PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
Question
Book Icon
Chapter 3, Problem 110P

(a)

To determine

The velocity of the particle.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The position of the particle at t=0 is r1=(4.0m)i^+(3.0m)j^.

The position of the particle at t=2 is r2=(10m)i^(2m)j^ .

The velocity of the particle is v2=(5.0m/s)i^(6.0m/s)j^ .

Formula used:

Write the expression for the average velocity.

  vav=ΔrΔt

Here, Δr is the change in the position vector and Δt is the change in time.

  vav=r2r1t2t1   ...... (1)

Here, r2 is the final position vector of the particle, r1 is the initial position vector of the particle, t2 is the final time of the particle and t1 is the initial time of the particle.

Write the expression for the average velocity of the particle.

  vav=v1+v22

Here, vav is the average velocity, v1 is the initial velocity of the particle and v2 is the final velocity.

Solve the above equation for v1 .

  v1=2vavv2   ...... (2)

Calculation:

Substitute (10m)i^(2m)j^ for r2 , (4.0m)i^+(3.0m)j^ for r1 , 0 for t1 and 2s for t2 in equation (1).

  vav=( 10m)i^( 2m)j^( 4.0m)i^+( 3.0m)j^2svav=(3.0m/s)i^(2.5m/s)j^

Substitute (3.0m/s)i^(2.5m/s)j^ for vav and (5.0m/s)i^(6.0m/s)j^ for v2 in equation (2).

  v1=2[(3.0m/s)i^(2.5m/s)j^][(5.0m/s)i^(6.0m/s)j^]v1=(1.0m/s)i^+(1.0m/s)j^

Conclusion:

Thus, the initial velocity is (1.0m/s)i^+(1.0m/s)j^ .

(b)

To determine

The acceleration of the particle.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The position of the particle at t=0 is r1=(4.0m)i^+(3.0m)j^.

The position of the particle at PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 3, Problem 110P , additional homework tip  1is r2=(10m)i^(2m)j^ .

The velocity of the particle is v2=(5.0m/s)i^(6.0m/s)j^ .

Formula used:

Write the expression for the acceleration of the particle.

  a=ΔvΔt

Here, Δv

is the change in the velocity vector and PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 3, Problem 110P , additional homework tip  2is the change in time.

  a=v2v1t2t1   ...... (3)

Here, v2 is the final velocity, v1 is the initial velocity, t2 is the final time and t1 is the initial time.

Calculation:

Substitute (5.0m/s)i^(6.0m/s)j^ for v2 , (1.0m/s)i^+(1.0m/s)j^ for v1 , 2s for t2 and 0 for t1 in equation (3).

  a=( 5.0m/s )i^( 6.0m/s )j^( ( 1.0m/s ) i ^ +( 1.0m/s ))j^2sa=(2.0m/ s 2)i^(3.5m/ s 2)j^

Conclusion:

The acceleration of the particle is (2.0m/s2)i^(3.5m/s2)j^ .

(c)

To determine

The velocity of the particle as the function of time.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The position of the particle at PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 3, Problem 110P , additional homework tip  3isPHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 3, Problem 110P , additional homework tip  4.

The position of the particle at PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 3, Problem 110P , additional homework tip  5isPHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 3, Problem 110P , additional homework tip  6

The velocity of the particle is v2=(5.0m/s)i^(6.0m/s)j^ .

Formula used:

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

  v(t)=v1+at   ...... (4)

Here, v(t) is the velocity of the particle as the function of time.

Calculation:

Substitute (1.0m/s)i^+(1.0m/s)j^ for v1 and (2.0m/s2)i^(3.5m/s2)j^ for a in equation (4).

  v(t)=(1.0m/s)i^+(1.0m/s)j^+(( 2.0m/ s 2 )i^( 3.5m/ s 2 )j^)tv(t)=(1.0m/s+( 2.0m/ s 2 )t)i^+(1.0m/s( 3.5m/ s 2 )t)j^

Conclusion:

Thus, the velocity of the particle as the function of timeis (1.0m/s+(2.0m/ s 2)t)i^+(1.0m/s(3.5m/ s 2)t)j^ .

(d)

To determine

The position vector of the particle as the function of time.

(d)

Expert Solution
Check Mark

Explanation of Solution

Given:

The position of the particle at PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 3, Problem 110P , additional homework tip  7isPHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 3, Problem 110P , additional homework tip  8.

The position of the particle at PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 3, Problem 110P , additional homework tip  9isPHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 3, Problem 110P , additional homework tip  10

The velocity of the particle is v2=(5.0m/s)i^(6.0m/s)j^ .

