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.
College Physics Chapter 3 Problems

College Physics 2nd Edition Solutions Table of Contents
Chapter 1: Introduction: The Nature of Science and Physics
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 10: Rotational Motion and Angular Momentum
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
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