Describing Motion: Displacement, Velocity, and Acceleration

Historical Context: The Evolution of Thought on Motion

• Ancient Greeks (Aristotle):

  • Primarily focused on the question of 'why' things happen, rather than a quantitative description of motion.

  • Conceptualized their world as composed of four basic elements: earth, water, air, and fire, each with a 'natural place' in the cosmos.

  • Believed objects moved towards their 'natural locations' (e.g., smoke, a mixture of air and fire, would naturally rise to the sky; rocks, primarily earth, would fall to the ground).

  • This qualitative mindset, emphasizing 'why' over 'how' or 'how fast', dominated scientific thought for approximately 1800 years, hindering quantitative analysis and experimental verification.

• The Renaissance (Galileo Galilei):

  • Galileo, a revolutionary thinker, fundamentally shifted the scientific inquiry from 'why' things move to 'how' things move and 'how fast'.

  • He employed an empirical approach, conducting experiments (e.g., with inclined planes and falling objects) to observe and measure motion, focusing on mathematical descriptions rather than teleological explanations.

  • This seemingly simple change in inquiry, from qualitative reasoning to quantitative experimentation, sparked a scientific revolution and laid the groundwork for modern physics.

• Significance: The questions we ask profoundly shape our understanding, scientific progress, and how we describe physical phenomena like motion. Galileo's shift allowed for the development of precise mathematical models.

The Concept of Displacement: 'How Far Away is It?'

• Initial Question - 'How far has it gone?':

  • Illustrated with a drone example: flying at 30 \text{ miles/hour} for 2 \text{ minutes}, it covers 1 \text{ mile}.

  • This quantity, often referred to as distance, is the total path length traveled. However, it doesn't necessarily mean the drone is 1 \text{ mile} away from its starting point; it could have flown in circles or reversed direction.

  • This question is often poorly formed for precisely retrieving an object or determining its final position relative to its start.

• Improved Question - 'How far away is it, and in what direction?':

  • This focuses on the change in position from a starting point to an ending point, making it a vector quantity with both magnitude and direction.

  • This is the concept of displacement.

• Definition of Displacement (\Delta \vec{s} or \vec{s}):

  • The change in an object's placement, defined as the shortest straight-line distance from the initial position (\vec{s}i) to the final position (\vec{s}f).

  • Mathematically, \Delta \vec{s} = \vec{s}f - \vec{s}i.

  • It only depends on the initial and final positions, not on the actual path taken between them.

  • If an object starts and ends at the same place, its net displacement is zero, regardless of the distance traveled or the complexity of the path.

  • Displacement is a vector, meaning it has both magnitude (how far) and direction (where).

• Components and Trigonometry:

  • Displacement can be broken down into perpendicular components along chosen axes, such as a change in x-position (\Delta x) and a change in y-position (\Delta y).

  • These components form a right-angled triangle where the magnitude of the total displacement (|\vec{s}| or \Delta s) is the hypotenuse.

  • Trigonometric Relationships (where \theta is the angle concerning the x-axis, for example):

    • \Delta y = |\vec{s}| \sin(\theta)

    • \Delta x = |\vec{s}| \cos(\theta)

  • Pythagorean Theorem (for the magnitude of displacement):

    • |\vec{s}|^2 = (\Delta x)^2 + (\Delta y)^2

    • Or |\vec{s}| = \sqrt{(\Delta x)^2 + (\Delta y)^2}

Defining Velocity: 'How Fast Did Displacement Occur?'

• Definition (\vec{v}):

  • Velocity is a vector quantity that describes the rate at which displacement occurs, including both the magnitude (speed) and direction of motion.

  • Mathematically expressed as the change in displacement divided by the change in time.

  • \vec{v} = \frac{\Delta \vec{s}}{\Delta t} (where \Delta \vec{s} is displacement and \Delta t is the time interval).

  • The SI unit for velocity is meters per second (\text{m/s}).

  • Note: It is crucial to distinguish velocity (\vec{v}) from speed. Speed is the magnitude of velocity, a scalar quantity that only indicates 'how fast' an object is moving (e.g., 50 \text{ km/h}). Velocity tells us 'how fast' and 'in what direction' (e.g., 50 \text{ km/h} North).

• Components of Velocity:

  • Just like displacement, velocity can be resolved into perpendicular components, representing the rate of change of position along each axis:

    • v_x = \frac{\Delta x}{\Delta t}

    • v_y = \frac{\Delta y}{\Delta t}

  • These components maintain their trigonometric relationships with the total velocity magnitude (|\vec{v}|), similar to displacement.

• Example Problem: Running Back and Fumbled Football

  • Scenario: A football is 3 \text{ meters} from the sideline, heading towards it at an angle of 30^\circ from straight downfield (meaning 30^\circ relative to the y-axis, if y is downfield and x is across the field), at a speed of 4 \text{ m/s}.

