EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 2, Problem 111P

(a)

To determine

The velocity as the function of time for the given time interval.

(a)

Expert Solution
Check Mark

Answer to Problem 111P

The velocity as the function of time interval is v(t)=(0.10m/s3)t2+9.5m/s .

Explanation of Solution

Given:

The acceleration of the particle is given by ax=(0.20m/s3)t .

The velocity of particle at t=0 is 9.5m/s .

The position of particle at t=0 is 5.0m .

Formula used:

Write expression for the acceleration of the particle.

  dv(t)dt=a(t)

Here, a is the acceleration of the particle and dv(t)dt is the rate of change of velocity.

  dv(t)=a(t)dt

Substitute (0.20m/s3) for a(t) in above expression and integrate.

  dv(t)=(0.20m/ s 3 )tdt

Simplify above expression.

  v(t)=(0.20m/ s 3)2t2+C............. (1)

Calculation:

Substitute 0 for t in equation (1).

  v(0)=( 0.20m/ s 3 )2(0)2+CC=v(0)

Substitute 9.5m/s for v(0) in above expression.

  C=9.5m/s

Substitute 9.5m/s for C in equation (1).

  v(t)=( 0.20m/ s 3 )2t2+9.5m/s=(0.10m/ s 3)t2+9.5m/s

Conclusion:

Thus, the velocity as the function of time interval is v(t)=(0.10m/s3)t2+9.5m/s .

(b)

To determine

The position as function of time for the given time interval.

(b)

Expert Solution
Check Mark

Answer to Problem 111P

The position as function of time for the given interval is x(t)=(0.10m/ s 3)t33+(9.5m/s)t5.0m .

Explanation of Solution

Given:

The acceleration of the particle is given by ax=(0.20m/s3)t .

The velocity of particle at t=0 is 9.5m/s .

The position of particle at t=0 is 5.0m .

Formula used:

Write expression for the velocity of the particle as function of time for given time period.

  v(t)=(0.10m/s3)t2+9.5m/s

Write expression for velocity of the particle.

  dx(t)dt=v(t)

Here, v is the velocity of the particle and dxdt is the rate of change of position of the particle.

Rearrange above expression for dx .

  dx(t)=v(t)dt  ........(2)

Calculation:

Substitute (0.10m/s3)t2+9.5m/s for v(t) in equation (2) and integrate.

  dx(t)=[( 0.10m/ s 3 )t2+9.5m/s]dt

Simplify above expression.

  x(t)=(0.10m/ s 3)t33+(9.5m/s)t+D

  ........(3)

Substitute 0 for t in above expression.

  x(0)=( 0.10m/ s 3 ) ( 0 )33+(9.5m/s)(0)+Dx(0)=D

Substitute 5m for x(0) in above expression.

  D=5.0m

Substitute 5.0m for D in equation (3).

  x(t)=( 0.10m/ s 3 )t33+(9.5m/s)t+(5.0m)x(t)=( 0.10m/ s 3 )t33+(9.5m/s)t5.0m

Conclusion:

Thus, the position as function of time for the given interval is x(t)=(0.10m/ s 3)t33+(9.5m/s)t5.0m .

(c)

To determine

The average velocity for the given time interval and compare to the average of instantaneous velocities of starting and ending times.

(c)

Expert Solution
Check Mark

Answer to Problem 111P

The average velocity for the given time interval is 12.8m/s and it is not equal to the average of instantaneous velocities of the start and ending times.

Explanation of Solution

Given:

The acceleration of the particle is given by ax=(0.20m/s3)t .

The velocity of particle at t=0 is 9.5m/s .

The position of particle at t=0 is 5.0m .

Formula used:

Write expression for average velocity of the particle.

  vav=1Δtt=t1t2v(t)dt............. (4)

Write expression for instantaneous velocity of the particle.

  v(t)=(0.10m/s3)t2+9.5m/s

  ........(5)

Write expression for average of instantaneous velocities for t=0 and t=10 .

  vav=v(10)+v(0)2............. (6)

Calculation:

Substitute 0s for t1 , 10s for t2 , 10s for Δt and (0.10m/s3)t2+9.5m/s for v(t) in equation (4).

  vav=110t=010[( 0.10m/ s 3 ) t 2+9.5m/s]dtvav=110( ( 0.10m/ s 3 ) t 3 3+( 9.5m/s )t)010vav=110[( 0.10m/ s 3 ) ( 10 )33+(9.5m/s)(10)]vav=12.8m/s

Substitute 0s for t in equation (5).

  v(0)=(0.10m/ s 3)(0)2+9.5m/sv(0)=9.5m/s

Substitute 10s for t in equation (5).

  v(10)=(0.10m/ s 3)(10s)2+9.5m/sv(10)=[10+9.5]m/sv(10)=19.5m/s

Substitute 19.5m/s for v(10) and 0m/s for v(0) in equation (6).

  vav=19.5m/s+9.5m/s2vav=14.5m/s

Conclusion:

