College Physics by Openstax Chapter 3 Problem 19: Comparing Vector Subtraction Paths to Verify Displacement Consistency


Problem:

Do Exercise 3.16 again using analytical techniques and change the second leg of the walk to 25.0 m straight south. (This is equivalent to subtracting B from A — that is, finding R’=A – B ) (b) Repeat again, but now you first walk 25.0 m north and then 18.0 m east. (This is equivalent to subtract A from B —that is, to find A=B+C . Is that consistent with your result?)


Solution:

Part A

From the given statement, you first walk 18.0 m straight west and then 25.0 straight south. These vectors are represented by the graph shown below.

To solve for the resultant, we simply need to use the Pythagorean theorem to solve for the hypotenuse of the right triangle formed. That is,

\begin{align*}
R & = \sqrt{\left( 18.0 \ \text{m} \right)^2+\left( 25.0\ \text{m} \right)^2} \\
R & = 30.8058 \ \text{m} \\
R & = 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

To solve for the angle, θ, we shall use the tangent function.

\begin{align*}
\theta & = \arctan \left( \frac{25.0 \ \text{m}}{18.0 \ \text{m}} \right) \\
\theta & = 54.2461^\circ \\
\theta & = 54.2^\circ
\end{align*}

Therefore, the compass direction of the resultant is 54.2° South of West.

Part B

From the statement, you walk 25.0 m north first and then 18.0 m east. This is represented by the figure shown below.

\begin{align*}
R & = \sqrt{\left( 18.0 \ \text{m} \right)^2+\left( 25.0\ \text{m} \right)^2} \\
R & = 30.8058 \ \text{m} \\
R & = 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

So, we have a right triangle with legs 25.0 m and 18.0 m. We are tasked to solve for the value of R, and the angle θ for the compass direction. The value of R can be solved using the Pythagorean Theorem as in Part A.

To solve for the angle, θ, we shall use the tangent function.

\begin{align*}
\theta & = \arctan \left( \frac{18.0 \ \text{m}}{25.0 \ \text{m}} \right) \\
\theta & = 35.7539^\circ \\
\theta & = 35.8^\circ
\end{align*}

Therefore, the compass direction of the resultant is 35.8° East of North.

This result is consistent with the previous results.


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College Physics 2nd Edition Solutions Table of Contents

Chapter 1: Introduction: The Nature of Science and Physics

Chapter 2: Kinematics

Chapter 3: Two-Dimensional Kinematics

Chapter 4: Dynamics: Force and Newton’s Law of Motion

Chapter 5: Further Applications of Newton’s Laws: Friction, Drag, and Elasticity

Chapter 6: Uniform Circular Motion and Gravitation

Chapter 7: Work, Energy, and Energy Resources

Chapter 8: Linear Momentum and Collisions

Chapter 9: Statics and Torque

Chapter 10: Rotational Motion and Angular Momentum

Chapter 11: Fluid Mechanics

Chapter 12: Fluid Dynamics and Its Biological and Medical Applications

Chapter 13: Temperature, Kinetic Theory, and the Gas Laws

Chapter 14: Heat and Heat Transfer Methods

Chapter 15: Thermodynamics

Chapter 16: Oscillatory Motion and Waves

Chapter 17: Physics of Hearing

Chapter 18: Electric Charge and Electric Field

Chapter 19: Electric Potential and Electric Field

Chapter 20:
Electric Current, Resistance, and Ohm’s Law

Chapter 21: Circuits and DC Instruments

Chapter 22: Magnetism

Chapter 23: Electromagnetic Induction, AC Circuits, and Electrical Technologies

Chapter 24: Electromagnetic Waves

Chapter 25: Geometric Optics

Chapter 26: Vision and Optical Instrument

Chapter 27: Wave Optics

Chapter 28: Special Relativity

Chapter 29: Introduction to Quantum Physics

Chapter 30: Atomic Physics

Chapter 31: Radioactivity and Nuclear Physics

Chapter 32: Medical Applications of Nuclear Physics

Chapter 33: Particle Physics

Chapter 34: Frontiers of Physics