Section 1

Physics I: Kinematics in 1D, 2D, & 3D

STUDY GUIDE

🎓 Physics I - Study Guide

📋 Course Structure


code
📚 Physics I
├── 📖 Chapter 1: Describing Motion: Kinematics in One Dimension
│   ├── 🔹 Reference Frames and Displacement
│   ├── 🔹 Average Velocity
│   ├── 🔹 Instantaneous Velocity
│   ├── 🔹 Acceleration
│   ├── 🔹 Motion at Constant Acceleration
│   └── 🔹 Freely Falling Objects
├── 📖 Chapter 2: Kinematics in Two or Three Dimensions; Vectors
│   ├── 🔹 Vectors and Scalars
│   ├── 🔹 Addition of Vectors - Graphical Methods
│   ├── 🔹 Subtraction of Vectors, and Multiplication of a Vector by a Scalar
│   ├── 🔹 Adding Vectors by Components
│   ├── 🔹 Unit Vectors
│   ├── 🔹 Vector Kinematics
│   ├── 🔹 Projectile Motion
│   ├── 🔹 Solving Problems Involving Projectile Motion
│   └── 🔹 Relative Velocity
├── 📖 Chapter 3: Measurement, Estimating, and Units
│   ├── 🔹 Measurement and Uncertainty
│   ├── 🔹 Significant Figures
│   ├── 🔹 Unit Systems and Conversions
│   └── 🔹 Dimensional Analysis

Section 2

📖 Chapter 1: Describing Motion: Kinematics in One Dimension

What this chapter covers: This chapter introduces the fundamental concepts of motion in one dimension. It defines displacement, velocity (average and instantaneous), and acceleration. It explores motion with constant acceleration and the specific case of freely falling objects. The chapter also emphasizes problem-solving techniques using kinematic equations.

🔑 Essential Concepts & Formulas

Concept/Formula	Definition/Equation	When to Use	Quick Check
Displacement	$\Delta x = x_2 - x_1$	Calculating change in position	Ensure $x_2$ and $x_1$ are in the same reference frame
Average Velocity	$v = \frac{\Delta x}{\Delta t}$	Calculating average speed over a time interval	Check units: m/s
Instantaneous Velocity	$v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$	Finding velocity at a specific time	Use derivative of position function
Average Acceleration	$a = \frac{\Delta v}{\Delta t}$	Calculating average change in velocity	Check units: m/s $^2$
Instantaneous Acceleration	$a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}$	Finding acceleration at a specific time	Use derivative of velocity function
Constant Acceleration: Velocity	$v = v_0 + at$	Constant acceleration, solving for final velocity	Check if acceleration is constant
Constant Acceleration: Position	$x = x_0 + v_0t + \frac{1}{2}at^2$	Constant acceleration, solving for final position	Check if acceleration is constant
Constant Acceleration: Velocity-Position	$v^2 = v_0^2 + 2a(x - x_0)$	Constant acceleration, relating velocity and position	Check if acceleration is constant
Free Fall Acceleration	$g = 9.80 \, \text{m/s}^2$	Objects falling near Earth's surface (neglecting air resistance)	Ensure air resistance is negligible

🛠️ Problem Types

Type A: Constant Acceleration Problems

Setup: "When you encounter a scenario where an object moves with constant acceleration, such as a car accelerating uniformly or an object falling freely (neglecting air resistance)."

Method: "Apply the kinematic equations for constant acceleration: $v = v_0 + at$ , $x = x_0 + v_0t + \frac{1}{2}at^2$ , and $v^2 = v_0^2 + 2a(x - x_0)$ . Identify knowns and unknowns, and choose the appropriate equation to solve for the desired quantity."

Final Answer

Example: A car accelerates from rest at a constant rate of 3.0 m/s

^2

. How far does it travel in 4.0 s? Solution:

x = x_0 + v_0t + \frac{1}{2}at^2 = 0 + 0(4.0) + \frac{1}{2}(3.0)(4.0)^2 = 24

Type B: Free Fall Problems

Setup: "If presented with a situation where an object is falling vertically under the influence of gravity (neglecting air resistance)."

Method: "Use the kinematic equations for constant acceleration, with $a = -g = -9.80 \, \text{m/s}^2$ . Choose a coordinate system with the positive direction pointing upwards. Solve for quantities such as time of flight, maximum height, or final velocity."

