📄 Print pdf
00971504825082
Motion Representation
Click here to view learning outcomes
Types of Motion
When an object changes its position, we say the object has moved and there are different types of motion
There is circular motion - linear motion - and vibrational motion
What concerns us now is motion in a straight line where the object may move forward and backward or up and down
We call this type motion in a straight line or one-dimensional motion
Motion Diagram
The motion of an object can be described through sequential images taken of the object at equal time intervals
The runner moves in a straight line at constant speed because he covers equal distances in equal times
The ball follows a parabolic path and moves in two dimensions
The motion of an object can be described by considering the object as a point taken from the center of the object, and through the particle model we determine the object's motion
The particle model indicates constant speed because it covers equal distances in equal times and moves in a straight line
The particle model indicates increasing speed because it covers increasing distances in equal times and moves in a straight line
The particle model indicates decreasing speed because it covers decreasing distances in equal times and moves in a straight line
Position of an Object
When describing the motion of an object, we determine it relative to a fixed point considered as a reference point or origin within a reference frame
The distance of the object from this point is called position
The position of a particle is only determined by choosing a reference point we consider as the origin
Position: A vector quantity that can be positive or negative
Position to the right of the reference point is positive
Position to the left of the reference point is negative
Physical quantities are divided into two types
Vector quantities
Scalar quantities
Have magnitude and direction
Have magnitude but no direction
Examples:
Velocity
Displacement
Acceleration
Examples:
Time
Temperature
Distance
Concept of Displacement and Distance
Distance
A positive scalar quantity representing "the distance traveled by the object during its motion". In straight-line motion, distance is numerically equal to the distance covered by the object
Example: Ahmed moved from position -2 to position +6 then changed direction and moved to position +3. The distance traveled equals:
\[s=8+3=11 m\]
Displacement
A vector quantity representing the difference between the object's position at the end of its motion and the starting point of motion
In other words, displacement is the change in position of an object during its motion.
When an object moves from its initial position
\[X_i\]
to its final position
\[X_f\]
Displacement is given by:
\[\Delta {\mathbf{X} }= X_f – X_i \]
Example: Ahmed moved from position -2 to position +6 then changed direction and moved to position +3. The displacement equals:
\[\Delta X = X_f – X_i=+3 - (- 2)=+5\] The positive sign indicates displacement is in the positive direction
In this simulation we will learn about distance and displacement in one dimension
Displacement and Distance Calculation
Simulation of Object Motion in One Dimension
Solved Example
A car moved from position
\[150\;m\] to position
\[60\;m\]. The displacement equals:
Solution \[\Delta X = X_f – X_i=60 - (150)=- 90 m \]
Solved Example
Calculate the displacement and distance if Ahmed moved according to the following diagram
\[A\Rightarrow D\Rightarrow B \Rightarrow C \]
Solution
Choose the reference point for motion wherever you want
( B ) Let it be at position
Solution
Choose the reference point for motion wherever you want
( B ) Let it be at position
\[S=140+40 +140 =320 m \]
\[X_i=-180 m , X_f=-140 m \]
\[\Delta X = X_f – X_i=(-140 )- (-180)=+40 m \]
1
Majed moved from Ajman to Umm Al Quwain then changed direction and headed to Sharjah passing through Ajman
Using the following figure, the distance and displacement traveled
Click here to show solution method
Choose the correct answer
\[ s=42 m\;\; ,\;\; ∆X= 12 m \;\;\;\;\;\;-C\]
Displacement to the West
\[ s=42 m\;\; ,\;\; ∆X= 12 m \;\;\;\;\;\;-A\]
Displacement to the East
\[ s=27 m \;\;, \;\; ∆X= 15 m\;\;\;\;\;\;-D\]
Displacement to the West
\[ s=27 m \;\;, \;\; ∆X= 15 m\;\;\;\;\;\;-B\]
Displacement to the East
Position-Time Graph
The following table shows a student's position
relative to time
Draw the graph
that shows the relationship between position and time
Click here to show solution method
From the drawn graph, answer the following questions
At what time was the student moving at constant speed in the positive direction
\[.........................................\]
At what time was the student moving at constant speed in the negative direction
\[.........................................\]
At what time was the student stationary
\[.........................................\]
What is the distance traveled by the student
\[.........................................\]
What is the displacement of the student
\[.........................................\]
Click here to show solution method
Learn, Draw and Calculate
Enter position and time values to plot the graph and calculate distance and displacement traveled
Position-Time Graph
Position (m)
Time (s)
Experiment of Motion at Constant Speed and Drawing Position-Time Graph
In this simulation, adjust the speed indicator to an appropriate value
Solved Example
In a running race, sequential photos were taken of a runner every two seconds as shown below, resulting in the following images and the position-time relationship was plotted
resulting in the following graph
From which position was the runner's motion study started and at which position did it end \[x_i= 4 m , x_f= 26 m \]
What is the distance traveled by the runner \[s=6+10=16 m \]
What is the displacement \[\Delta 𝑥= 𝑥_𝑓−𝑥_𝑖=26-4=22 m\]
Note distance and displacement have the same value because the runner didn't change direction
What is the reason for the change in the graph's slope?
