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Position, velocity, acceleration and time

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Intros
Lessons
  2. Displacement Vs. Distance
  3. Velocity Vs. Speed
  4. Acceleration
  5. Position Vs. Time Graph
  6. Velocity Vs. Time Graph
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Examples
Lessons
  1. A taxi driver travels across the city through the map as shown below. The total time taken to cover the whole journey is 30min.
    Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs
    1. What is the displacement of the driver from the start point to point A?
    2. What is the total distance covered form start point to point A?
    3. What is the displacement of the driver from the start point to the end point?
    4. What is the total distance covered by the driver from the start point to the end point?
    5. If the driver goes back to the start point, what would be the total distance and displacement?
    6. Calculate the average speed and average velocity of the journey from the start point to the end point.
    7. What would be the average speed and average velocity if the driver goes back to the start point?
  2. A person walks 20m west, then 50m north and stops. After 5 minutes of resting, he resumes his walk towards the east for 20m. The entire trip took 10 minutes.
    1. Calculate the total distance and displacement in 10 minutes.
    2. Calculate the average speed and average velocity in 10 minutes
  3. The following graph represents position Vs. time graph of a car as a function of time.
    Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs
    1. What is the car average velocity from 20s to 30s?
    2. What is the car average velocity from 30s to 40s?
    3. What is the car average velocity from 10s to 20s?
    4. What is the instantaneous velocity at 40s?
    5. What is the displacement of the car after 50s?
    6. What is the total distance traveled by the car in 50s?
  4. The following graph represents position Vs. time graph for an object as a function of time.
    Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs
    1. What is the car average velocity from 5s to 15s?
    2. What is the instantaneous velocity at 10s?
    3. What is the car average velocity from 0s to 20s?
  5. The following graph represent the Velocity Vs. Time graph of an object moving with constant acceleration.
    Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs
    1. Find the acceleration of the object during the following intervals.
      [0s, 40s], [40s, 70s], [70s ,90s], [90s, 100s]
    2. What is the total displacement of the object?
    3. What is the distance covered during 100s?
    4. Describe the motion of the object.
  6. The following represents the motion of a particle with a variable acceleration.
    Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs
    1. What is the car average acceleration from 5s to 15s?
    2. What is the instantaneous acceleration at 10s?
    3. What is the car average acceleration from 0s to 20s?
Topic Notes
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Introduction

Welcome to our exploration of fundamental physics concepts: position, velocity, acceleration, and time. These interconnected ideas form the backbone of motion analysis in physics. Position tells us where an object is, velocity describes how quickly it's moving and in what direction, while acceleration reveals how its velocity changes over time. Time, of course, is the crucial dimension through which we measure these changes. Our introduction video serves as an excellent starting point, offering clear explanations and visual aids to help you grasp these concepts. It's designed to make these sometimes challenging ideas more accessible and relatable. As we dive deeper into each concept, you'll see how they work together to describe the world around us. Understanding these basics is essential for tackling more complex physics problems later on. So, let's begin this exciting journey into the world of motion!

Position and Distance: Understanding the Basics

Defining Position

Position is a fundamental concept in physics that describes the location of an object in space. It is typically measured relative to a reference point or origin. For example, you might describe the position of a book as "2 meters to the right of the desk." Position is crucial for understanding an object's placement in a coordinate system.

Understanding Distance

Distance, on the other hand, refers to the total length of the path traveled by an object. It is a scalar quantity, meaning it only has magnitude and no direction. For instance, if you walk from your home to a store and back, the total distance you've traveled is the sum of the lengths of both journeys, regardless of your final position.

The Difference Between Position and Distance

The key difference between position and distance lies in what they measure. Position tells us where an object is located, while distance tells us how far it has traveled. Position can change even if distance remains constant, and vice versa. For example, if you run around a circular track, your position changes continuously, but you may end up at the same position you started from. The distance you've run, however, would be the length of the entire track.

Introducing Displacement

Displacement is a vector quantity that describes the shortest distance between two points, taking into account both magnitude and direction. Unlike distance, which measures the total path length, displacement only considers the straight-line distance from the starting point to the endpoint. This makes displacement a crucial concept in physics, especially when dealing with motion and forces.

