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Motion in One Dimension
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Motion in One Dimension |
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What is Kinematics?
It is a branch of mechanics that describes the physical concept of motion of objects without any consideration of masses or forces that cause the motion. Thus, it is the opposite of dynamics.
Therefore, it contrasts with dynamics or kinetics.
Types of Motion
In mechanics, motion is defined as "the movement of an object or a material point from one place to another in a given time."
The movements of objects are classified into three types: Rotational Motion – Oscillatory Motion – Linear Motion.
When describing the motion of an object, we define it relative to a fixed point (reference point or origin within a reference frame).
The distance of the object from this point is called the 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.
\[S=150+70+100=320 m \]
\[ΔX=X_f-X_i=150−60=90𝑚\]
\[ΔX=X_f-X_i=60−150=−90𝑚\]
\[S=140+40+140=320 m\]
To calculate displacement, we choose the reference point for the motion. Let’s take the reference point at position D:
\[X_i=-140 m , X_f=-100 \]\[ΔX=X_f-X_i=-100-(-140)=40 𝑚\]
\[.............................................\]\[.............................................\]\[.............................................\]
Calculate the distance traveled:
\[.............................................\]
Calculate the displacement:
\[.............................................\] \[.............................................\]
We determine the reference point, let’s take position A:
\[X_i=0 , X_f= 140 \] \[∆𝑥= 𝑥_𝑓−𝑥_𝑖=140-0=140 m\]\[t= 3 min=3×60=180 S\]\[\vec v_{avrg}=\frac{\vec X_f-\vec X_i}{t_f-t_i}=\frac{140-0}{180-0}=0.78 m/s\]
When an object moves in one dimension, some objects move at constant speed, while others move at uniformly varying speed. Object moving with uniformly varying speed Object moving with constant speed Acceleration is non-zero \[𝑎≠0\] Acceleration is zero \[𝑎=0\] \[v_f=v_i+𝑎.t\]\[{v_f}^2={v_i}^2 + 2.𝑎(x_f-x_i)\]\[x_f=x_i+v_i.t+\frac{1}{2}.𝑎.t^2\]\[x_f=x_i+\frac{1}{2}({v_f}+{v_i}).t\] \[X_f=X_i+v.t\] Acceleration \[𝑎=\frac{2∆Y}{t^2}\] Time \[t (s)\] Displacement \[∆Y(m)\] Selected Location \[𝑎=...........\] \[t=...........\] \[∆Y=..........\] Earth \[𝑎=...........\] \[t=..........\] \[∆Y=..........\] Sun \[𝑎=..........\] \[t=...........\] \[∆Y=...........\] Moon \[𝑎=...........\] \[t=...........\] \[∆Y=...........\] Mars \[𝑎=...........\] \[t=..........\] \[∆Y=..........\] Jupiter \[𝑎=...........\] \[t=...........\] \[∆Y=...........\] Venus
Distance: A positive scalar quantity representing the path length traveled by the object during its motion.
In straight-line motion, the distance is the numerical value of the length traveled.
The unit of distance in the SI system is meters (m). It can also be expressed in centimeters (cm), kilometers (km), or miles.
\[m\;\;,cm\;\;,Km\;\;,mile\]
Example
A car moved according to the following diagram:
\[A\Rightarrow B\Rightarrow C \Rightarrow D \]
The distance traveled is:
Displacement: A vector quantity representing the difference between the object's final position and its initial position.
It is the shortest distance from the starting point to the endpoint.
The unit of displacement in the SI system is meters (m). It can also be expressed in centimeters (cm), kilometers (km), or miles.
\[m\;\;,cm\;\;,Km\;\;,mile\] When an object moves from initial position
\[X_i\]
to final position
\[X_f\]
the displacement is given by:
\[ΔX=X_f-X_i\]
Displacement is a vector quantity, so both magnitude and direction must be specified.
If the car moves to the right, the displacement vector is positive.
If the car moves to the left, the displacement vector is negative.
In a running race, consecutive photos of a runner were taken every two seconds, resulting in the following images. The position-time graph was plotted, resulting in the following line 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? \[s=12+10=22 m \]
What is the displacement? \[∆𝑥= 𝑥_𝑓−𝑥_𝑖=26-4=22 m\]
Note: Distance and displacement have the same value because the runner did not change direction.
What is the reason for the change in the slope of the graph?
The reason is the change in the runner's speed. In the sixth second, the runner changed speed but remained constant afterward.
Question
Rashid moved, and the motion graph was plotted, determining his position over time.
How did Rashid's position change over time?
Click here to see the solution.
Average velocity is a vector quantity representing the rate of change of an object's position.
It can also be defined as the displacement of the object per unit time.
When an object moves in linear motion, it covers equal distances in equal time intervals, and we say the object is moving with uniform velocity.