Final Answer

Example: An object is thrown vertically upwards with an initial velocity of 15.0 m/s. What is the maximum height reached by the object? Solution:

v^2 = v_0^2 + 2a(y - y_0) \implies 0 = (15.0)^2 + 2(-9.80)(y - 0) \implies y = \frac{(15.0)^2}{2(9.80)} \approx 11.5

🧮 Solved Example

Problem: A ball is thrown vertically upward with an initial velocity of 20.0 m/s. (a) How long does it take to reach its maximum height? (b) What is its maximum height? (c) Determine the velocity of the ball when it returns to its starting point.

Given: Initial velocity $v_0 = 20.0 \, \text{m/s}$ , acceleration due to gravity $g = -9.80 \, \text{m/s}^2$ (negative since it opposes the upward motion).

Steps:

(a) Time to reach maximum height: At maximum height, the velocity is 0. Use $v = v_0 + at$ to solve for $t$ .
(b) Maximum height: Use $v^2 = v_0^2 + 2a(y - y_0)$ to solve for $y$ , where $y_0 = 0$ .
(c) Velocity at starting point: Use $v^2 = v_0^2 + 2a(y - y_0)$ again, but this time $y = 0$ .

"

✅
Answer: (a) $t = \frac{-v_0}{a} = \frac{-20.0}{-9.80} \approx 2.04 \, \text{s}$ (b) $y = \frac{v^2 - v_0^2}{2a} = \frac{0 - (20.0)^2}{2(-9.80)} \approx 20.4 \, \text{m}$ (c) $v = \sqrt{v_0^2 + 2a(y - y_0)} = \sqrt{(20.0)^2 + 2(-9.80)(0 - 0)} = -20.0 \, \text{m/s}$ (downward)

⚠️ Common Mistakes

❌ Mistake 1: Forgetting the sign of acceleration due to gravity.

✅ How to avoid: Always choose a consistent coordinate system and assign the correct sign to $g$ (usually negative if upward is positive).

❌ Mistake 2: Using constant acceleration equations when acceleration is not constant.

✅ How to avoid: Check if the problem states constant acceleration. If not, use calculus-based approaches.

💡 Study Tip

Visualize the motion by drawing diagrams. This helps in understanding the direction of velocity and acceleration, and in choosing the correct kinematic equation.

📖 Chapter 2: Kinematics in Two or Three Dimensions; Vectors

What this chapter covers: This chapter extends the concepts of kinematics to two and three dimensions by introducing vectors. It covers vector addition, subtraction, and multiplication, along with components and unit vectors. Projectile motion and relative velocity are analyzed using vector principles.

🔑 Essential Concepts & Formulas

Concept/Formula	Definition/Equation	When to Use	Quick Check
Vector Components	$V_x = V \cos \theta$ , $V_y = V \sin \theta$	Resolving a vector into x and y components	Ensure $\theta$ is measured from the x-axis
Magnitude of Vector	$V = \sqrt{V_x^2 + V_y^2}$	Finding the magnitude of a vector from its components	Ensure $V_x$ and $V_y$ are in consistent units
Direction of Vector	$\theta = \tan^{-1} \left( \frac{V_y}{V_x} \right)$	Finding the direction of a vector from its components	Check the quadrant of the angle
Vector Addition (Components)	$R_x = A_x + B_x$ , $R_y = A_y + B_y$	Adding vectors using components	Ensure all vectors are resolved into components first
Unit Vector Notation	$\mathbf{V} = V_x \mathbf{i} + V_y \mathbf{j} + V_z \mathbf{k}$	Expressing a vector using unit vectors	$\mathbf{i}$ , $\mathbf{j}$ , and $\mathbf{k}$ are unit vectors along x, y, and z axes
Projectile Motion: Horizontal	$x = x_0 + v_{0x}t$	Analyzing horizontal motion of a projectile (constant velocity)	Neglecting air resistance
Projectile Motion: Vertical	$y = y_0 + v_{0y}t - \frac{1}{2}gt^2$	Analyzing vertical motion of a projectile (constant acceleration)	Neglecting air resistance
Relative Velocity	$\mathbf{v}_{AB} = \mathbf{v}_{AC} + \mathbf{v}_{CB}$	Relating velocities in different reference frames	Ensure subscripts are consistent

🛠️ Problem Types

Type A: Projectile Motion Problems

Setup: "When you encounter a problem involving an object launched into the air, such as a ball thrown at an angle or a projectile fired from a cannon."

