Position, velocity, acceleration and time

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Intros
Lessons
  1. Displacement Vs. Distance
  2. Velocity Vs. Speed
  3. Acceleration
  4. Position Vs. Time Graph
  5. 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?
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Topic Notes
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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, Ξ”X=Xfβˆ’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 (Ξ”X):Ξ”X=Xfβˆ’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 (Ξ”X):Ξ”X=Xfβˆ’Xi=Xcβˆ’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

    Ξ”X=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 (Ξ”X):Ξ”X=Xfβˆ’Xi=XDβˆ’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 (Ξ”X):Ξ”X=Xfβˆ’Xi=XAβˆ’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=Ξ”XΞ”t=Xfβˆ’Xitfβˆ’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=Ξ”VΞ”t=Vfβˆ’Vitfβˆ’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} = x2βˆ’x1t2βˆ’t1=Ξ”XΞ”t\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} = 50βˆ’105βˆ’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 = 20βˆ’02βˆ’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} = x2βˆ’x1t2βˆ’t1=Ξ”XΞ”t\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=x2βˆ’x1t2βˆ’t1=25βˆ’7.55βˆ’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=x2βˆ’x1t2βˆ’t1=3.75βˆ’1.252βˆ’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=V2βˆ’V1t2βˆ’t1=50βˆ’105βˆ’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=V2βˆ’V1t2βˆ’t1=25βˆ’55βˆ’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=V2βˆ’V1t2βˆ’t1=3.75βˆ’1.252βˆ’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.