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An Olympic-class sprinter starts a race with an

College Physics | 1st Edition | ISBN: 9781938168000 | Authors: Paul Peter Urone, Roger Hinrichs ISBN: 9781938168000 42

Solution for problem 20PE Chapter 2

College Physics | 1st Edition

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College Physics | 1st Edition | ISBN: 9781938168000 | Authors: Paul Peter Urone, Roger Hinrichs

College Physics | 1st Edition

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Problem 20PE

Problem 20PE

An Olympic-class sprinter starts a race with an acceleration of 4.50 m/s2 . (a) What is her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period.

Step-by-Step Solution:
Step 1 of 3

Solution 20PE

An Olympic-class sprinter starts a race with an acceleration of 4.50

a.)Her speed at 2.4 s later can be calculated using the equation . Now, ,. So,

 =  10.8

b.) A graph of her position vs. time for this period can be plotted as shown below.

Assuming that the acceleration is constant ,the position is given by  the equation . Now, ,. So, . Plot of time period vs position is as follows

Time period (in     Position (in

0

0

1

2.25

2

9

3

20.25

4

36

5

56.25


Step 2 of 3

Chapter 2, Problem 20PE is Solved
Step 3 of 3

Textbook: College Physics
Edition: 1
Author: Paul Peter Urone, Roger Hinrichs
ISBN: 9781938168000

College Physics was written by and is associated to the ISBN: 9781938168000. The full step-by-step solution to problem: 20PE from chapter: 2 was answered by , our top Physics solution expert on 07/07/17, 04:39PM. This textbook survival guide was created for the textbook: College Physics , edition: 1. The answer to “An Olympic-class sprinter starts a race with an acceleration of 4.50 m/s2 . (a) What is her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period.” is broken down into a number of easy to follow steps, and 33 words. This full solution covers the following key subjects: acceleration, class, graph, later, olympic. This expansive textbook survival guide covers 34 chapters, and 3125 solutions. Since the solution to 20PE from 2 chapter was answered, more than 2739 students have viewed the full step-by-step answer.

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