The Vector Nature of Displacement

As a vector quantity, displacement has both magnitude and direction. This means that to fully describe displacement, you need to specify not just how far an object has moved, but also in which direction. For example, a displacement of "5 meters north" provides complete information about the change in position. The vector nature of displacement makes it particularly useful in calculations involving velocity, acceleration, and other vector quantities in physics.

Examples Illustrating Position, Distance, and Displacement

Let's consider a few examples to clarify these concepts:

  1. Circular Motion: Imagine a runner completing one lap around a 400-meter track. Their final position is the same as their starting position, so the displacement is zero. However, the total distance traveled is 400 meters.
  2. City Walking: If you walk 3 blocks east, then 4 blocks north, your displacement would be the straight-line distance from your starting point to your endpoint (5 blocks in a northeastern direction, calculated using the Pythagorean theorem). The total distance walked, however, would be 7 blocks (3 + 4).
  3. Mountain Hike: When hiking up a winding trail to the summit of a mountain, your total distance traveled might be several kilometers. However, your displacement would be much less the shortest distance between two points from the base to the summit.

Practical Applications

Understanding the differences between position, distance, and displacement is crucial in various fields:

  • Navigation: GPS systems use position to locate you and calculate the shortest route (displacement) to your destination, even though the actual distance traveled may be longer due to roads and obstacles.
  • Physics: In calculations involving motion, displacement is often more useful than distance because it accounts for direction, allowing for more precise predictions of an object's future position.
  • Engineering: When designing transportation systems or planning construction projects, engineers must consider both the total distance of a route and the magnitude and direction in displacement between points to optimize efficiency.

Conclusion

Position, distance, and displacement are interconnected concepts that play vital roles in understanding motion and spatial relationships. While position describes location, distance measures the total path length, and displacement accounts for the shortest path between two points, including direction. Recognizing the distinctions between these concepts is essential for accurately analyzing movement.

Velocity and Speed: Exploring Motion

Understanding Velocity and Speed

In the realm of physics, velocity and speed are fundamental concepts that describe how objects move. While often used interchangeably in everyday language, these terms have distinct scientific meanings. Understanding the difference between velocity and speed is crucial for accurately analyzing motion in various scenarios.

Defining Speed

Speed is a scalar quantity that measures how fast an object is moving, regardless of its direction. It is calculated by dividing the total distance traveled by the time taken. For example, if a car travels 100 kilometers in 2 hours, its speed would be 50 kilometers per hour (km/h). The formula for speed is:

Speed = Distance / Time

Defining Velocity

Velocity, on the other hand, is a vector quantity that describes both the speed and direction of an object's motion. It is expressed as the rate of change of position with respect to time. Velocity takes into account not just how fast an object is moving, but also in which direction it is traveling. The formula for velocity is similar to speed, but includes direction:

Velocity = Displacement / Time

Key Differences Between Velocity and Speed

The primary distinction between velocity and speed lies in their nature as vector and scalar quantities, respectively. Speed tells us only how fast something is moving, while velocity provides information about both speed and direction. For instance, a car traveling at 60 km/h north has a different velocity from one traveling at 60 km/h south, even though their speeds are identical.

Average Velocity

Average velocity is a measure of the overall rate of change of position over a given time interval. It is calculated by dividing the total displacement (change in position) by the total time taken. The formula for average velocity is:

Average Velocity = (Final Position - Initial Position) / (Final Time - Initial Time)

Example of Average Velocity Calculation

Consider a runner who starts at the 0-meter mark of a track, runs 400 meters clockwise, and finishes at the 100-meter mark after 2 minutes. The average velocity would be:

Average Velocity = (100 m - 0 m) / 2 minutes = 50 m/min in the clockwise direction

Note that even though the runner covered 400 meters, the displacement (change in position) is only 100 meters.

Instantaneous Velocity

While average velocity gives us an overall picture of motion, instantaneous velocity describes the velocity at a specific moment in time. It is the velocity of an object at a particular instant or a point on its path. Mathematically, instantaneous velocity is the limit of the average velocity as the time interval approaches zero.

