Introduction to Motion

Physics helps us understand and describe how objects move through space. When we study motion, we're looking at one of the most fundamental aspects of the physical world.

Motion describes how an object's position changes over time. It's something you observe every day - from a ball you throw to vehicles on the road.

In General Physics 1, we'll start by building a foundation for understanding motion, which will help us analyze more complex physical phenomena later.

Motion in Two or Three Dimensions

When objects move in space, we need a way to track their positions accurately. We use a coordinate system with x, y, and z axes to pinpoint exactly where something is.

A position vector (r⃗) tells us exactly where a particle is located using components in each direction: r⃗ = xi⃗ + yj⃗ + zk⃗. Think of this as directions to find a point in 3D space.

When an object moves from one position to another, we calculate the displacement vector (Δr⃗) by subtracting the initial position from the final position: Δr⃗ = r⃗f - r⃗i. For example, moving from r⃗1 = -3i⃗ + 2j⃗ + 5k⃗ to r⃗2 = 9i⃗ + 2j⃗ + 8k⃗ gives a displacement of 12i⃗ + 0j⃗ + 3k⃗.

💡 Remember that displacement only cares about the starting and ending positions - it doesn't matter what path the object took to get there!

Velocity in Multiple Dimensions

Average velocity measures how quickly displacement occurs over a time interval: v⃗av = Δr⃗/Δt. It points in the same direction as the displacement.

Instantaneous velocity tells us the object's velocity at a specific moment. It's found by taking the limit as the time interval approaches zero: v⃗ = dr⃗/dt. This velocity vector is always tangent to the path.

The components of instantaneous velocity come from the rate of change of position in each direction: vx = dx/dt, vy = dy/dt, and vz = dz/dt. These components tell us how quickly the object moves in each direction.

Think of instantaneous velocity as a snapshot of the object's motion at a specific moment, while average velocity summarizes the entire journey between two points.

Acceleration in Multiple Dimensions

Acceleration describes how velocity changes over time. The average acceleration over a time interval is: a⃗av = Δv⃗/Δt, showing how much the velocity vector changed.

Instantaneous acceleration gives us the acceleration at an exact moment: a⃗ = dv⃗/dt. This tells us both how the speed is changing and how the direction is changing.

The components of acceleration can be written in two equivalent ways:

• As the rate of change of velocity: ax = dvx/dt, ay = dvy/dt, az = dvz/dt
• As the second derivative of position: ax = d²x/dt², ay = d²y/dt², az = d²z/dt²

🔑 Unlike in one dimension, acceleration in multiple dimensions can change an object's speed, direction, or both simultaneously!

Analyzing Motion with Position Functions

When we know the position functions x(t) and y(t) for an object, we can calculate everything about its motion. For the rabbit problem with x = -0.31t² + 7.2t + 28 and y = 0.22t² – 9.1t + 30, we follow three steps:

First, find the position vector at any time by plugging in the value of t. At t = 15s, we calculate the exact coordinates and find how far the rabbit is from the origin.

Second, find velocity by taking the first derivative of position functions. The velocity vector v⃗ = vxi⃗ + vyj⃗ tells us how fast the rabbit moves in each direction, with vx = dx/dt and vy = dy/dt.

Third, find acceleration by taking the second derivative of position functions (or the first derivative of velocity). The acceleration vector a⃗ = axi⃗ + ayj⃗ reveals how the rabbit's velocity is changing.

💡 Being able to analyze motion at any specific time gives us complete understanding of the rabbit's journey, even without watching it move!

Projectile Motion Basics

Projectile motion is a special type of two-dimensional motion where an object is launched and then moves under the influence of gravity only. Think of a football pass or a thrown baseball.

The key insight is that projectile motion can be separated into two independent parts:

• Horizontal motion with constant velocity (vx remains unchanged throughout)
• Vertical motion with constant acceleration due to gravity $object accelerates downward at g = 9.8 m/s²$

When an object is launched at an angle θi with initial velocity vi, its initial velocity components are:

• Horizontal: vxi = vi cos(θi)
• Vertical: vyi = vi sin(θi)

As the projectile moves through points A to E on its path, its horizontal velocity never changes, while its vertical velocity continuously decreases, reaches zero at the peak (point C), and then increases downward.

Analyzing Projectile Drop Problems

When solving a problem like the plane dropping a package, we use the separate horizontal and vertical motion equations. The package has an initial horizontal velocity (from the plane) and zero initial vertical velocity.

For horizontal motion: Since there's no acceleration in this direction, the package keeps moving at 40.0 m/s horizontally. The horizontal distance it travels equals this velocity multiplied by the time it takes to hit the ground.

For vertical motion: We use free-fall equations with initial velocity vy = 0 and acceleration g = 9.8 m/s². The time to hit the ground can be found from y = vyi·t + ½gt².

The final velocity when the package hits the ground combines both components: the unchanged horizontal velocity and the vertical velocity that developed during the fall $vy = gt$. This gives us both the speed and direction of impact.

💡 Even though the package was simply "dropped," it follows a parabolic path because of its initial horizontal velocity from the plane!

Projectile Motion: Long Jump Analysis

In the long jump example, we need to find both the maximum height and horizontal distance when someone jumps at 20.0° with a speed of 11.0 m/s.

First, break down the initial velocity into components:

• Horizontal: vx = 11.0 m/s × cos(20.0°) = 10.3 m/s
• Vertical: vy = 11.0 m/s × sin(20.0°) = 3.76 m/s

To find the maximum height, we need the time when vertical velocity becomes zero $vy = 0$. Using vy = vyi - gt, we find when the jumper reaches the peak. Then we use y = vyit - ½gt² to calculate the maximum height.

For the horizontal distance (range), we need the total time in the air. Since landing happens at the same level as takeoff, we can find when y = 0. The horizontal distance is simply x = vx × total time.

This analysis helps us understand why long jumpers need both speed and proper takeoff angle to maximize their performance.

Projectile Motion with Different Heights

When a projectile is thrown from an elevated position, like the stone thrown from a building, the analysis gets more interesting but follows the same principles.

To find the time to reach the ground, we use the vertical motion equation with a negative initial height (measured from ground level): y = -45.0 m + vy₀t - ½gt². Since vy₀ = 20.0 m/s × sin(30.0°) = 10.0 m/s, we can solve for t.

The range (horizontal distance) is calculated using the horizontal velocity and the time in the air: x = vx₀t, where vx₀ = 20.0 m/s × cos(30.0°) = 17.3 m/s.

For the final velocity, we combine two components:

• Horizontal: vx remains constant at 17.3 m/s
• Vertical: vy = vy₀ - gt (will be negative, indicating downward motion)

🚀 The stone travels farther than if thrown from ground level because it has more time in the air due to starting from height!

Practice Problems for Projectile Motion

These problems help you apply projectile motion concepts to real-world scenarios. In each case, break the motion into horizontal and vertical components.

For the beer mug sliding off the counter, we need to work backward from the known horizontal distance and height to find the initial velocity. Since we know how far it landed from the counter (1.40 m) and the height it fell from (0.860 m), we can determine both the speed and direction it had when leaving the counter.

The bullet problem requires finding three key aspects of projectile motion: maximum height $using vertical motion equations when vy = 0$, time of flight (when the bullet returns to its original height), and range (how far horizontally it travels during that time).

The archer problem combines concepts from previous examples, with the added complexity of starting from height. By analyzing the vertical and horizontal components separately, we can determine how high the arrow goes above ground level, how far it travels horizontally, and its impact velocity.

💡 Remember that projectile motion analysis always works the same way: break into components, apply constant velocity horizontally and free-fall vertically!