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Motion in Two and Three Dimensions
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When an object moves in two or three dimensions, it has a position
\[\overrightarrow r =(X,Y,Z)=( X \widehat x , Y \widehat y ,Z \widehat z )\]
Therefore, the object has velocity on the three axes
\[\overrightarrow v =(v_x,v_y,v_z)=(v_x \widehat x , v_y \widehat y ,v_z \widehat z )\]
\[\overrightarrow v =\frac{d\vec r}{dt}=d \frac{( X \widehat x , Y \widehat y ,Z \widehat z )}{dt}=\frac{d X }{dt}\widehat x+\frac{d Y }{dt}\widehat y+\frac{d Z }{dt}\widehat z\]
Cartesian components of velocity
\[𝑣_x=\frac{d X }{dt}\;\;\;\;,\;\;\;\;\;𝑣_y=\frac{d Y }{dt}\;\;\;\;,\;\;\;\;\;𝑣_z=\frac{d Z }{dt}\;\;\;\;,\;\;\;\;\;\]
Therefore, the object has acceleration on the three axes
\[\vec a=\frac{d\vec v}{dt}=\frac{d 𝑣_x }{dt}\widehat x+\frac{d 𝑣_y }{dt}\widehat y+\frac{d 𝑣_z }{dt}\widehat z\]
Cartesian components of acceleration
\[a_x=\frac{d 𝑣_x }{dt}\;\;\;\;,\;\;\;\;\;a_y=\frac{d 𝑣_y }{dt}\;\;\;\;,\;\;\;\;\;a_z=\frac{d 𝑣_z }{dt}\;\;\;\;,\;\;\;\;\;\]
Velocity and Acceleration in Two and Three Dimensions
Motion in two or three dimensions: changing the magnitude or direction of velocity results in acceleration
\[\vec a=\frac {\Delta\vec v}{\Delta t}=\frac{𝑣_2-𝑣_2}{t_2 -t_1}\]
Solved Example
A deer runs in a park in two dimensions
\[X(t)=0.5.t^2+12t+15\;\;\;\;\;\;Y(t)=-0.3.t^2+6t+20\]
where the units are \[X(m)\;\;\;Y(m)\;\;\;t(s)\]
Calculate the deer's position at time \[t=8\;\;s\]
\[X(8)=0.5× 8^2+12× 8+15=143 \;\;m \]\[Y(t)=-0.3× 8^2+6× 8+20=48.8\;\;m\]
\[r=\sqrt {{r_x}^2+{r_y}^2}=\sqrt {{143}^2+{48.8}^2}=151.1 \;\;m\]
Direction
\[𝜃=tan^{-1}\frac {Y}{X}=tan^{-1}\frac {48.8}{143}=18.8^0\]
Calculate the deer's velocity at time
\[t=8\;\;s\]
\[V_x=\frac {dX}{dt} =d\frac {0.5.t^2+12t+15}{dt}=0.5×2×t+12×1=0.5×2×8+12×1=20\;\;m/s\]
\[V_y=\frac {dY}{dt} =d\frac {-0.3.t^2+6t+20}{dt}=-0.3×2×t+6×1=-0.3×2×8+6×1=1.2\;\;m/s\]
\[v=\sqrt {{v_x}^2+{v_y}^2}=\sqrt {{20}^2+{1.2}^2}=20.03 \;\;m/s\]
Direction
\[𝜃=tan^{-1}\frac {v_y}{v_x}=tan^{-1}\frac {1.2}{20}=3.4^0\]
Calculate the deer's acceleration at time
\[t=8\;\;s\]
\[a_x=\frac {dv_x}{dt} =d\frac {1t+12}{dt}=1×1+0=1\;\;m/s^2\]
\[a_y=\frac {dv_Y}{dt} =d\frac {-0.6t+6}{dt}=1×-0.6+0=-0.6\;\;m/s^2\]
\[a=\sqrt {{a_x}^2+{a_y}^2}=\sqrt {{1}^2+{-0.6}^2}=1.17 \;\;m/s^2\]
Direction
\[𝜃=tan^{-1}\frac {a_y}{a_x}=tan^{-1}\frac {-0.6}{1}=-31^0\]
Ideal Projectile Motion
An ideal projectile is any object launched with an initial velocity and moves under the influence of its weight
while neglecting air resistance, wind speed, and object rotation, thus the projectile moves in two dimensions
When studying motion in two dimensions, it's easier to study the motion on each axis separately
The initial velocity has two components: horizontal and vertical
\[v_x=v_0cos (𝜃) \] \[v_y=v_0sin (𝜃 )\]
Projectile at an angle experiment
Run the experiment and set the projectile height either from ground level or from a height within limits, set an appropriate initial velocity, specify the launch angle, and make the motion slow so you can take readings accurately. Take displacement readings on both axes every 0.5 seconds by clicking the pause button, then record the readings in the table below
Projectile Motion Simulation
Time (s)
Horizontal Distance (m)
Height (m)
Vertical Velocity (m/s)
Show Horizontal Acceleration
Show Vertical Acceleration
The following graph shows the relationship between velocity and time for a projectile on the X-axis
What is the acceleration value?