If it covers unequal distances in equal time intervals, we say the object is moving with variable velocity.
The direction of the average velocity vector is the direction of displacement.
\[\vec v_{avrg}=\frac{\vec X_f-\vec X_i}{t_f-t_i}\]
The unit of velocity in the SI system is:
\[m/s\]
Solved Example
Calculate the average velocity if Ahmed moved according to the following diagram:
\[A\Rightarrow B\Rightarrow C \Rightarrow D \]
in a time of:
\[t=3\;\;min\]
Direction: To the right.
Calculating Average Velocity from Position-Time Graph
Calculate the average velocity from the position-time graph.
During the time interval:
\[t=4\;\; s \Rightarrow t=6\;\;s\]
Instantaneous velocity: The velocity of an object at a specific moment. In calculus, it is the derivative of displacement with respect to time when time approaches zero.
\[\vec v_x=\lim_{∆t \rightarrow 0} \frac {∆𝑥}{∆t} = \frac {d𝑥}{dt}\]
Solved Example
The position of a bicycle relative to the X-axis was determined by the equation:
\[X=-2t^2+15t+4 \]
The velocity of the bicycle at the fourth second is:
\[\vec v_x= \frac {d𝑥}{dt}=\frac {d(-2t^2+15t+4)}{dt}=-4t+15=-4×4+15=-1 m/s\]
Calculate the instantaneous velocity from the position-time graph.
At the moment:
\[t=4\;\;s\]
Speed: The distance traveled per unit time.
It is a scalar quantity.
Measured in units of:
\[v=\frac{s}{t}\]
\[m/s\]
Solved Example
The position-time graph of a bicycle was plotted. The speed of the bicycle during the entire period is:
\[S=8+15=23 m \]\[t=18 S\]
\[v=\frac {23}{18}=1.28 m/s\]
Average acceleration is a vector quantity representing the rate of change of an object's velocity.
Or the change in velocity vector over a time interval.
\[\vec a=\frac{∆\vec v}{∆t}\]
The position equation of a car is:
\[X=12-10 t^2+t^3\]
The acceleration of the car at the fifth second is:
\[\vec v_x= \frac {dx}{dt}=-20.t+3.t^2\] \[\vec a_x= \frac {dv}{dt}=-20+6.t=-20+6×5=-20+30=10 m/s^2\]
Calculate the average acceleration from the velocity-time graph.
During the time interval:
\[t=4\;\; s\Rightarrow t=6\;\;s\]
Calculate the instantaneous acceleration from the velocity-time graph.
At the moment:
\[t=4\;\;s\]
A plane lands on a runway and touches the ground at a speed of:
\[45\;\;m/s\]
It travels a distance of:
\[540\;\;m\]
until it stops. Calculate the average acceleration of the plane on the runway and the time it took to stop.
\[{v_f}^2={v_i}^2 + 2.𝑎(x_f-x_i)\]
\[{0}^2={45}^2 + 2.𝑎(540)\]
\[{0}^2=2025 + 1080 a\]
\[a=\frac {-2025}{1080}=-1.87 m/s^2\]
\[v_f=v_i+𝑎.t\]
\[0=45+(-1.87).t\]
\[t=\frac{-45}{-1.87}=24 s\]
It is the motion of an object without initial velocity under the influence of its weight only, neglecting air resistance.
Objects fall toward Earth due to gravity.
Gravity causes objects to accelerate downward during free fall when air resistance is negligible.
In this simulation, we will determine the acceleration due to free fall of a ball in different regions of the universe.
Free Fall Experiment:
Objects in free fall move with an acceleration equal to the acceleration due to gravity.
In the absence of air resistance, all objects reach the ground at the same time if dropped from the same height.
\[ g= -9.81 m/s^2 , v_i=0\]
The equations of motion become:
\[v_f=-g.t\]
\[{v_f}^2=- 2.g(Y_f-Y_i)\]
\[Y_f=Y_i-\frac{1}{2}.g.t^2\]
We call the motion of objects moving upward or downward under the influence of gravity "vertical projectile motion."
If an object is projected upward with a certain initial velocity, the magnitude of the velocity in the vertical direction decreases until it stops at a certain moment.
The moment when the velocity becomes zero is called the maximum height reached by the object.
After that, the object begins to fall downward. It is very important to remember that the acceleration is constant.
However, the velocity vector may change in magnitude and direction. At maximum height, where the velocity is zero, the acceleration remains equal to the acceleration due to gravity.
\[g=-9.81 m/s^2\]
The equations of motion become:
\[v_f=v_i-g.t\]
\[{v_f}^2={v_i}^2 - 2.g(Y_f-Y_i)\]
\[Y_f=x_i+v_i.t-\frac{1}{2}.g.t^2\]
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