Homework Help is Here – Start Your Trial Now!
Engineering
Electrical Engineering
From t=0 tot = 3.46 min, a man stands still, and from t = 3.46 min to t = 6.92 min, he walks briskly in a straight line at a constant speed of 2.09 m/s. What are (a) his average velocity Vag and (b) his average acceleration dag in the time interval 1.00 min to 4.46 min? What are (c) Vavg and (d) avg in the time interval 2.00 min to 5.46 min?

From t=0 tot = 3.46 min, a man stands still, and from t = 3.46 min to t = 6.92 min, he walks briskly in a straight line at a constant speed of 2.09 m/s. What are (a) his average velocity Vag and (b) his average acceleration dag in the time interval 1.00 min to 4.46 min? What are (c) Vavg and (d) avg in the time interval 2.00 min to 5.46 min?

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
See similar textbooks
icon
Related questions
Question

First 3 parts is already solved 

Plz solve d part       

 

 

 

From t=0 tot 3.46 min, a man stands still, and from t = 3.46 min to t = 6.92 min, he walks briskly in a straight line at a constant speed of 2.09 m/s. What are (a) his average velocity Vag and (b) his average acceleration og in the time interval 1.00 min to 4.46 min? What are (c) va and (daag in the time interval 2.00 min to 5.46 min? (a) Number (b) Number i (c) Number i (d) Number Units m/s H !Units m/s^2 Units m/s !Units m/s^2 The tolerance for answers is 1 in the 3rd significant digit. V V V
Transcribed Image Text:From t=0 tot 3.46 min, a man stands still, and from t = 3.46 min to t = 6.92 min, he walks briskly in a straight line at a constant speed of 2.09 m/s. What are (a) his average velocity Vag and (b) his average acceleration og in the time interval 1.00 min to 4.46 min? What are (c) va and (daag in the time interval 2.00 min to 5.46 min? (a) Number (b) Number i (c) Number i (d) Number Units m/s H !Units m/s^2 Units m/s !Units m/s^2 The tolerance for answers is 1 in the 3rd significant digit. V V V
We are given velocities at various intervals of time. On one interval object is at rest. On other, the object is moving at constant speed. The average acceleration is given as change in velocity davg = time taken We use this to find average acceleration. Calculations: Part a) Time interval is t = 1 to 4. 46 mins The constant velocity is v= 2.09m/s acts from t = 3.46 to 4. 46 mins in our interval. Hence displacement is s=vx (4.46 3.46) × 60 s = 2.09 x 60 s = 125.4 m Hence average velocity is d Vavg 4.46-1 = 125.4 Vavg 3.46x60 = Vavg = 0.604 m/s Hence average velocity is 0.604 m/s. Part b) Initial velocity isu = 0 Average acceleration is v-u davg 4.46-1 2.09 davg = 3.46x60 davg = 0.01 m/s² Average acceleration is 0.01 m/s². = Part c) Time interval is t=2 to 5. 46 mins The constant velocity is v= 2.09m/s acts from t = 3.46 to 5. 46 mins in our interval. Hence displacement is svx (4.46 - 3.46) × 60 s = 2.09 x 60 × 2 s = 250.8 m Hence average velocity is d Vavg = 5.46-2 Vavg Vavg Hence average velocity is 1.208 m/s. 250.8 3.46x60 1.208 m/s
Transcribed Image Text:We are given velocities at various intervals of time. On one interval object is at rest. On other, the object is moving at constant speed. The average acceleration is given as change in velocity davg = time taken We use this to find average acceleration. Calculations: Part a) Time interval is t = 1 to 4. 46 mins The constant velocity is v= 2.09m/s acts from t = 3.46 to 4. 46 mins in our interval. Hence displacement is s=vx (4.46 3.46) × 60 s = 2.09 x 60 s = 125.4 m Hence average velocity is d Vavg 4.46-1 = 125.4 Vavg 3.46x60 = Vavg = 0.604 m/s Hence average velocity is 0.604 m/s. Part b) Initial velocity isu = 0 Average acceleration is v-u davg 4.46-1 2.09 davg = 3.46x60 davg = 0.01 m/s² Average acceleration is 0.01 m/s². = Part c) Time interval is t=2 to 5. 46 mins The constant velocity is v= 2.09m/s acts from t = 3.46 to 5. 46 mins in our interval. Hence displacement is svx (4.46 - 3.46) × 60 s = 2.09 x 60 × 2 s = 250.8 m Hence average velocity is d Vavg = 5.46-2 Vavg Vavg Hence average velocity is 1.208 m/s. 250.8 3.46x60 1.208 m/s
Expert Solution
steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Mathematical Modeling of Mechanical System
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,
SEE MORE TEXTBOOKS