Calculating Instantaneous Velocity

Determining instantaneous velocity often involves calculus, as it requires finding the derivative of the position function with respect to time. In simpler terms, it's the slope of the tangent line to the position-time graph at a specific point. For example, if a car's position is given by the function x(t) = 3t² + 2t, where x is in meters and t is in seconds, the instantaneous velocity at any time t would be:

v(t) = dx/dt = 6t + 2 meters per second

Practical Applications of Velocity and Speed

Understanding velocity and speed is crucial in various fields, including physics, engineering, and sports science. In transportation, these concepts help in designing efficient routes and calculating fuel consumption. In sports, coaches use velocity measurements to improve athletes' performance. Weather forecasters rely on velocity calculations to predict wind patterns and storm movements.

Conclusion

Velocity and speed are essential concepts in understanding motion. While speed simply tells us how fast an object is moving, velocity provides a more complete picture by

Acceleration: Understanding Changes in Velocity

What is Acceleration?

Acceleration is a fundamental concept in physics that describes how quickly an object's velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction. Understanding acceleration is crucial for analyzing motion and predicting the behavior of objects in various scenarios.

The Significance of Acceleration in Motion

Acceleration plays a vital role in our daily lives and in numerous scientific and engineering applications. From the smooth acceleration of a car to the rapid deceleration of a spacecraft entering the atmosphere, acceleration is omnipresent. It helps us understand how forces affect motion and is essential in fields such as automotive design, sports science, and space exploration.

Average Acceleration

Average acceleration is the rate of change in velocity over a specific time interval. It provides a general idea of how an object's speed and direction change during a given period. The formula for average acceleration is:

a = (vf - vi) / t

Where 'a' is the average acceleration, 'vf' is the final velocity, 'vi' is the initial velocity, and 't' is the time interval. For example, if a car increases its speed from 0 to 60 km/h in 10 seconds, its average acceleration would be 6 km/h/s or 1.67 m/s2.

Instantaneous Acceleration

While average acceleration gives us a broad picture, instantaneous acceleration provides a more precise measure of how velocity changes at a specific moment in time. It is the limit of the average acceleration as the time interval approaches zero. Mathematically, it is expressed as the derivative of velocity with respect to time:

a = dv/dt

Instantaneous acceleration is particularly useful when dealing with non-uniform motion, where the rate of change in velocity varies continuously.

The Relationship Between Velocity and Acceleration

Velocity and acceleration are closely related but distinct concepts. While velocity describes how position changes over time, acceleration describes how velocity changes over time. This relationship can be visualized in several ways:

  • Constant velocity means zero acceleration
  • Positive acceleration indicates increasing velocity
  • Negative acceleration (deceleration) indicates decreasing velocity
  • Acceleration can change the direction of velocity, not just its magnitude

Examples and Calculations

Let's consider a few examples to illustrate the concepts of acceleration:

  1. A car accelerates from 0 to 100 km/h in 8 seconds. Calculate its average acceleration.
    Solution: a = (100 km/h - 0 km/h) / 8 s = 12.5 km/h/s or 3.47 m/s2
  2. A ball is thrown vertically upward with an initial velocity of 20 m/s. Assuming the acceleration due to gravity is -9.8 m/s2, how long will it take for the ball to reach its maximum height?
    Solution: At the highest point, final velocity = 0 m/s
    Using v = u + at, we get: 0 = 20 + (-9.8)t
    Solving for t: t = 20 / 9.8 = 2.04 seconds

Practical Applications of Acceleration

Understanding acceleration is crucial in various real-world scenarios:

  • Vehicle Design: Engineers use acceleration principles to design safer, more efficient vehicles.
  • Sports Performance: Athletes and coaches analyze acceleration to improve performance in spr

    Graphical Analysis: Position, Velocity, and Acceleration

    Understanding Motion Graphs

    Graphical analysis is a powerful tool in physics for understanding the relationships between position, velocity, and acceleration. These graphs provide visual representations of an object's motion, allowing us to derive important information about its behavior over time. Let's explore how to interpret and analyze position vs. time, velocity vs. time, and acceleration vs. time graphs.

    Position vs. Time Graphs

    A position vs. time graph shows how an object's position changes over time. The slope of this graph represents the object's velocity. Here's how to interpret it:

    • A horizontal line indicates the object is stationary.
    • A positively sloped line shows the object moving away from the origin.
    • A negatively sloped line indicates movement towards the origin.
    • A curved line suggests changing velocity (acceleration or deceleration).