The reason is the runner changed his speed at the sixth second but remained constant
Play and Learn
2
The following graph shows the relationship between position and time
for a bicycle's motion. The distance and displacement traveled equal:
\[ s=34 m\;\; ,\;\; ∆X= 12 m \;\;\;\;\;\;-C\]
Displacement to the West
\[ s=34 m\;\; ,\;\; ∆X= 12 m \;\;\;\;\;\;-A\]
Displacement to the East
\[ s=30 m \;\;, \;\; ∆X= 0 m\;\;\;\;\;\;-D\]
\[ s=30 m \;\;, \;\; ∆X= 12 m\;\;\;\;\;\;-B\]
Displacement to the East
Click here to show solution method
Choose the correct answer
3
One of the following graphs shows the motion of an object with variable velocity in the positive direction
Click here to show solution Choose the correct answer
Velocity
Speed
Average Velocity
Distance traveled per unit time
Change in position divided by time
\[ v =\frac{d}{t_f-t_i}\]
\[\vec v =\frac{\vec X_f-\vec X_i}{t_f-t_i}\]
Scalar quantity (magnitude only)
Vector quantity (direction is same as displacement)
Unit: \[\frac{m}{s}\]
Unit: \[\frac{m}{s}\]
Solved Example
The position-time graph for a bicycle is shown below
Total distance traveled:
The bicycle moved from position 4 to position 12 (distance = 8 m), then changed direction to position 0 (distance = 12 m), then continued in negative direction to position -8 (distance = 8 m)
L = 8 + 12 + 8 = 28 m
Total displacement:
Initial position (xi = 4 m)
Final position (xf = -8 m)
\[\Delta 𝑥= 𝑥_𝑓−𝑥_𝑖=-8-4=-12 m\]
Average velocity:
\[\vec v =\frac{\vec X_f-\vec X_i}{t_f-t_i}= \frac{-12}{16-0}=-0.75 \frac{m}{S}\]
Speed:
\[ v =\frac{L}{t_f-t_i}=\frac{28}{16-0}=1.75\frac{m}{S}\]
(Acceleration-Time) and (Velocity-Time) Graphs - Constant Velocity Motion Experiment and Position-Time Graph
Relationship Between Average Velocity and Slope of Position-Time Graph
The graph below shows the position-time relationship for a bicycle moving at constant velocity
\[ X_i= 0.0, X_f = 24 \]
\[\Delta 𝑥= X_𝑓−X_𝑖=24-0=24 m\]
\[\vec v =\frac{\vec X_f-\vec X_i}{t_f-t_i}= \frac{24}{12-0}=2 \frac{m}{S}\]
Calculate the slope of the graph and write the motion equation from the graph
Click here to show solution
The following position-time graph shows a bicycle's motion. The average velocity equals
\[ 𝑣= -5\;\; m/s \;\;\;\;\;\;-C\]
\[𝑣= +5\;\; m/s \;\;\;\;\;\;-A\]
\[ 𝑣= -8\;\; m/s \;\;\;\;\;\;-D\]
\[ 𝑣= +8\;\; m/s \;\;\;\;\;\;-B\]
Click here to show solution Choose the correct answer
Motion Representation |
Click here to view learning outcomes
Types of Motion
When an object changes its position, we say the object has moved and there are different types of motion
There is circular motion - linear motion - and vibrational motion
What concerns us now is motion in a straight line where the object may move forward and backward or up and down
We call this type motion in a straight line or one-dimensional motion
Motion Diagram
The motion of an object can be described through sequential images taken of the object at equal time intervals
The motion of an object can be described by considering the object as a point taken from the center of the object, and through the particle model we determine the object's motion
When describing the motion of an object, we determine it relative to a fixed point considered as a reference point or origin within a reference frame
Vector quantities
Scalar quantities
Have magnitude and direction
Have magnitude but no direction Examples:
Velocity
Displacement
Acceleration
Examples:
Time
Temperature
Distance
Displacement
In other words, displacement is the change in position of an object during its motion.