Method: "Separate the motion into horizontal and vertical components. Use the kinematic equations for constant velocity in the horizontal direction and constant acceleration in the vertical direction. Solve for quantities such as range, maximum height, and time of flight."

Final Answer

Example: A projectile is launched with an initial velocity of 30.0 m/s at an angle of 40.0° above the horizontal. What is the range of the projectile? Solution:

v_{0x} = 30.0 \cos(40.0^\circ)

v_{0y} = 30.0 \sin(40.0^\circ)

. Time of flight:

t = \frac{2v_{0y}}{g}

. Range:

R = v_{0x} t

Type B: Relative Velocity Problems

Setup: "If presented with a situation involving observers in different reference frames, such as a boat moving in a river or cars moving on a highway."

Method: "Use the relative velocity equation $\mathbf{v}_{AB} = \mathbf{v}_{AC} + \mathbf{v}_{CB}$ , where $\mathbf{v}_{AB}$ is the velocity of object A relative to frame B, $\mathbf{v}_{AC}$ is the velocity of object A relative to frame C, and $\mathbf{v}_{CB}$ is the velocity of object C relative to frame B. Pay attention to the direction of the velocities."

Final Answer

Example: A boat is traveling east across a river at 4.0 m/s relative to the water. The river is flowing south at 3.0 m/s. What is the velocity of the boat relative to the shore? Solution:

\mathbf{v}_{BS} = \mathbf{v}_{BW} + \mathbf{v}_{WS}

\mathbf{v}_{BW} = 4.0 \mathbf{i}

\mathbf{v}_{WS} = -3.0 \mathbf{j}

\mathbf{v}_{BS} = 4.0 \mathbf{i} - 3.0 \mathbf{j}

. Magnitude:

\sqrt{4.0^2 + (-3.0)^2} = 5.0

m/s.

🧮 Solved Example

Problem: A ball is thrown with an initial velocity of 15 m/s at an angle of 30 degrees above the horizontal from the top of a 20 m high building. (a) What is the time it takes for the ball to hit the ground? (b) What is the range of the ball?

Given: $v_0 = 15 \, \text{m/s}$ , $\theta = 30^\circ$ , $h = 20 \, \text{m}$ , $g = 9.8 \, \text{m/s}^2$

Steps:

(a) Find initial velocities: $v_{0x} = v_0 \cos \theta$ , $v_{0y} = v_0 \sin \theta$ .
(b) Vertical motion: Use $y = y_0 + v_{0y}t - \frac{1}{2}gt^2$ to solve for $t$ . Here, $y = 0$ , $y_0 = 20 \, \text{m}$ .
(c) Range: Use $x = v_{0x}t$ to find the range.

"

✅
Answer: (a) $v_{0x} = 15 \cos 30^\circ \approx 13 \, \text{m/s}$ , $v_{0y} = 15 \sin 30^\circ = 7.5 \, \text{m/s}$ . $0 = 20 + 7.5t - 4.9t^2$ . Solving for $t$ (using quadratic formula) gives $t \approx 3.0 \, \text{s}$ (positive root). (b) $x = 13 \cdot 3.0 \approx 39 \, \text{m}$

⚠️ Common Mistakes

❌ Mistake 1: Incorrectly resolving vectors into components.

✅ How to avoid: Draw a clear diagram and use trigonometric functions correctly. Ensure the angle is measured from the correct axis.

❌ Mistake 2: Not separating horizontal and vertical motion in projectile motion problems.

✅ How to avoid: Analyze the horizontal and vertical components separately, using the appropriate kinematic equations for each.

💡 Study Tip

Practice vector addition and subtraction using both graphical and component methods. This will help you develop a strong understanding of vector principles.

📖 Chapter 3: Measurement, Estimating, and Units

What this chapter covers: This chapter focuses on the principles of measurement, uncertainty, significant figures, unit systems, and dimensional analysis. It emphasizes the importance of accurate measurements, proper use of units, and estimation techniques in physics.

