Problem

The sweep angle of a windshield wiper arm is $115…

Problem 53 Easy Difficulty

"Lightning Bolt " In 2010, Usain Bolt of Jamaica held the world record for the 100 meters and the 200 meters sprints. His maximum stride angle shown below is $5^{\circ}$ less than 1.5 times its supplement. Find his maximum stride angle. You may need to refer to problem 52 to review the geometry involved.
(IMAGE CANNOT COPY)

Answer

Maximum stride angle: $106^{\circ}$

Related Courses

Precalculus

Algebra

Elementary and Intermediate Algebra

Chapter 2

Equations, Inequalities, and Problem Solving

Section 5

Problem Solving

Discussion

You must be signed in to discuss.

Top Algebra Educators

Watch More Solved Questions in Chapter 2

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Video Transcript

so we can say for party first three seconds we have that The maximum velocity is equaling the initial velocity, which we know to be zero at that. At the beginning, this would be plus the maximum acceleration times T of 3.0 seconds. And so the maximum velocity would be equaling 3.0 times the maximum acceleration. Now the distance traveled. Within those three seconds, we can say X sub one. This would be equal to 1/2 times the maximum acceleration times 3.0 seconds squared. And so this would be equaling 1/2 times the maximum velocity divided by three, multiplied by nine. And this is equaling 1.50 times the maximum velocity now, because in the next 6.69 seconds, bull did not ex ah bolt the not accelerate. So we can say, except to is simply equaling 6.69 times the maximum velocity. Since he keeps his velocity for the next 6.69 seconds. We know that the total distance x of one plus except to should eat well 100 meters. Yeah, that's the 100 meter dash and this is equaling 1.5. These maximum 1.50 times of velocity max. Most fly by 6.69 times of velocity max and so this would be equaling God. Seven. So 8.19 times the maximum velocity and so to solve. For the maximum velocity, this would be 100 divided by 8.1912 point two meters per second. And so the maximum acceleration is equaling 12.2 meters per second, divided by three. And this is equaling 4.7 meters per second squared. This would be our final answer for the maximum acceleration during those 1st 3 seconds. That would be your answer for part A For part B. We have that here again. For the 1st 3 seconds we have that V Max equals the initial plus the acceleration max times 3.0 seconds. This is again three times the maximum acceleration and the distance traveled within those 1st 3 seconds weakened then say that X sub one is equaling again. 1.5 times piece of max. We're using the exact same relationship in part a now for, except to, However, this is a different time. So we have the sub max multiplied by 15 my values 16 0.30 seconds and so this is equaling 16.3 zero velocity Max. And we can then say that the total distance is again equal in 100 meters. And here we have that 1.5 plus 16.3 so 17.80 times the maximum velocity equals 100 meters. And so the maximum velocity is equaling 5.62 meters, her second. And we can then say that the maximum acceleration here with the equaling 5.62 meters per second, divided by 3.0 And this is equaling 1.87 meters per second squared. This would be our final acceleration. Our final maximum acceleration for part B. That is the end of the solution. Thank you.

Get More Help with this Textbook