v1v2  ........(7)

Calculation:

Substitute 29m/s for v1 and 6m/s for v2 and 0.8s for tr in equation (5)

  s2=360(23)0.8s2=341.6m .

Substitute 23m/s for vir , 0.75m/s2 for a and 341.6m for s2 in equation (6)

  vfr= 2322(0.75)(341.6)vfr=4.07m/s .

Substitute 4.07m/s for vfr and 6m/s for v2 in equation (7)

  4.07=v16v1=10.07m/s

Conclusion:

Thus, the velocity of passenger train at minimum acceleration to avoid collision is 10.07m/s .

(c)

To determine

Distance covered by passenger train between sighting and collision

(c)

Expert Solution
Check Mark

Answer to Problem 88P

The distance covered by passenger train between sighting and collision is 516.16m .

Explanation of Solution

Formula used:

Write the expression for time when train just collides.

  s2=virt12(0.75t2)  ........(8)

Write the expression for distance covered by passenger train between sighting and collision

  s3=v1(t+tr)12at2  ........(9)

Here, s3 is distance covered by passenger train between sighting and collision, t is time when train just collides.

Calculation:

Substitute 341.6m for s2 and 23m/s for vir in equation (8)

  341.6=23t12(0.75t2)t=25.23s .

Substitute 29m/s for v1 , 25.23s for t and 0.8s for tr in equation (9).

  s3=29(25.23+0.8)12(0.75)(25.23)2s3=516.16m

Conclusion:

Distance covered by passenger train between sighting and collision is 516.16m .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Question 3 of 17 L X L L T 0.5/ In the figure above, three uniform thin rods, each of length L, form an inverted U. The vertical rods each have a mass m; the horizontal rod has a mass 3m. NOTE: Express your answer in terms of the variables given. (a) What is the x coordinate of the system's center of mass? xcom L 2 (b) What is the y coordinate of the system's center of mass? Ycom 45 L X Q Search MD bp N
Sketch the harmonic on graphing paper.
Exercise 1: (a) Using the explicit formulae derived in the lectures for the (2j+1) × (2j + 1) repre- sentation matrices Dm'm, (J/h), derive the 3 × 3 matrices corresponding to the case j = 1. (b) Verify that they satisfy the so(3) Lie algebra commutation relation: [D(Î₁/ħ), D(Î₂/h)]m'm₁ = iƊm'm² (Ĵ3/h). (c) Prove the identity 3 Dm'm,(β) = Σ (D(Ρ)D(Ρ))m'¡m; · i=1

Chapter 2 Solutions

Physics for Scientists and Engineers, Vol. 1

Ch. 2 - Prob. 11PCh. 2 - Prob. 12PCh. 2 - Prob. 13PCh. 2 - Prob. 14PCh. 2 - Prob. 15PCh. 2 - Prob. 16PCh. 2 - Prob. 17PCh. 2 - Prob. 18PCh. 2 - Prob. 19PCh. 2 - Prob. 20PCh. 2 - Prob. 21PCh. 2 - Prob. 22PCh. 2 - Prob. 23PCh. 2 - Prob. 24PCh. 2 - Prob. 25PCh. 2 - Prob. 26PCh. 2 - Prob. 27PCh. 2 - Prob. 28PCh. 2 - Prob. 29PCh. 2 - Prob. 30PCh. 2 - Prob. 31PCh. 2 - Prob. 32PCh. 2 - Prob. 33PCh. 2 - Prob. 34PCh. 2 - Prob. 35PCh. 2 - Prob. 36PCh. 2 - Prob. 37PCh. 2 - Prob. 38PCh. 2 - Prob. 39PCh. 2 - Prob. 40PCh. 2 - Prob. 41PCh. 2 - Prob. 42PCh. 2 - Prob. 43PCh. 2 - Prob. 44PCh. 2 - Prob. 45PCh. 2 - Prob. 46PCh. 2 - Prob. 47PCh. 2 - Prob. 48PCh. 2 - Prob. 49PCh. 2 - Prob. 50PCh. 2 - Prob. 51PCh. 2 - Prob. 52PCh. 2 - Prob. 53PCh. 2 - Prob. 54PCh. 2 - Prob. 55PCh. 2 - Prob. 56PCh. 2 - Prob. 57PCh. 2 - Prob. 58PCh. 2 - Prob. 59PCh. 2 - Prob. 60PCh. 2 - Prob. 61PCh. 2 - Prob. 62PCh. 2 - Prob. 63PCh. 2 - Prob. 64PCh. 2 - Prob. 65PCh. 2 - Prob. 66PCh. 2 - Prob. 67PCh. 2 - Prob. 68PCh. 2 - Prob. 69PCh. 2 - Prob. 70PCh. 2 - Prob. 71PCh. 2 - Prob. 72PCh. 2 - Prob. 73PCh. 2 - Prob. 74PCh. 2 - Prob. 75PCh. 2 - Prob. 76PCh. 2 - Prob. 77PCh. 2 - Prob. 78PCh. 2 - Prob. 79PCh. 2 - Prob. 80PCh. 2 - Prob. 81PCh. 2 - Prob. 82PCh. 2 - Prob. 83PCh. 2 - Prob. 84PCh. 2 - Prob. 85PCh. 2 - Prob. 86PCh. 2 - Prob. 87PCh. 2 - Prob. 88PCh. 2 - Prob. 89PCh. 2 - Prob. 90PCh. 2 - Prob. 91PCh. 2 - Prob. 92PCh. 2 - Prob. 93PCh. 2 - Prob. 94PCh. 2 - Prob. 95PCh. 2 - Prob. 96PCh. 2 - Prob. 97PCh. 2 - Prob. 98PCh. 2 - Prob. 99PCh. 2 - Prob. 100PCh. 2 - Prob. 101PCh. 2 - Prob. 102PCh. 2 - Prob. 103PCh. 2 - Prob. 104PCh. 2 - Prob. 105PCh. 2 - Prob. 106PCh. 2 - Prob. 107PCh. 2 - Prob. 108PCh. 2 - Prob. 109PCh. 2 - Prob. 110PCh. 2 - Prob. 111PCh. 2 - Prob. 112PCh. 2 - Prob. 113PCh. 2 - Prob. 114PCh. 2 - Prob. 115PCh. 2 - Prob. 116PCh. 2 - Prob. 117PCh. 2 - Prob. 118PCh. 2 - Prob. 119PCh. 2 - Prob. 120PCh. 2 - Prob. 121PCh. 2 - Prob. 122P
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Relative Velocity - Basic Introduction; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=_39hCnqbNXM;License: Standard YouTube License, CC-BY