    To calculate displacement, find the difference between the final and initial positions on the graph. The steepness of the slope indicates the magnitude of velocity.

    Velocity vs. Time Graphs

    Velocity vs. time graphs display how an object's velocity changes over time. The slope of this graph represents acceleration. Key points to remember:

    • A horizontal line shows constant velocity (zero acceleration).
    • A positively sloped line indicates increasing velocity (positive acceleration).
    • A negatively sloped line suggests decreasing velocity (negative acceleration or deceleration).

    The area under a velocity-time curve represents displacement. To calculate it, find the area between the curve and the time axis. For constant velocity, this is simply the rectangle's area (velocity × time).

    Acceleration vs. Time Graphs

    Acceleration vs. time graphs show how an object's acceleration changes over time. Interpreting these graphs:

    • A horizontal line indicates constant acceleration.
    • A positively sloped line shows increasing acceleration.
    • A negatively sloped line represents decreasing acceleration.

    The area under an acceleration-time curve represents the change in velocity. To find this, calculate the area between the curve and the time axis.

    Deriving Information from Graphs

    Understanding the relationships between position, velocity, and acceleration graphs is crucial for graphical analysis:

    • The slope of a position-time graph gives instantaneous velocity.
    • The slope of a velocity-time graph provides instantaneous acceleration.
    • The area under a velocity-time graph yields displacement.
    • The area under an acceleration-time graph gives the change in velocity.

    Calculating Displacement, Velocity, and Acceleration

    To calculate displacement from a position-time graph, find the change in position between two points. For velocity, calculate the slope of the position-time graph or use the formula: velocity = displacement / time. Acceleration can be found by calculating the slope of the velocity-time graph or using: acceleration = change in velocity / time.

    Position-Velocity and Velocity-Acceleration Relationships

    Understanding the connections between position, velocity, and acceleration is essential for comprehensive graphical analysis:

    • Position-velocity: Velocity is the rate of change of position.
    • Velocity-acceleration: Acceleration is the rate of change of velocity.

    These relationships allow us to predict an object's behavior and understand how changes in one quantity affect the others.

    Practical Applications

    Graphical analysis of position vs velocity vs acceleration has numerous real-world applications, including:

    Relationships Between Position, Velocity, and Acceleration

    Understanding Position, Velocity, and Acceleration

    In physics, the concepts of position, velocity, and acceleration are fundamental to understanding motion. These three quantities are closely related and form the basis for describing how objects move through space and time. The relationship between position to velocity to acceleration is crucial in various fields, from basic mechanics to advanced physics.

    Position: The Starting Point

    Position refers to an object's location in space, typically measured relative to a reference point. It's usually denoted by x in one-dimensional motion or (x, y, z) in three-dimensional space. Position is the most basic descriptor of an object's motion and serves as the foundation for understanding velocity and acceleration.

    Velocity: The Rate of Change of Position

    Velocity is the rate of change of velocity at which an object's position changes with respect to time. It's a vector quantity, meaning it has both magnitude (speed) and direction. The relationship between position and velocity is fundamental in physics. Velocity can be calculated by taking the derivative of position with respect to time:

    v = dx/dt

    Where v is velocity, x is position, and t is time. This equation shows how velocity vs acceleration vs speed are related, with velocity incorporating both the speed and direction of motion.

    Acceleration: The Rate of Change of Velocity

    Acceleration is the rate of change of velocity at which an object's velocity changes over time. Like velocity, it's a vector quantity. The relationship between velocity and acceleration is similar to that between position and velocity. Acceleration can be derived from velocity by taking its derivative with respect to time:

    a = dv/dt

    Where a is acceleration and v is velocity. This equation highlights the key difference in acceleration vs velocity, showing that acceleration describes how quickly velocity is changing.

    From Position to Velocity to Acceleration

    The progression from position to velocity to acceleration demonstrates a clear mathematical relationship. Each quantity is the derivative of the previous one with respect to time:

    • Position (x) Velocity (v = dx/dt)
    • Velocity (v) Acceleration (a = dv/dt)

    This chain of derivatives illustrates how these quantities are interconnected in describing motion.