When an object moves from its initial position
\[X_i\]
to its final position
\[X_f\]
Displacement is given by:
\[\Delta {\mathbf{X} }= X_f – X_i \]
Majed moved from Ajman to Umm Al Quwain then changed direction and headed to Sharjah passing through Ajman
Using the following figure, the distance and displacement traveled
Choose the correct answer \[ s=42 m\;\; ,\;\; ∆X= 12 m \;\;\;\;\;\;-C\]
Displacement to the West
\[ s=42 m\;\; ,\;\; ∆X= 12 m \;\;\;\;\;\;-A\]
Displacement to the East
\[ s=27 m \;\;, \;\; ∆X= 15 m\;\;\;\;\;\;-D\]
Displacement to the West
\[ s=27 m \;\;, \;\; ∆X= 15 m\;\;\;\;\;\;-B\]
Displacement to the East
The following table shows a student's position
relative to time
Draw the graph
that shows the relationship between position and time
From the drawn graph, answer the following questions
At what time was the student moving at constant speed in the positive direction
\[.........................................\]
At what time was the student moving at constant speed in the negative direction
\[.........................................\]
At what time was the student stationary
\[.........................................\]
What is the distance traveled by the student
\[.........................................\]
What is the displacement of the student
\[.........................................\]
The following graph shows the relationship between position and time
for a bicycle's motion. The distance and displacement traveled equal:
\[ s=34 m\;\; ,\;\; ∆X= 12 m \;\;\;\;\;\;-C\]
Displacement to the West
\[ s=34 m\;\; ,\;\; ∆X= 12 m \;\;\;\;\;\;-A\]
Displacement to the East
\[ s=30 m \;\;, \;\; ∆X= 0 m\;\;\;\;\;\;-D\]
\[ s=30 m \;\;, \;\; ∆X= 12 m\;\;\;\;\;\;-B\]
Displacement to the East Choose the correct answer One of the following graphs shows the motion of an object with variable velocity in the positive direction Choose the correct answer Speed Average Velocity Distance traveled per unit time Change in position divided by time \[ v =\frac{d}{t_f-t_i}\] \[\vec v =\frac{\vec X_f-\vec X_i}{t_f-t_i}\] Scalar quantity (magnitude only) Vector quantity (direction is same as displacement) Unit: \[\frac{m}{s}\] Unit: \[\frac{m}{s}\] The graph below shows the position-time relationship for a bicycle moving at constant velocity Calculate the slope of the graph and write the motion equation from the graph The following position-time graph shows a bicycle's motion. The average velocity equals \[ 𝑣= -5\;\; m/s \;\;\;\;\;\;-C\]
\[𝑣= +5\;\; m/s \;\;\;\;\;\;-A\]
\[ 𝑣= -8\;\; m/s \;\;\;\;\;\;-D\]
\[ 𝑣= +8\;\; m/s \;\;\;\;\;\;-B\]
Choose the correct answer
The runner moves in a straight line at constant speed because he covers equal distances in equal times
The ball follows a parabolic path and moves in two dimensions
The particle model indicates constant speed because it covers equal distances in equal times and moves in a straight line
The particle model indicates increasing speed because it covers increasing distances in equal times and moves in a straight line
The particle model indicates decreasing speed because it covers decreasing distances in equal times and moves in a straight line
Position of an Object
The distance of the object from this point is called position
The position of a particle is only determined by choosing a reference point we consider as the origin
Position: A vector quantity that can be positive or negative
Position to the right of the reference point is positive
Position to the left of the reference point is negative
Physical quantities are divided into two types
Concept of Displacement and Distance
