Physics For Scientists And Engineers Student Solutions Manual, Vol. 1
Physics For Scientists And Engineers Student Solutions Manual, Vol. 1
6th Edition
ISBN: 9781429203029
Author: David Mills
Publisher: W. H. Freeman
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 For Scientists And Engineers Student Solutions Manual, Vol. 1, 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 For Scientists And Engineers Student Solutions Manual, Vol. 1, 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 For Scientists And Engineers Student Solutions Manual, Vol. 1, Chapter 3, Problem 110P , additional homework tip  3isPhysics For Scientists And Engineers Student Solutions Manual, Vol. 1, Chapter 3, Problem 110P , additional homework tip  4.

The position of the particle at Physics For Scientists And Engineers Student Solutions Manual, Vol. 1, Chapter 3, Problem 110P , additional homework tip  5isPhysics For Scientists And Engineers Student Solutions Manual, Vol. 1, 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 For Scientists And Engineers Student Solutions Manual, Vol. 1, Chapter 3, Problem 110P , additional homework tip  7isPhysics For Scientists And Engineers Student Solutions Manual, Vol. 1, Chapter 3, Problem 110P , additional homework tip  8.

The position of the particle at Physics For Scientists And Engineers Student Solutions Manual, Vol. 1, Chapter 3, Problem 110P , additional homework tip  9isPhysics For Scientists And Engineers Student Solutions Manual, Vol. 1, 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
At t = 0, a particle leaves the origin with a velocity of 9.0 m/s in the positive ydirection and moves in the xy plane with a constant acceleration of (2.0i - 4.0j)m/s2. At the instant the x coordinate of the particle is 15 m, what is the speedof the particle?
at t=0, a particle leaves the origin with a velocity of 12m/s in the positive x-direction and moves in the xy plane with a constant acceleration of (-2.0i + 4.0j)m/s2. at the instant the y coordinate is 18m, what is the x coordinate of the particle?
A particle starts from the origin at t = 0 with a velocity of 6.0 m/s and moves in the xy plane with a constant acceleration of (−2.0î + 4.0ĵ  ) m/s2 . At the instant the particle achieves its maximum positive x coordinate, how far is it from the origin?

Chapter 3 Solutions

Physics For Scientists And Engineers Student Solutions Manual, Vol. 1

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