    Deriving Quantities: From Acceleration to Velocity to Position

    Interestingly, we can also work backwards, deriving position from velocity and velocity from acceleration through integration:

    • Acceleration Velocity: v = a dt
    • Velocity Position: x = v dt

    These relationships allow us to calculate any of these quantities if we know one of them and have the appropriate initial conditions.

    Practical Examples and Calculations

    Let's consider a simple example to illustrate the relationship between position to velocity to acceleration:

    Imagine a car moving in a straight line. Its position is given by the equation x(t) = 2t² + 3t + 1, where x is in meters and t is in seconds.

    1. To find velocity, we differentiate position with respect to time: v(t) = dx/dt = 4t + 3 m/s
    2. To find acceleration, we differentiate velocity with respect to time: a(t) = dv/dt = 4 m/s²

    This example shows how we can derive velocity and acceleration from a given position function, demonstrating the practical application of these relationships.

    Velocity vs Acceleration vs Speed: Clarifying the Differences

    While closely related, velocity, acceleration, and speed have distinct meanings:

    • Speed is a scalar quantity, measuring only

    Conclusion

    In this article, we've explored the fundamental concepts of position, velocity, acceleration, and time in physics. We've seen how these quantities are interconnected and how they can be used to describe and predict motion. Understanding these relationships is crucial for solving problems in kinematics and dynamics. The graphs and equations we've discussed provide powerful tools for analyzing motion in various scenarios. Remember that velocity is the rate of change of position, while acceleration is the rate of change of velocity. Time plays a critical role in all these calculations. We encourage you to further explore these concepts through practice problems and real-world applications. The introduction video mentioned earlier serves as an excellent visual resource to reinforce these ideas. By mastering these foundational principles, you'll be well-equipped to tackle more advanced topics in physics and engineering. Keep practicing and exploring the fascinating world of motion!

Displacement Vs. Distance

Understanding the concepts of displacement and distance is crucial when studying motion in physics. This guide will walk you through the definitions, differences, and calculations of displacement and distance using a step-by-step approach.

Step 1: Definition of Displacement and Distance

Displacement is defined as the change in position of an object. It is a vector quantity, which means it has both magnitude and direction. To calculate displacement, you need to know the initial and final positions of the object. The difference between these positions gives you the displacement.

Distance, on the other hand, is the total length of the path covered by the object. It is a scalar quantity, meaning it only has magnitude and no direction. Distance is the sum of all the lengths of the paths taken by the object.

Step 2: Displacement Calculation Example

Consider an object moving from north to east and then to south. To find the displacement, you need to determine the initial and final positions. The difference between these positions gives you the displacement. For example, if the object moves 80 meters east and then 30 meters west, the displacement is the difference between the initial and final positions, which is 50 meters east.

Step 3: Distance Calculation Example

Using the same example, the distance covered by the object is the sum of all the paths taken. The object moves 80 meters east and then 30 meters west, so the total distance covered is 80 + 30 = 110 meters.

Step 4: Comparing Displacement and Distance

Displacement is a vector quantity and includes both magnitude and direction, while distance is a scalar quantity and only includes magnitude. For instance, in the example above, the displacement is 50 meters east, while the distance is 110 meters.

Step 5: Example with a Rectangular Field

Let's consider an athlete running around a rectangular field. The athlete moves from point A to B to C and then back to A. If AB is 240 centimeters and BC is 120 centimeters, we can calculate the distance and displacement for different paths.

- From A to B: The distance and displacement are both 240 centimeters. - From A to B to C: The distance is 240 + 120 = 360 centimeters. The displacement is calculated using the Pythagorean theorem, resulting in approximately 268.33 centimeters. - From A to B to C to D: The distance is 240 + 120 + 240 = 600 centimeters. The displacement is 120 centimeters. - From A to B to C to D back to A: The distance is 240 + 120 + 240 + 120 = 720 centimeters. The displacement is 0 because the initial and final positions are the same.

Step 6: Summary

In summary, displacement and distance are two different ways to measure the movement of an object. Displacement is a vector quantity that considers the change in position and direction, while distance is a scalar quantity that only considers the total length of the path covered. Understanding these concepts is essential for solving problems related to motion in physics.

