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Exploring Position, Velocity, and Acceleration in Physics
Dive into the fundamental concepts of motion. Understand how position changes over time, how velocity represents this change, and how acceleration describes the rate of velocity change. Master these key physics principles today!

Now Playing:Position velocity acceleration and time – Example 0a
Intros
  1. Displacement Vs. Distance
  2. Velocity Vs. Speed
Examples
  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?

Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs
Jump to:NotesConceptFAQsPrerequisites
Notes
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.
Concept

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!

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.

Prerequisites

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.