  • Goal: Calculate the time until it goes out of bounds (sideline).

  • Strategy: To determine when the ball crosses the sideline, we only need to consider the component of velocity perpendicular to the sideline (typically the y-component in our defined coordinate system, if the sideline is parallel to the x-axis). The x-component of velocity does not contribute to crossing the sideline.

  • Calculation:

    • The angle given is 30^\circ from straight downfield. If 'straight downfield' is our y-axis, the velocity component perpendicular to the sideline (towards the sideline) would be v_y.

    • v_y = v \sin(\theta) (where \theta is the angle with respect to the downfield direction)

    • v_y = (4 \text{ m/s}) \sin(30^\circ)

    • v_y = (4 \text{ m/s}) (0.5) = 2 \text{ m/s}

    • Using the definition of velocity: v_y = \frac{\Delta y}{\Delta t}.

    • Rearranging for time: \Delta t = \frac{\Delta y}{v_y}

    • \Delta t = \frac{3 \text{ m}}{2 \text{ m/s}} = 1.5 \text{ seconds}

    • So, the football will go out of bounds in 1.5 seconds.

Average vs. Instantaneous Velocity

• Graphical Representation:

  • On a position versus time graph (s-t graph), velocity is represented by the slope of the line.

• Average Velocity:

  • The average velocity (v_{\text{avg}}) is the slope of the secant line connecting two points on the position-time graph over a chosen time interval (\Delta t).

  • v{\text{avg}} = \frac{\text{total displacement}}{\text{total time}} = \frac{\Delta \vec{s}}{\Delta t} = \frac{\vec{s}f - \vec{s}i}{tf - t_i}

  • Example: For a position change of 2.75 \text{ meters} over 10 \text{ seconds}, the average velocity is 0.275 \text{ m/s}. This value represents the constant velocity that would be required to achieve the same displacement in the same time.

  • Can be calculated over any specified interval (e.g., 6 to 10 \text{ seconds}), providing a measure of the overall rate of change during that period.

• Instantaneous Velocity:

  • The instantaneous velocity (\vec{v}) is the velocity at a precise moment in time, representing how fast and in what direction an object is moving at that exact instant.

  • Graphically, it's the slope of the tangent line to the position-time curve at that specific point.

  • Conceptually, it's found by taking infinitesimally small time intervals (\Delta t approaching zero) around the point of interest. In calculus, this is the derivative of position with respect to time (v = \frac{ds}{dt}).

• Conceptual Challenge: Instantaneous vs. Average Velocity

  • Given a curved position-time graph (concave up, indicating increasing velocity), the instantaneous velocity at 4 \text{ seconds} is greater than the average velocity over the first 4 \text{ seconds}.

  • Reasoning: A concave-up curve signifies that the slope is continually increasing. The tangent line at 4 \text{ seconds} (representing instantaneous velocity) will be steeper than the secant line connecting the origin (0,0) to the point (4, s(4)) (representing average velocity over the first 4 seconds). This indicates that the object is speeding up, and its velocity at 4 seconds is higher than its average velocity over the preceding interval.

Introducing Acceleration: 'How Fast is 'How Fast' Changing?'

• The Need for Acceleration:

  • Illustrated by Elliot vs. Darlene: Elliot claims he is faster than a train because he can reach the end of a platform first, focusing on who can change speed faster (acceleration). Darlene focuses on comparing constant speeds (velocity).

  • This highlights that describing motion requires understanding not just how fast an object is moving (velocity), but also how its velocity itself changes over time.

• Definition (\vec{a}):

  • Acceleration is a vector quantity that describes the change in velocity over a given time interval.

  • It can involve a change in speed, a change in direction, or both.

  • \vec{a} = \frac{\text{change in velocity}}{\text{change in time}} = \frac{\Delta \vec{v}}{\Delta t} = \frac{\vec{v}f - \vec{v}i}{tf - ti}

• Units of Acceleration:

  • Always contain two units of time in the denominator, reflecting a 'rate of a rate'.

  • Example: For a car accelerating from 0 to 60 \text{ miles/hour} in 5 \text{ seconds}, the units are 'miles per hour per second'.

  • The standard metric (SI) unit is 'meters per second per second', or more commonly, \text{m/s}^2. This means velocity changes by so many meters per second, every second.

• Applicability of Previous Concepts:

  • The ideas of average vs. instantaneous acceleration and trigonometric components (for acceleration in 2D or 3D) apply directly, mirroring how they were used for velocity.

  • Average acceleration is the slope of the secant line on a velocity-time graph, while instantaneous acceleration is the slope of the tangent line on a velocity-time graph (a = \frac{dv}{dt}).