FAQs

Here are some frequently asked questions about position, velocity, and acceleration:

1. What is the difference between velocity and acceleration?

Velocity is the rate of change of position with respect to time, indicating how fast an object is moving and in what direction. Acceleration, on the other hand, is the rate of change of velocity with respect to time, showing how quickly an object's velocity is changing. While velocity is measured in units like meters per second (m/s), acceleration is measured in units like meters per second squared (m/s²).

2. How is acceleration related to velocity?

Acceleration is the derivative of velocity with respect to time. In other words, acceleration describes how velocity changes over time. Positive acceleration increases velocity, while negative acceleration (deceleration) decreases velocity. Constant velocity means zero acceleration, as the velocity is not changing.

3. What is the difference between position, velocity, and acceleration?

Position describes an object's location in space. Velocity is the rate of change of position, indicating how quickly the position is changing. Acceleration is the rate of change of velocity, showing how quickly the velocity is changing. These three quantities form a chain of derivatives: velocity is the derivative of position, and acceleration is the derivative of velocity.

4. How do you find acceleration with velocity and position?

To find acceleration from velocity and position, you need to take two derivatives. First, differentiate the position function to get velocity. Then, differentiate the resulting velocity function to get acceleration. If you have discrete data points, you can calculate average acceleration by finding the change in velocity over a given time interval.

5. What is the correct relationship between acceleration, velocity, and position?

The relationship between these quantities can be expressed mathematically as follows:
Velocity (v) = dx/dt (where x is position and t is time)
Acceleration (a) = dv/dt = d²x/dt² (the second derivative of position with respect to time)
This relationship allows us to derive any of these quantities if we know one of them and have the appropriate initial conditions.

Prerequisite Topics

Understanding the fundamental concepts that lay the groundwork for more advanced topics is crucial in mastering physics and mathematics. When delving into the study of "Position, velocity, acceleration and time," it's essential to have a solid grasp of several prerequisite topics. These foundational concepts not only provide the necessary context but also enhance your ability to comprehend and apply more complex principles.

One of the key prerequisite topics is the rate of change. This concept is fundamental to understanding how position changes over time, which is the essence of velocity. The rate of change of position with respect to time gives us velocity, while the rate of change of velocity gives us acceleration. By mastering this concept, students can more easily grasp the relationships between position, velocity, and acceleration.

Another crucial prerequisite is understanding operations on vectors in magnitude and direction form. This topic is particularly important because position, velocity, and acceleration are all vector quantities. They have both magnitude and direction, and understanding how to perform operations on vectors is essential for solving problems involving motion in multiple dimensions. This knowledge allows students to analyze and predict the motion of objects in more complex scenarios.

The relationship between position, velocity, and acceleration is another critical prerequisite topic. Understanding how these quantities are related through derivatives and integrals is fundamental to solving problems involving motion. For instance, knowing that velocity is the derivative of position with respect to time, and acceleration is the derivative of velocity, allows students to calculate one quantity when given information about another.

By thoroughly grasping these prerequisite topics, students will find themselves better equipped to tackle more advanced concepts in kinematics and dynamics. The rate of change provides the foundation for understanding how quantities change over time. Vector operations allow for the analysis of motion in multiple dimensions. And the relationships between position, velocity, and acceleration tie everything together, enabling students to solve complex problems involving motion.

In conclusion, mastering these prerequisite topics is not just about ticking boxes on a curriculum. It's about building a strong foundation that will support your understanding of more advanced concepts in physics and mathematics. By investing time in these fundamental ideas, students will find that more complex topics become more accessible and intuitive, leading to a deeper and more comprehensive understanding of the subject matter.

In this lesson, we will learn:
  • The definition and difference between Displacement and Distance
  • The definition and difference between Speed and Velocity
  • Definition of Acceleration
  • Position Vs. Time graph
  • Velocity Vs. Time graph
  • Acceleration Vs. Time graph


Notes:

Displacement Vs. Distance

Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs


Displacement: it is defined as change in position, ΔX=Xf−Xi \Delta X = X_{f}-X_{i}.
It is a vector quantity, it has both magnitude and direction.

Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs

Distance: the length of the path covered between two points. It is also defined as the magnitude of displacement between two positions.
It is a scalar quantity, it has only magnitude.

Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs


Example:

An athlete is running around a rectangular field as shown below with a length of 240 cm and width of 120cm.

Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs


Calculate distance and displacement covered for the following paths;

  1.  A  \,A \, → B \, B
    Distance (d): total distance covered form A to B is 240cm.
    Displacement (ΔX):ΔX=Xf−Xi=AB(\Delta X): \Delta X= X_{f} - X_{i} = AB = 240cm


  2.  A  \,A \, → B  \, B \, → C \, C
    Distance (d): total distance covered: 240cm + 120cm = 360cm
    Displacement (ΔX):ΔX=Xf−Xi=Xc−XA=AC(\Delta X): \Delta X= X_{f} - X_{i} = X_{c} - X_{A} = AC

    Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs

    To find the length of AC, we have to use Pythagoras theorem:

    (AC)2=(AB)2+(BC)2 (AC)^{2}=(AB)^{2}+(BC)^{2}
    (AC)2=(240cm)2+(120cm)2 (AC)^{2}=(240cm)^{2}+(120cm)^{2}
    (AC)2=57600cm2+14400cm2 (AC)^{2}=57600cm^{2} + 14400cm^{2}
    (AC) (AC) = 268.33 cm

    ΔX=AC=268.33 cm \Delta X = AC = 268.33 \, cm


  3.  A  \,A \, → B  \, B \, → C  \, C \, → D \, D

    Distance (d): AB+BC+CDAB + BC + CD = 240cm + 120cm + 240cm = 600cm
    Displacement (ΔX):ΔX=Xf−Xi=XD−XA=AD (\Delta X):\Delta X = X_{f} - X_{i}=X_{D}- X_{A}= A_{D} = 120cm


  4.  A  \,A \, → B  \, B \, → C  \, C \, → D  \, D \, → A \, A

    Distance (d): AB+BC+CD+ADAB + BC + CD + AD = 240cm + 120cm + 240cm + 120cm = 720cm
    Displacement (ΔX):ΔX=Xf−Xi=XA−XA= (\Delta X):\Delta X=X_{f}-X_{i}=X_{A}-X_{A}= 0


Velocity Vs. Speed


Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs


Average Velocity: change position divided by the time taken; Vavg=ΔXΔt=Xf−Xitf−ti \large V_{avg} = \frac{\Delta X} {\Delta t} = \frac{X_{f} - X_{i}} {t_{f} - t_{i}}
It is a vector quantity which has both magnitude and direction.

Average Speed: total distance travelled divided by the time taken; Savg=dt \large S_{avg} = \frac{d} {t}
It is a scalar quantity which has only magnitude.

Instantaneous Velocity: velocity at an instant of time.

Instantaneous Speed: speed at an instant of time.


Acceleration


An object whose velocity is changing is said to be accelerating, it specifies how rapidly the velocity of the object is changing.

Average Acceleration: it is defined as the change in velocity divided by the time taken.
It is a vector quantity which has both magnitude and direction.

aavg=ΔVΔt=Vf−Vitf−ti \large a_{avg} = \frac{\Delta V} {\Delta t} = \frac{V_{f} - V_{i}} {t_{f} - t_{i}}


Instantaneous Acceleration: it is defined as the acceleration of the object at instant of time.

Standard unit of acceleration: ms2 \large \frac{m} {s^{2}}


Graphical Analysis


Position Vs. Time Graph

The following graph represents the position Vs. time graph for an object moving with constant velocity.

Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs


Using position Vs. time graph we can find the following quantitates;

  1. Average velocity
    To find VavgV_{avg} , find the slope of the graph.

    Vavg=slope=RiseRun=V_{avg} = slope = \frac{Rise} {Run} = x2−x1t2−t1=ΔXΔt\large \frac{x_{2} - x_{1}} {t_{2} - t_{1}} = \frac{\Delta X} {\Delta t}

    Example;

    Vavg=slope=RiseRun=V_{avg} = slope = \frac{Rise} {Run} = 50−105−1=404\large \frac{50 - 10} {5 - 1} = \frac{40} {4} = 10m/s


    Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs

  2. Instantaneous Velocity
    3. To find instantaneous velocity from the graph, calculate the slope of the graph at that instant of time.

    Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs

Vinst V_{inst} at t=2s,Vinst=slope=t = 2s, V_{inst} = slope = 20−02−0=\large \frac{20- 0} {2- 0} = 10 m/s

Note: for bodies moving with constant velocity; Vinst=VavgV_{inst}=V_{avg}

The following graph represents position Vs. time graph for an object moving with variable velocity (velocity is NOT constant).

Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs


  1. Average Velocity
    To find VavgV_{avg}, find the slope of the graph.

    Vavg=slope=RiseRun=V_{avg} = slope = \frac{Rise} {Run} = x2−x1t2−t1=ΔXΔt\large \frac{x_{2} - x_{1}} {t_{2} - t_{1}} = \frac{\Delta X} {\Delta t}


    Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs


    VV avg (2.8 s → 5.0s) = slope =RiseRun=x2−x1t2−t1=25−7.55−2.8\large \frac{Rise} {Run} = \frac{x_{2} - x_{1}} {t_{2} - t_{1}} = \frac{25 - 7.5} {5-2.8} = 7.95 m/s


  2. Instantaneous Velocity
    3. To find instantaneous velocity from the graph, calculate the slope of the graph at that instant of time. To do that, draw the tangent at that instant of time and calculate the slope of the tangent.


    Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs


    VinstV_{inst} at t=2st = 2s

    VinstV_{inst} = slope of the tangent =RiseRun=x2−x1t2−t1=3.75−1.252−1\large \frac{Rise} {Run} = \frac{x_{2} - x_{1}} {t_{2} - t_{1}} = \frac{3.75 - 1.25} {2-1} = 2.5 m/s

    Note: for bodies moving with variable velocity; Vinst≠Vavg\large V_{inst} \neq V_{avg}


Velocity Vs. Time Graph

Using Velocity Vs, Time graph you can find;
  1. Acceleration (slope)
  2. Displacement (Area under graph)


Constant Acceleration

To find average acceleration from Velocity Vs. Time graph, calculate the slope of the graph.

For bodies moving with constant acceleration, aavg=ainst\large a_{avg} = a_{inst}


Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs


aavg=ainst \large a_{avg}= a_{inst} = slope=RiseRun=V2−V1t2−t1=50−105−1=404\large \frac{Rise} {Run} = \frac{V_{2} - V_{1}} {t_{2} - t_{1}} = \frac{50 - 10} {5-1} = \frac{40} {4} = 10 ms2 \large \frac{m} {s^{2}}



Variable Acceleration

Average Acceleration
To find average acceleration from the graph, calculate the slope of the graph during the specific interval of time.


Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs


aavg=  \large a_{avg} = \, slope=RiseRun=V2−V1t2−t1=25−55−2=203\large \frac{Rise} {Run} = \frac{V_{2} - V_{1}} {t_{2} - t_{1}} = \frac{25 - 5} {5-2} = \frac{20} {3} = 6.67 m/s2


Instantaneous Acceleration
To find instantaneous acceleration from the graph, calculate the slope of the graph at that instant of time. To do that, draw the tangent at that instant of time and calculate the slope of the tangent.


Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs


ainsta_{inst} at t=2st = 2s;

ainst=  \large a_{inst} = \, slope of the tangent =RiseRun=V2−V1t2−t1=3.75−1.252−1=\large \frac{Rise} {Run} = \frac{V_{2} - V_{1}} {t_{2} - t_{1}} = \frac{3.75 - 1.25} {2-1} = 2.5 m/s2

Note: for bodies moving with variable acceleration; ainst≠aavg a_{inst} \neq a_{avg}


Displacement as the Area Under the Graph
Using Velocity Vs. Time graph you can find the displacement by calculating the area under the graph.


Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs


Displacement from 1s to 5s;

Area of a triangle = 12\large \frac{1}{2} base × height = 12\large \frac{1}{2} × 40 × 4 = 80m

Note: Since the motion is along a straight line, the displacement is equal to distance covered.

Let’s consider the following Velocity Vs. Time graph, where the object changes direction as it moves.


Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs


Total displacement; area under the graph
Area of a triangle = 12\large \frac{1}{2} base × height = 12\large \frac{1}{2} × 10 × 50 = 250m

Total distance covered; distance covered from 0s to 5s + distance covered from 5s to 10s
Total distance = 50m + 50m = 100m

Note: Distance and Displacement are different, since the object changed direction.