🔑 Essential Concepts & Formulas

Concept/Formula	Definition/Equation	When to Use	Quick Check
Percent Uncertainty	$\frac{\Delta A}{A} \times 100\%$	Quantifying the relative uncertainty in a measurement	Ensure $\Delta A$ and $A$ have the same units
Significant Figures: Multiplication/Division	Result has the same number of significant figures as the least precise factor	Determining the number of significant figures in a calculated result	Round the final answer appropriately
Significant Figures: Addition/Subtraction	Result has the same number of decimal places as the least precise term	Determining the number of significant figures in a calculated result	Round the final answer appropriately
Unit Conversion	Multiplying by conversion factors	Converting between different units	Ensure units cancel correctly
Dimensional Analysis	Checking the consistency of equations by verifying that the dimensions on both sides are the same	Verifying the correctness of a formula	Ensure all terms have the same dimensions

🛠️ Problem Types

Type A: Uncertainty Calculations

Setup: "When you encounter a measurement with an associated uncertainty and need to determine the relative or percent uncertainty."

Method: "Calculate the absolute uncertainty ( $\Delta A$ ) and divide it by the measured value ( $A$ ). Multiply by 100% to express the uncertainty as a percentage."

Final Answer

Example: A length is measured as 2.50 ± 0.05 cm. What is the percent uncertainty? Solution:

\frac{0.05}{2.50} \times 100\% = 2\%

Type B: Unit Conversions

Setup: "If presented with a quantity in one unit system and need to convert it to another unit system."

Method: "Use appropriate conversion factors to multiply the quantity by a ratio of units that equals 1. Ensure that the original units cancel out, leaving the desired units."

Final Answer

Example: Convert 15 miles per hour to meters per second. Solution:

15 \frac{\text{miles}}{\text{hour}} \times \frac{1609 \text{ m}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ s}} \approx 6.7 \text{ m/s}

🧮 Solved Example

Problem: A rectangular plate has a length of 15.5 cm ± 0.2 cm and a width of 8.4 cm ± 0.1 cm. Calculate the area of the plate and its uncertainty.

Given: Length $l = 15.5 \, \text{cm} \pm 0.2 \, \text{cm}$ , Width $w = 8.4 \, \text{cm} \pm 0.1 \, \text{cm}$

Steps:

Calculate the area: $A = l \times w$ .
Calculate the percent uncertainty in length and width.
Add the percent uncertainties to find the percent uncertainty in the area.
Calculate the absolute uncertainty in the area.

"

✅
Answer: $A = 15.5 \times 8.4 = 130.2 \, \text{cm}^2$ Percent uncertainty in length: $\frac{0.2}{15.5} \times 100\% \approx 1.3\%$ Percent uncertainty in width: $\frac{0.1}{8.4} \times 100\% \approx 1.2\%$ Percent uncertainty in area: $1.3\% + 1.2\% = 2.5\%$ Absolute uncertainty in area: $0.025 \times 130.2 \approx 3.3 \, \text{cm}^2$ Area: $130 \pm 3 \, \text{cm}^2$ (rounded to significant figures)

⚠️ Common Mistakes

❌ Mistake 1: Incorrectly applying significant figure rules.

✅ How to avoid: Follow the rules for multiplication/division and addition/subtraction separately. Remember that the least precise measurement determines the precision of the result.

❌ Mistake 2: Forgetting to propagate uncertainties correctly in calculations.

✅ How to avoid: Use the rules for adding percent uncertainties in multiplication/division and adding absolute uncertainties in addition/subtraction.

💡 Study Tip

Practice unit conversions and dimensional analysis regularly. This will help you develop a strong understanding of units and their relationships.

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Physics I: Kinematics in 1D, 2D, & 3D

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Physics I: Kinematics in 1D, 2D, & 3D

🎓 Physics I - Study Guide

📋 Course Structure

📖 Chapter 1: Describing Motion: Kinematics in One Dimension

🔑 Essential Concepts & Formulas

🛠️ Problem Types

🧮 Solved Example

⚠️ Common Mistakes

💡 Study Tip

📖 Chapter 2: Kinematics in Two or Three Dimensions; Vectors

🔑 Essential Concepts & Formulas

🛠️ Problem Types

🧮 Solved Example

⚠️ Common Mistakes

💡 Study Tip

📖 Chapter 3: Measurement, Estimating, and Units

🔑 Essential Concepts & Formulas

🛠️ Problem Types

🧮 Solved Example

⚠️ Common Mistakes

💡 Study Tip

2 more sections