Distance
A positive scalar quantity representing "the distance traveled by the object during its motion". In straight-line motion, distance is numerically equal to the distance covered by the object
Example: Ahmed moved from position -2 to position +6 then changed direction and moved to position +3. The distance traveled equals:
\[s=8+3=11 m\]
A vector quantity representing the difference between the object's position at the end of its motion and the starting point of motion
Example: Ahmed moved from position -2 to position +6 then changed direction and moved to position +3. The displacement equals:
\[\Delta X = X_f – X_i=+3 - (- 2)=+5\] The positive sign indicates displacement is in the positive direction
In this simulation we will learn about distance and displacement in one dimension
Simulation of Object Motion in One Dimension
Solved Example
A car moved from position
\[150\;m\] to position
\[60\;m\]. The displacement equals:
Solution \[\Delta X = X_f – X_i=60 - (150)=- 90 m \]
Solved Example
Calculate the displacement and distance if Ahmed moved according to the following diagram
\[A\Rightarrow D\Rightarrow B \Rightarrow C \]
Solution
Choose the reference point for motion wherever you want
( B ) Let it be at position
Solution
Choose the reference point for motion wherever you want
( B ) Let it be at position
\[S=140+40 +140 =320 m \]
\[X_i=-180 m , X_f=-140 m \]
\[\Delta X = X_f – X_i=(-140 )- (-180)=+40 m \]
Click here to show solution method
Position-Time Graph
Click here to show solution method
Click here to show solution method
Learn, Draw and Calculate
Enter position and time values to plot the graph and calculate distance and displacement traveled
Position (m)
Time (s)
Experiment of Motion at Constant Speed and Drawing Position-Time Graph
In this simulation, adjust the speed indicator to an appropriate value
Solved Example
In a running race, sequential photos were taken of a runner every two seconds as shown below, resulting in the following images and the position-time relationship was plotted
resulting in the following graph
From which position was the runner's motion study started and at which position did it end \[x_i= 4 m , x_f= 26 m \]
What is the distance traveled by the runner \[s=6+10=16 m \]
What is the displacement \[\Delta 𝑥= 𝑥_𝑓−𝑥_𝑖=26-4=22 m\]
Note distance and displacement have the same value because the runner didn't change direction
What is the reason for the change in the graph's slope?
The reason is the runner changed his speed at the sixth second but remained constant
Play and Learn
Click here to show solution method
Click here to show solution
Velocity
Solved Example
The position-time graph for a bicycle is shown below
Total distance traveled:
The bicycle moved from position 4 to position 12 (distance = 8 m), then changed direction to position 0 (distance = 12 m), then continued in negative direction to position -8 (distance = 8 m)
L = 8 + 12 + 8 = 28 m
Total displacement:
Initial position (xi = 4 m)
Final position (xf = -8 m)
\[\Delta 𝑥= 𝑥_𝑓−𝑥_𝑖=-8-4=-12 m\]
Average velocity:
\[\vec v =\frac{\vec X_f-\vec X_i}{t_f-t_i}= \frac{-12}{16-0}=-0.75 \frac{m}{S}\]
Speed:
\[ v =\frac{L}{t_f-t_i}=\frac{28}{16-0}=1.75\frac{m}{S}\]
(Acceleration-Time) and (Velocity-Time) Graphs - Constant Velocity Motion Experiment and Position-Time Graph
Relationship Between Average Velocity and Slope of Position-Time Graph
\[ X_i= 0.0, X_f = 24 \]
\[\Delta 𝑥= X_𝑓−X_𝑖=24-0=24 m\]
\[\vec v =\frac{\vec X_f-\vec X_i}{t_f-t_i}= \frac{24}{12-0}=2 \frac{m}{S}\]
Click here to show solution
Click here to show solution
0 Comments