Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
9th Edition
ISBN: 9781305932302
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 3, Problem 46P

(a)

To determine

The position vector of point A in terms of fraction f .

(a)

Expert Solution
Check Mark

Answer to Problem 46P

The position vector of point A in terms of fraction f is A=[(5+11f)i^+(3+9f)j^]m .

Explanation of Solution

Section 1:

To determine: The horizontal position of point A that must be at a fraction f of the destination.

Answer: The horizontal position of point A that must be at a fraction f of the destination is (5+11f)m .

Given information:

The initial position vector is (5i^+3j^)m and final position vector is (16i^+12j^)m .

Formula to calculate the x-position of point A that must be at a fraction f of the destination is,

x=x1+f(x2x1)

  • x is the position of the point A in x-direction.
  • x1 is the initial position in x-direction.
  • x2 is the final position in x-direction.
  • f is the fraction of the destination path.

Substitute 5m for x1 and 16m for x2 .

x=5m+f(16m5m)=5m+f(11m)=(5+11f)m

Section 2:

To determine: The vertical position of point A that must be at a fraction f of the destination.

Answer: The vertical position of point A that must be at a fraction f of the destination is (3+9f)m .

Given information:

The initial position vector is (5i^+3j^)m and final position vector is (16i^+12j^)m .

Formula to calculate the y-position of point A that must be at a fraction of f of the destination is,

y=y1+f(y2y1)

  • y is the position of the point A in y-direction.
  • y1 is the initial position in y-direction.
  • y2 is the final position in y-direction.
  • f is the fraction of the destination path.

Substitute 3m for y1 and 12m for y2 .

y=3m+f(13m3m)=3m+f(9m)=(3+9f)m

Section 3:

To determine: The position vector of the point A that must lie at a fraction f on the destination path.

Answer: The position vector of the point A that must lie at a fraction f on the destination path is A=[(5+11f)i^+(3+9f)j^]m .

Given information:

The initial position vector is (5i^+3j^)m and final position vector is (16i^+12j^)m .

Formula to calculate the position vector of the point A that must lie at a fraction f on the destination path is,

A=xi^+yj^

Substitute (5+11f)m for x and (3+9f)m for y to find A .

A=[(5+11f)i^+(3+9f)j^]m

Conclusion:

Therefore, position vector of point A in terms of fraction f is A=[(5+11f)i^+(3+9f)j^]m .

(b)

To determine

The position vector of point A for f=0 .

(b)

Expert Solution
Check Mark

Answer to Problem 46P

The position vector of point A for f=0 is A=[5i^+3j^]m .

Explanation of Solution

Given information:

The initial position vector is (5i^+3j^)m and final position vector is (16i^+12j^)m .

The position vector of point A in terms of fraction f is,

A=[(5+11f)i^+(3+9f)j^]m .

Substitute 0 for f in the above expression.

A=[(5+11×0)i^+(3+9×0)j^]m=[(5+0)i^+(3+0)j^]=5i^+3j^

Conclusion:

Therefore, position vector of point A for f=0 is A=[5i^+3j^]m .

(c)

To determine

Whether the position vector of point A for f=0 is reasonable.

(c)

Expert Solution
Check Mark

Explanation of Solution

Introduction: The position vector described the position of an object relative to a fixed reference point.

Given information:

The initial position vector is (5i^+3j^)m and final position vector is (16i^+12j^)m .

The position vector of point A in terms of fraction f is,

A=[(5+11f)i^+(3+9f)j^]m

When the value of f is zero then the position vector of point A becomes,

A=[5i^+3j^]m

This position vector is similar to the initial position vector of the destination path. So, if the value of f is zero then the point A move to the initial point that really exist.

Conclusion:

Therefore, position vector of point A for f=0 is reasonable because the position of point A is shift to the starting point of the destination path.

(d)

To determine

The position vector of point A for f=1 .

(d)

Expert Solution
Check Mark

Answer to Problem 46P

The position vector of point A for f=1 is 16i^+12j^ .

Explanation of Solution

Given information:

The initial position vector is (5i^+3j^)m and final position vector is (16i^+12j^)m .

The position vector of point A in terms of fraction f is,

A=[(5+11f)i^+(3+9f)j^]m

Substitute 1 for f in the above expression.

A=[(5+11×1)i^+(3+9×1)j^]m=[(5+11)i^+(3+9)j^]=16i^+12j^

Conclusion:

Therefore, position vector of point A for f=1 is 16i^+12j^ .

(e)

To determine

Whether the position vector of point A for f=1 is reasonable.

(e)

Expert Solution
Check Mark

Explanation of Solution

Introduction: The position vector described the position of an object relative to a fixed reference point.

Given information:

The initial position vector is (5i^+3j^)m and final position vector is (16i^+12j^)m .

The position vector of point A in terms of fraction f is,

A=[(5+11f)i^+(3+9f)j^]m

When the value of f is 1 , then the position vector of point A becomes,

A=16i^+12j^

This position vector is similar to the final position vector of the destination path. So, if the value of f is 1 then the point A move to the final point that really exist.

Conclusion:

Therefore, position vector of point A for f=1 is reasonable because the position of point A is shifted to the final point of the destination path.

