Basic Biomechanics
Basic Biomechanics
7th Edition
ISBN: 9780073522760
Author: Susan J Hall
Publisher: McGraw-Hill Education
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Textbook Question
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Chapter 10, Problem 1IP

A runner completes 6½ laps around a 400 m track during a 12 min (720 s) run test. Calculate the following quantities:

a. The distance the runner covered

b. The runner’s displacement at the end of 12 min

c. The runner’s average speed

d. The runner’s average velocity

e. The runner’s average pace (Answers: a. 2.6 km; b. 160 m; c. 3.6 m/s; d. 0.22 m/s; e. 4.6 min/km)

a)

Expert Solution
Check Mark
Summary Introduction

To determine: The distance runner covered.

Answer to Problem 1IP

The distance covered by the runner is 2.6km.

Explanation of Solution

Calculation:

Express the distance covered by the runner.

Distancecoveredbytherunner=Numberoflaps×distanceinsinglelap

Substitute 6.5 for number of laps and 400 m for distance in single lap to find the distance covered by the runner.

Distancecoveredbytherunner=6.5×400m=2600m(1km1,000m)=2.6km

Therefore, distance covered by the runner is 2.6km.

Conclusion

Therefore, distance covered by the runner is 2.6km.

b)

Expert Solution
Check Mark
Summary Introduction

To determine: The runner’s displacement at the end of 12 min.

Answer to Problem 1IP

Displacement by the runner at the end of 12 min is 141.4m.

Explanation of Solution

Calculation:

At 12 min runner completes 6.5 laps and the runner will be at the position of 0.5 lap.

Consider each side of the track is 100 m.

Use Pythagoras theorem to express the runner’s displacement at the end of 12 min.

Runner’sdisplacementattheendof12min}=(Lengthoftrack)2+(Breadthoftrack)2

Substitute 100m for length and breadth to find the displacement by the runner at the end of 12 min.

Runner’sdisplacementattheendof12min}=(100)2+(100)2=141.4m

Therefore, displacement by the runner at the end of 12 min is 141.4m.

Conclusion

Therefore, displacement by the runner at the end of 12 min is 141.4m.

c)

Expert Solution
Check Mark
Summary Introduction

To determine: The runner’s average speed.

Answer to Problem 1IP

The average speed of the runner is 3.6m/s.

Explanation of Solution

Calculation:

Express the average speed of the runner.

Averagespeed=Totaldistancetotaltime

Substitute 2600 m for total distance and 12 min to find the average speed.

Averagespeed=2,600m12min=2,600m12min(60sec1min)=3.611m/s3.6m/s

Therefore, average speed of the runner is 3.6m/s.

Conclusion

Therefore, average speed of the runner is 3.6m/s.

d)

Expert Solution
Check Mark
Summary Introduction

To determine: The runner’s average velocity.

Answer to Problem 1IP

The average velocity of the runner is 0.2m/s.

Explanation of Solution

Calculation:

Express the average velocity of the runner.

Averagevelocity=Totaldisplacementtotaltime

Substitute 141.4 m for total displacement and 12 min to find the average velocity.

Averagevelocity=141.4m12min=141.4m12min(60sec1min)=0.19m/s0.2m/s

Therefore, average velocity of the runner is 0.2m/s.

Conclusion

Therefore, average velocity of the runner is 0.2m/s.

e)

Expert Solution
Check Mark
Summary Introduction

To determine: The runner’s average pace.

Answer to Problem 1IP

The average pace of the runner is 4.6min/km.

Explanation of Solution

Calculation:

Express the average pace of the runner.

Averagepace=1Averagespeed

Substitute 3.6m/s for average speed to find the average pace.

Averagepace=13.6m/s(1km1,000m)(60sec1min)=10.2166km/min=4.61min/km4.6min/km

Therefore, average pace of the runner is 4.6min/km.

Conclusion

Therefore, average pace of the runner is 4.6min/km.

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Students have asked these similar questions
1. What is the MET level of Joan who runs at 5.1 mph, 2.7% grade? 2. Joan (in problem #1) lowers her speéd to 3.0 mph. In order to maintain the same MET level, what should the grade be on the treadmill?
These figures show that: S Vor (1-min¹) 30 A 2.5 20 1.5 1.0 0.5 (50) 10 Concentric Eccentric Exercise rate (100) 20 (150) Integrated EMG (counts min) B 600 400 200 (50) 30 10 Mean torque on pedals (Nm) (100) Concentric 20 Exercise rate (200) (150) Eccentric 30 O concentric contractions require more oxygen and more EMG than eccentric contractions O concentric contractions are easier to perform than eccentric contractions O eccentric contractions place a greater demand on the cardiovascular and nervous systems compared to concentric contractions O eccentric contractions use less oxygen but more EMG than concentric contractions
Exercise intensity can be estimated using a. perceived exertion (RPE) O b. heart rate O c. talk test O d. All of these are ways to estimate exercise intensity
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