Thus, the average velocity for the given time interval is 12.8m/s and it is not equal to the average of instantaneous velocities of the start and ending times.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A piece of metal is placed on top of a 2.0 - kg wooden block (mass density = 562 kg/m³) piece. UseArchimedes' principle to calculate the mass (in kg) of copper if the top of the wood surface is exactly at thewater's surface?
A filmmaker wants to achieve an interesting visual effect by filming a scene through a converging lens with a focal length of 50.0 m. The lens is placed betwen the camera and a horse, which canters toward the camera at a constant speed of 7.9 m/s. The camera starts rolling when the horse is 36.0 m from the lens. Find the average speed of the image of the horse (a) during the first 2.0 s after the camera starts rolling and (b) during the following 2.0 s.
Answer the question (Physics)

Chapter 2 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

Ch. 2 - Prob. 11PCh. 2 - Prob. 12PCh. 2 - Prob. 13PCh. 2 - Prob. 14PCh. 2 - Prob. 15PCh. 2 - Prob. 16PCh. 2 - Prob. 17PCh. 2 - Prob. 18PCh. 2 - Prob. 19PCh. 2 - Prob. 20PCh. 2 - Prob. 21PCh. 2 - Prob. 22PCh. 2 - Prob. 23PCh. 2 - Prob. 24PCh. 2 - Prob. 25PCh. 2 - Prob. 26PCh. 2 - Prob. 27PCh. 2 - Prob. 28PCh. 2 - Prob. 29PCh. 2 - Prob. 30PCh. 2 - Prob. 31PCh. 2 - Prob. 32PCh. 2 - Prob. 33PCh. 2 - Prob. 34PCh. 2 - Prob. 35PCh. 2 - Prob. 36PCh. 2 - Prob. 37PCh. 2 - Prob. 38PCh. 2 - Prob. 39PCh. 2 - Prob. 40PCh. 2 - Prob. 41PCh. 2 - Prob. 42PCh. 2 - Prob. 43PCh. 2 - Prob. 44PCh. 2 - Prob. 45PCh. 2 - Prob. 46PCh. 2 - Prob. 47PCh. 2 - Prob. 48PCh. 2 - Prob. 49PCh. 2 - Prob. 50PCh. 2 - Prob. 51PCh. 2 - Prob. 52PCh. 2 - Prob. 53PCh. 2 - Prob. 54PCh. 2 - Prob. 55PCh. 2 - Prob. 56PCh. 2 - Prob. 57PCh. 2 - Prob. 58PCh. 2 - Prob. 59PCh. 2 - Prob. 60PCh. 2 - Prob. 61PCh. 2 - Prob. 62PCh. 2 - Prob. 63PCh. 2 - Prob. 64PCh. 2 - Prob. 65PCh. 2 - Prob. 66PCh. 2 - Prob. 67PCh. 2 - Prob. 68PCh. 2 - Prob. 69PCh. 2 - Prob. 70PCh. 2 - Prob. 71PCh. 2 - Prob. 72PCh. 2 - Prob. 73PCh. 2 - Prob. 74PCh. 2 - Prob. 75PCh. 2 - Prob. 76PCh. 2 - Prob. 77PCh. 2 - Prob. 78PCh. 2 - Prob. 79PCh. 2 - Prob. 80PCh. 2 - Prob. 81PCh. 2 - Prob. 82PCh. 2 - Prob. 83PCh. 2 - Prob. 84PCh. 2 - Prob. 85PCh. 2 - Prob. 86PCh. 2 - Prob. 87PCh. 2 - Prob. 88PCh. 2 - Prob. 89PCh. 2 - Prob. 90PCh. 2 - Prob. 91PCh. 2 - Prob. 92PCh. 2 - Prob. 93PCh. 2 - Prob. 94PCh. 2 - Prob. 95PCh. 2 - Prob. 96PCh. 2 - Prob. 97PCh. 2 - Prob. 98PCh. 2 - Prob. 99PCh. 2 - Prob. 100PCh. 2 - Prob. 101PCh. 2 - Prob. 102PCh. 2 - Prob. 103PCh. 2 - Prob. 104PCh. 2 - Prob. 105PCh. 2 - Prob. 106PCh. 2 - Prob. 107PCh. 2 - Prob. 108PCh. 2 - Prob. 109PCh. 2 - Prob. 110PCh. 2 - Prob. 111PCh. 2 - Prob. 112PCh. 2 - Prob. 113PCh. 2 - Prob. 114PCh. 2 - Prob. 115PCh. 2 - Prob. 116PCh. 2 - Prob. 117PCh. 2 - Prob. 118PCh. 2 - Prob. 119PCh. 2 - Prob. 120PCh. 2 - Prob. 121PCh. 2 - Prob. 122P
Knowledge Booster
Background pattern image
Physics
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.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill
Text book image
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Position/Velocity/Acceleration Part 1: Definitions; Author: Professor Dave explains;https://www.youtube.com/watch?v=4dCrkp8qgLU;License: Standard YouTube License, CC-BY