• Acceleration of Falling Objects:

  • Falling objects in free fall (neglecting air resistance) accelerate; their velocity continuously changes due to gravity.

  • The acceleration due to gravity on Earth is constant and approximately g = 9.80 \text{ m/s}^2 (downward).

  • This means that for every second an object is in free fall, its downward velocity increases by 9.80 \text{ m/s}. (e.g., if dropped from rest, after 1 second it's moving at 9.8 \text{ m/s} down, after 2 seconds at 19.6 \text{ m/s} down, etc.).

Case Study: Softball Toss and Key Distinctions

• Scenario: A softball is tossed upward above a 10 \text{ ft} fence, reaches its maximum height, and then falls back down, eventually passing the height of the fence again.

  • Point 1: Emerging above the fence (moving upward).

  • Point 2: At the maximum height.

  • Point 3: Falling back down, level with point 1.

• Analysis of Velocity (\vec{v}) and Acceleration (\vec{a}):

  • Assume upward direction is positive, so the acceleration due to gravity constantly acts downward and is negative (-g).

• Explanation:

  • Velocity: At point 1, the ball is thrown upward, so its velocity is positive (+v0). At point 2 (the peak of its trajectory), it momentarily stops before changing direction; thus, its instantaneous velocity is zero. At point 3, the ball is falling downward, so its velocity is negative (-v0). Due to symmetry and neglecting air resistance, the magnitude of its downward velocity at point 3 will be the same as its upward velocity at point 1 (|v_0|).

  • Acceleration: Gravity acts consistently downward throughout the ball's flight, causing an acceleration of -g at all three points (and indeed, at every point in its trajectory, assuming constant gravitational field and negligible air resistance).

    • At point 1 (moving up), the negative acceleration -g means the ball is slowing down.

    • At point 3 (moving down), the negative acceleration -g means the ball is speeding up in the negative (downward) direction.

    • Crucial Insight (Point 2): Even though the velocity is zero at the peak, the acceleration is still -g. The velocity is zero for an instant, but it is changing from positive to negative at that very moment, indicating a non-zero, constant downward acceleration. This demonstrates a key concept in kinematics: an object can have zero velocity but non-zero acceleration.

    • Conversely, an object moving at a constant velocity (zero change in velocity) has zero acceleration.

Velocity vs. Time Graph Analysis

• Graph Description: A velocity-time graph showing a straight line with a constant negative slope. The velocity starts positive, crosses the time axis, and becomes negative.

• Conceptual Questions:

  • A: The acceleration is constant. True. The slope of a velocity-time graph represents acceleration. Since the graph is a straight line, its slope is constant, which means the acceleration is constant throughout the motion.

  • B: The object passes through the position x = 0. Cannot be determined. A velocity-time graph describes how velocity changes over time, but it does not provide information about the object's initial position (x_0) or absolute position. We know its velocity changes sign, meaning it could pass through x=0, but we don't have enough information to confirm its exact position without an initial condition.

  • C: The object has zero velocity at some instant. True. The graph clearly crosses the time axis (v=0) at a specific point in time. This indicates the moment when the object momentarily stops before reversing direction.

  • D: The object is always moving in the same direction. False. The velocity starts as positive, decreases to zero, and then becomes negative. A change in the sign of velocity (from positive to negative or vice-versa) signifies a change in the direction of motion.

• Connection to Softball Problem: This graph is a perfect representation of the softball's vertical motion after being tossed upward: starting with a high positive velocity, slowing down due to negative acceleration until its velocity becomes zero at the peak, and then speeding up in the negative (downward) direction, all under constant negative acceleration (-g).

Summary of New Vocabulary and Equations

• Displacement (\Delta \vec{s} or \vec{s}): A vector quantity representing the change in an object's placement from an initial to a final position, independent of the path taken. Distinguished from distance, which is the total path length traveled.

• Velocity (\vec{v}): A vector quantity representing the rate of change of displacement. \vec{v} = \frac{\Delta \vec{s}}{\Delta t}. Distinguished from speed, which is the scalar magnitude of velocity.

• Acceleration (\vec{a}): A vector quantity representing the rate of change of velocity. \vec{a} = \frac{\Delta \vec{v}}{\Delta t}. It describes how quickly velocity changes (either in magnitude or direction).

• Key Distinction: Velocity and acceleration are distinct concepts; one can be zero while the other is not (e.g., a ball at its maximum height has zero instantaneous velocity but constant non-zero acceleration due to gravity).

• Importance of Questions: The precise framing of questions (e.g., 'how far away' vs. 'how far traveled') is fundamental to accurately describing motion and advancing scientific understanding, moving from qualitative observations to quantitative models.