Formula used:

Write the expression for the position vector as the function of time.

  r1(t)=r1+v1(t)+12at2   ...... (5)

Calculation:

Substitute (4.0m)i^+(3.0m)j^ for r1 , (1.0m/s+(2.0m/ s 2)t)i^+(1.0m/s-(3.5m/ s 2)t)j^ for v1(t) and (2.0m/s2)i^(3.5m/s2)j^ for a in equation (5).

  r1(t)=( ( ( 4.0m )+( 1.0m/s )t+( 2.0m/ s 2 ) t 2 + 1 2 ( 2.0m/ s 2 ) t 2 ) i ^ +( ( 3.0m+( 1.0m/s )t( 3.5m/ s 2 ) t 2 )( 3.5m/ s 2 ) t 2 ) j ^ )r1(t)=(( 4.0m)+( 1.0m/s )t+( 1.0m/ s 2 )t2)i^+(3.0m+( 1.0m/s )t( 1.8m/ s 2 )t2)j^

Conclusion:

The position vector of the particle as the function of time is:

  ((4.0m)+(1.0m/s)t+(1.0m/ s 2)t2)i^+(3.0m+(1.0m/s)t(1.8m/ s 2)t2)j^

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
Question 3 of 17 L X L L T 0.5/ In the figure above, three uniform thin rods, each of length L, form an inverted U. The vertical rods each have a mass m; the horizontal rod has a mass 3m. NOTE: Express your answer in terms of the variables given. (a) What is the x coordinate of the system's center of mass? xcom L 2 (b) What is the y coordinate of the system's center of mass? Ycom 45 L X Q Search MD bp N
Sketch the harmonic on graphing paper.
Exercise 1: (a) Using the explicit formulae derived in the lectures for the (2j+1) × (2j + 1) repre- sentation matrices Dm'm, (J/h), derive the 3 × 3 matrices corresponding to the case j = 1. (b) Verify that they satisfy the so(3) Lie algebra commutation relation: [D(Î₁/ħ), D(Î₂/h)]m'm₁ = iƊm'm² (Ĵ3/h). (c) Prove the identity 3 Dm'm,(β) = Σ (D(Ρ)D(Ρ))m'¡m; · i=1

Chapter 3 Solutions

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS

Ch. 3 - Prob. 11PCh. 3 - Prob. 12PCh. 3 - Prob. 13PCh. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - Prob. 18PCh. 3 - Prob. 19PCh. 3 - Prob. 20PCh. 3 - Prob. 21PCh. 3 - Prob. 22PCh. 3 - Prob. 23PCh. 3 - Prob. 24PCh. 3 - Prob. 25PCh. 3 - Prob. 26PCh. 3 - Prob. 27PCh. 3 - Prob. 28PCh. 3 - Prob. 29PCh. 3 - Prob. 30PCh. 3 - Prob. 31PCh. 3 - Prob. 32PCh. 3 - Prob. 33PCh. 3 - Prob. 34PCh. 3 - Prob. 35PCh. 3 - Prob. 36PCh. 3 - Prob. 37PCh. 3 - Prob. 38PCh. 3 - Prob. 39PCh. 3 - Prob. 40PCh. 3 - Prob. 41PCh. 3 - Prob. 42PCh. 3 - Prob. 43PCh. 3 - Prob. 44PCh. 3 - Prob. 45PCh. 3 - Prob. 46PCh. 3 - Prob. 47PCh. 3 - Prob. 48PCh. 3 - Prob. 49PCh. 3 - Prob. 50PCh. 3 - Prob. 51PCh. 3 - Prob. 52PCh. 3 - Prob. 53PCh. 3 - Prob. 54PCh. 3 - Prob. 55PCh. 3 - Prob. 56PCh. 3 - Prob. 57PCh. 3 - Prob. 58PCh. 3 - Prob. 59PCh. 3 - Prob. 60PCh. 3 - Prob. 61PCh. 3 - Prob. 62PCh. 3 - Prob. 63PCh. 3 - Prob. 64PCh. 3 - Prob. 65PCh. 3 - Prob. 66PCh. 3 - Prob. 67PCh. 3 - Prob. 68PCh. 3 - Prob. 69PCh. 3 - Prob. 70PCh. 3 - Prob. 71PCh. 3 - Prob. 72PCh. 3 - Prob. 73PCh. 3 - Prob. 74PCh. 3 - Prob. 75PCh. 3 - Prob. 76PCh. 3 - Prob. 77PCh. 3 - Prob. 78PCh. 3 - Prob. 79PCh. 3 - Prob. 80PCh. 3 - Prob. 81PCh. 3 - Prob. 82PCh. 3 - Prob. 83PCh. 3 - Prob. 84PCh. 3 - Prob. 85PCh. 3 - Prob. 86PCh. 3 - Prob. 87PCh. 3 - Prob. 88PCh. 3 - Prob. 89PCh. 3 - Prob. 90PCh. 3 - Prob. 91PCh. 3 - Prob. 92PCh. 3 - Prob. 93PCh. 3 - Prob. 94PCh. 3 - Prob. 95PCh. 3 - Prob. 96PCh. 3 - Prob. 97PCh. 3 - Prob. 98PCh. 3 - Prob. 99PCh. 3 - Prob. 100PCh. 3 - Prob. 101PCh. 3 - Prob. 102PCh. 3 - Prob. 103PCh. 3 - Prob. 104PCh. 3 - Prob. 105PCh. 3 - Prob. 106PCh. 3 - Prob. 107PCh. 3 - Prob. 108PCh. 3 - Prob. 109PCh. 3 - Prob. 110PCh. 3 - Prob. 111PCh. 3 - Prob. 112PCh. 3 - Prob. 113PCh. 3 - Prob. 114PCh. 3 - Prob. 115PCh. 3 - Prob. 116PCh. 3 - Prob. 117PCh. 3 - Prob. 118PCh. 3 - Prob. 119PCh. 3 - Prob. 120PCh. 3 - Prob. 121PCh. 3 - Prob. 122P
Knowledge Booster
Background pattern image
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
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning