Homework Help is Here – Start Your Trial Now!
Science
Physics
Practice Problem: If the acceleration is 2.0 m/s² instead of 4.0 m/s. where is the cyclist, and how fast is he moving, 5.0 s after he passes the signpost if xo = 5.0 m and vox 15 m/s? Answers: 105 m, 25 m/s. =

Practice Problem: If the acceleration is 2.0 m/s² instead of 4.0 m/s. where is the cyclist, and how fast is he moving, 5.0 s after he passes the signpost if xo = 5.0 m and vox 15 m/s? Answers: 105 m, 25 m/s. =

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
See similar textbooks
icon
Related questions
Question
100%

I just need the practice problem solved with work shown. The rest of the image provides the information needed to solve the problem. The correct answer is also provided to check work. 

OSAGE O EXAMPLE 2.8 Constant acceleration on a motorcycle Here we will look at a more complex problem for which we will use all three of our primary constant- acceleration equations. A motorcyclist heading east through a small Iowa town accelerates after he passes a signpost at x = 0 marking the city limits. His acceleration is constant: ax = 4.0 m/s². At time t = 0, he is 5.0 m east of the signpost and has a velocity of vox = 15 m/s. (a) Find his position and velocity at time 1 = 2.0 s. (b) Where is the motorcyclist when his velocity is 25 m/s? SOLUTION SET UP Figure 2.21 shows our diagram. The problem tells us that x = 0 at the signpost, so that is the origin of coordinates. We point the x axis east, in the direction of motion. The (constant) acceleration is a = 4.0 m/s². At the initial time t = 0, the position is xo = 5.0 m and the initial velocity is vox = 15 m/s. SOLVE Part (a): We want to know the position and velocity (the val- ues of x and Ux, respectively) at the later time t =.2.0 s. Equation 2.10 gives the position x as a function of time t: x = xo + voxt + 1/a,t² = 5.0 m+ (15 m/s) (2.0 s) + (4.0 m/s²) (2.0 s)² = 43 m. Equation 2.6 gives the velocity Ux as a function of time t: Ux = Vox + axt = 15 m/s + (4.0 m/s²) (2.0 s) = 23 m/s. Vox xo = 5.0 m t = 0 FIGURE 2.21 = ax = 4.0 m/s² gen do the results change 15 m/s X = ? t = 2.0 s TO Ux = ? → x (east) Video Tutor Solut 25 m/s. Note that this occurs at a time later than 2.0 s and at a point farther than 43 m from Part (b): We want to know the value of x when ux the signpost. From Equation 2.11, we have = 2 v² = voz² + 2ax(x − xo), Ux (25 m/s)² = (15 m/s)2 + 2(4.0 m/s²)(x - 5.0 m), x = 55 m. Alternatively, we may use Equation 2.6 to find first the time when Ux = 25 m/s: Ux = Vox + axt, 25 m/s = 15 m/s + (4.0 m/s²)(t), t = 2.5 s. Then, from Equation 2.10, x = xo + voxt + 1⁄2 axt² = = 5.0 m+ (15 m/s) (2.5 s) + (4.0 m/s²) (2.5 s)² = 55 m. sors Rigodi pol oniworla A FIGUE the polic SOLUTION SET UP B the x axis x compor keep in m nent of The so we c objects The or both be th t. At the to E ini as aF REFLECT Do the results make sense? The cyclist accelerates from 15 m/s (about 34 mi/h) to 23 m/s (about 51 mi/h) in 2.0 s while travel- ing a distance of 38 m (about 125 ft). This is fairly brisk acceleration, but well within the realm of possibility for a high-performance bike. Practice Problem: If the acceleration is 2.0 m/s² instead of 4.0 m/s. where is the cyclist, and how fast is he moving, 5.0 s after he passes the signpost if xo = 5.0 m and vox = 15 m/s? Answers: 105 m, 25 m/s.
Transcribed Image Text:OSAGE O EXAMPLE 2.8 Constant acceleration on a motorcycle Here we will look at a more complex problem for which we will use all three of our primary constant- acceleration equations. A motorcyclist heading east through a small Iowa town accelerates after he passes a signpost at x = 0 marking the city limits. His acceleration is constant: ax = 4.0 m/s². At time t = 0, he is 5.0 m east of the signpost and has a velocity of vox = 15 m/s. (a) Find his position and velocity at time 1 = 2.0 s. (b) Where is the motorcyclist when his velocity is 25 m/s? SOLUTION SET UP Figure 2.21 shows our diagram. The problem tells us that x = 0 at the signpost, so that is the origin of coordinates. We point the x axis east, in the direction of motion. The (constant) acceleration is a = 4.0 m/s². At the initial time t = 0, the position is xo = 5.0 m and the initial velocity is vox = 15 m/s. SOLVE Part (a): We want to know the position and velocity (the val- ues of x and Ux, respectively) at the later time t =.2.0 s. Equation 2.10 gives the position x as a function of time t: x = xo + voxt + 1/a,t² = 5.0 m+ (15 m/s) (2.0 s) + (4.0 m/s²) (2.0 s)² = 43 m. Equation 2.6 gives the velocity Ux as a function of time t: Ux = Vox + axt = 15 m/s + (4.0 m/s²) (2.0 s) = 23 m/s. Vox xo = 5.0 m t = 0 FIGURE 2.21 = ax = 4.0 m/s² gen do the results change 15 m/s X = ? t = 2.0 s TO Ux = ? → x (east) Video Tutor Solut 25 m/s. Note that this occurs at a time later than 2.0 s and at a point farther than 43 m from Part (b): We want to know the value of x when ux the signpost. From Equation 2.11, we have = 2 v² = voz² + 2ax(x − xo), Ux (25 m/s)² = (15 m/s)2 + 2(4.0 m/s²)(x - 5.0 m), x = 55 m. Alternatively, we may use Equation 2.6 to find first the time when Ux = 25 m/s: Ux = Vox + axt, 25 m/s = 15 m/s + (4.0 m/s²)(t), t = 2.5 s. Then, from Equation 2.10, x = xo + voxt + 1⁄2 axt² = = 5.0 m+ (15 m/s) (2.5 s) + (4.0 m/s²) (2.5 s)² = 55 m. sors Rigodi pol oniworla A FIGUE the polic SOLUTION SET UP B the x axis x compor keep in m nent of The so we c objects The or both be th t. At the to E ini as aF REFLECT Do the results make sense? The cyclist accelerates from 15 m/s (about 34 mi/h) to 23 m/s (about 51 mi/h) in 2.0 s while travel- ing a distance of 38 m (about 125 ft). This is fairly brisk acceleration, but well within the realm of possibility for a high-performance bike. Practice Problem: If the acceleration is 2.0 m/s² instead of 4.0 m/s. where is the cyclist, and how fast is he moving, 5.0 s after he passes the signpost if xo = 5.0 m and vox = 15 m/s? Answers: 105 m, 25 m/s.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Centripetal force
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON
SEE MORE TEXTBOOKS