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
4B. Four electrons are located on the corners of a square, one on each corner, with the sides of the square being 25 cm long. a) Draw a sketch of the scenario and use your sketch to b)  Determine the total force (magnitude and direction) on one of the electrons from the other three?
Portfolio Problem 3. A ball is thrown vertically upwards with a speed vo from the floor of a room of height h. It hits the ceiling and then returns to the floor, from which it rebounds, managing just to hit the ceiling a second time. Assume that the coefficient of restitution between the ball and the floor, e, is equal to that between the ball and the ceiling. Compute e.
Portfolio Problem 4. Consider two identical springs, each with natural length and spring constant k, attached to a horizontal frame at distance 2l apart. Their free ends are attached to the same particle of mass m, which is hanging under gravity. Let z denote the vertical displacement of the particle from the hori- zontal frame, so that z < 0 when the particle is below the frame, as shown in the figure. The particle has zero horizontal velocity, so that the motion is one dimensional along z. 000000 0 eeeeee (a) Show that the total force acting on the particle is X F-mg k-2kz 1 (1. l k. (b) Find the potential energy U(x, y, z) of the system such that U x = : 0. = O when (c) The particle is pulled down until the springs are each of length 3l, and then released. Find the velocity of the particle when it crosses z = 0.

Chapter 3 Solutions

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term

Ch. 3 - Prob. 6OQCh. 3 - Yes or no: Is each of the following quantities a...Ch. 3 - What is the y component of the vector (3i-8k)m/s?...Ch. 3 - What is the x component of the vector shown in...Ch. 3 - What is the y component of the vector shown in...Ch. 3 - Prob. 11OQCh. 3 - A submarine dives from the water surface at an...Ch. 3 - Prob. 13OQCh. 3 - Is it possible to add a vector quantity to a...Ch. 3 - Can the magnitude of a vector have a negative...Ch. 3 - A book is moved once around the perimeter of a...Ch. 3 - Prob. 4CQCh. 3 - Prob. 5CQCh. 3 - Prob. 1PCh. 3 - Prob. 2PCh. 3 - Two points in the xy plane have Cartesian...Ch. 3 - Prob. 4PCh. 3 - The polar coordinates of a certain point are (r =...Ch. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 8PCh. 3 - Why is the following situation impossible? A...Ch. 3 - A force F1 of magnitude 6.00 units acts on an...Ch. 3 - The displacement vectors A and B shown in Figure...Ch. 3 - Three displacements are A=200m due south, B=250m...Ch. 3 - A roller-coaster car moves 200 ft horizontally and...Ch. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - A person walks 25.0 north of east for 3.10 km. How...Ch. 3 - Prob. 19PCh. 3 - Prob. 20PCh. 3 - While exploring a cave, a spelunker starts at the...Ch. 3 - Use the component method to add the vectors A and...Ch. 3 - Prob. 23PCh. 3 - A map suggests that Atlanta is 730 miles in a...Ch. 3 - Your dog is running around the grass in your back...Ch. 3 - Given the vectors A=2.00i+6.00j and B=3.00i2.00j,...Ch. 3 - A novice golfer on the green takes three strokes...Ch. 3 - A snow-covered ski slope makes an angle of 35.0...Ch. 3 - The helicopter view in Fig. P3.15 shows two people...Ch. 3 - In a game of American football, a quarterback...Ch. 3 - Prob. 31PCh. 3 - Vector A has x and y components of 8.70 cm and...Ch. 3 - The vector A has x, y, and z components of 8.00,...Ch. 3 - Prob. 34PCh. 3 - Vector A has a negative x component 3.00 units in...Ch. 3 - Given the displacement vectors A=(3i4j+4k)m and...Ch. 3 - Prob. 37PCh. 3 - Three displacement vectors of a croquet ball are...Ch. 3 - A man pushing a mop across a floor causes it to...Ch. 3 - Figure P3.28 illustrates typical proportions of...Ch. 3 - Express in unit-vector notation the following...Ch. 3 - radar station locates a sinking ship at range 17.3...Ch. 3 - Prob. 43PCh. 3 - Why is the following situation impossible? A...Ch. 3 - Review. You are standing on the ground at the...Ch. 3 - Prob. 46PCh. 3 - In an assembly operation illustrated in Figure...Ch. 3 - A fly lands on one wall of a room. The lower-left...Ch. 3 - As she picks up her riders, a bus driver traverses...Ch. 3 - A jet airliner, moving initially at 300 mi/h to...Ch. 3 - A person going for a walk follows the path shown...Ch. 3 - Find the horizontal and vertical components of the...Ch. 3 - Review. The biggest stuffed animal in the world is...Ch. 3 - An air-traffic controller observes two aircraft on...Ch. 3 - In Figure P3.55, a spider is resting after...Ch. 3 - The rectangle shown in Figure P3.56 has sides...Ch. 3 - Prob. 57APCh. 3 - A ferry transports tourists between three islands....Ch. 3 - Two vectors A and B have precisely equal...Ch. 3 - Two vectors A and B have precisely equal...Ch. 3 - Prob. 61APCh. 3 - Prob. 62APCh. 3 - Prob. 63APCh. 3 - Prob. 64APCh. 3 - A rectangular parallelepiped has dimensions a, b,...Ch. 3 - Vectors A and B have equal magnitudes of 5.00. The...Ch. 3 - A pirate has buried his treasure on an island with...
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
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
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
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
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
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Kinematics Part 3: Projectile Motion; Author: Professor Dave explains;https://www.youtube.com/watch?v=aY8z2qO44WA;License: Standard YouTube License, CC-BY