Engineering Fundamentals: An Introduction to Engineering
Engineering Fundamentals: An Introduction to Engineering
6th Edition
ISBN: 9780357391273
Author: Saeed Moaveni
Publisher: Cengage Learning US
Question
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Chapter 8, Problem 37P
To determine

Find the acceleration required to reach the speed of 10 mph from the rest position within the time period of 25 seconds. Calculate the total distance traveled by bicyclist and also calculate the average speed of cyclist during the first 25 s, 620 s, and 625 s.

Expert Solution & Answer
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Explanation of Solution

Calculate the acceleration required to reach the speed of 10 mph from the rest position within 25 s.

acceleration=changeinvelocitytime=10milesh25s=10milesh(1h3600s)25s{1h=3600s}=1.11×104ms2

During the initial 25 seconds, the speed increases from 0 to 10 mph. Therefore, the average speed is 10mph2=5mph.

Calculate the distance traveled by the bicyclist with the average speed of 5 miles/h during the initial 25 s as follows.

d1=(time)(averagespeed)=(25s)(5milesh)=(25s)(5milesh)(1h3600s)(5280ft1mile){1h=3600s,and1mile=5280ft}=183.33ft

Consider the constant average speed of 10 miles/h in the next 10 minutes.

Calculate the distance traveled by the bicyclist with the average speed of 10 miles/h during the next 10 minutes.

d2=(time)(averagespeed)=(600s)(10milesh)=(600s)(10milesh)(1h3600s)(5280ft1mile){1h=3600s,and1mile=5280ft}=8800ft

Here, the bicyclist decelerates at the constant rate from a speed of 10 mph to 0, the average speed of the bicyclist during the last 5 seconds is 5 mph.

Calculate the distance traveled during this period as below.

d3=(time)(averagespeed)=(5s)(5milesh)=(5s)(5milesh)(1h3600s)(5280ft1mile){1h=3600s,and1mile=5280ft}=36.66ft

The total distance traveled is d=d1+d2+d3.

Substitute 1833.33 ft for d1, 8800 ft for d2, and 36.66 ft for d3,

d=183.33ft+8800ft+36.66ft=9020ft=9020ft(1mile5280ft){1mile=5280ft}=1.71miles

Use the expression for the average speed of the bicyclist for the entire duration of travel.

Vaverage=distancetraveledtime

Substitute 9020 feet for distance traveled and 625 s for time.

Vaverage=9020ft625s=14.43fts=14.43fts(1mile5280ft)(3600s1h){1mile=5280ft, and1h=3600s}=9.84mph

Conclusion:

Thus, the acceleration required to reach the speed of 10 mph in 25 s is 1.11×104ms2_, the total distance traveled by bicyclist is 1.71miles_ and the average speed of cyclist during first 25 s, 620 s, and 625 s are 5milesh,10milesh,and5milesh_ respectively.

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