Slope
= a = .......
From the above, we find the equations of motion on the X-axis:
\[X=v_0.Cos{𝜃}.t\]
The following graph shows the relationship between velocity and time for a projectile on the Y-axis:
What is the acceleration value?
Slope
= a = .......
From the above, we find the equations of motion on the Y-axis:
\[𝜗_𝑓𝑦= 𝜗_{0}sin 𝜃- g t\] \[𝜗_{𝑓𝑦}^2= (𝜗_{0}Sin 𝜃 )^2 -2.g.∆y\]\[∆𝑦=𝜗_{0}Sin 𝜃.t -\frac{1}{2}.g.t^2\]
Notes: Use the experiment to verify the results
The time of flight equals twice the time to reach maximum height
The projectile reaches maximum range when launched at 45 degrees
At maximum height, the vertical velocity becomes zero
Two projectiles can reach the same range when launched at angles that add up to 90 degrees
Projectile motion follows a parabolic path. To prove this:
\[X=v_0.Cos{𝜃}.t\Rightarrow\;\; t=\frac{x}{v_0.cos{𝜃}}\]
\[Y=Y_0+𝜗_{0}Sin 𝜃.t -\frac{1}{2}.g.t^2\]
\[Y=Y_0+𝜗_{0}Sin 𝜃.\frac{x}{v_0.cos{𝜃}} -\frac{1}{2}.g.(\frac{x}{v_0.cos{𝜃}})^2\]
\[Y=Y_0+Sin 𝜃.\frac {x}{cos{𝜃}} -\frac {1}{2}.g.\frac{x^2}{v_0^2.cos^2{𝜃}}\]
\[Y=Y_0+tan(𝜃).X -\frac {g}{2v_0^2.cos^2{𝜃}}.X^2\]
This equation relates the projectile's position on the horizontal axis to its position on the vertical axis
Which is the same as the equation of a parabola:
\[Y=C+bX+aX^2\]
Solved Example
A soccer player kicks a ball with velocity
\[v_0 = 25 m/s \]
at an angle
\[𝜃=33.1^0\]
above the horizontal
from a distance
from the goal
\[X=40.6 m\]
Did the ball reach the goal or go over it?
Solution:
The ball is launched from the ground and returns to the ground
\[Y=Y_0=0 \]
\[Y=Y_0+tan(𝜃).X -\frac {g}{2v_0^2.cos^2{𝜃}}.X^2\]
\[0=0+tan(33.1).X -\frac {9.81}{2×25^2.cos^2{33.1}}.X^2\]
\[ 0.651 X=0.011 X^2\]
\[X=59.1 m\] The ball went over the goal
If the crossbar height is 2 meters, did the ball enter the goal or go over the crossbar?
We observe that:
\[Y_0=0\;\;\;\;\;v_0 = 25\;\; m/s\;\;\;\;\; 𝜃=33.1^0 \;\;\;\;\;X=40.6\;\; m\;\;\;\;\;Y=?\]
\[Y=Y_0+tan(𝜃).X -\frac {g}{2v_0^2.cos^2{𝜃}}.X^2\]
\[Y=0+tan(33.1)×40.6 -\frac {9.81}{2×25^2×cos^2{33.1}}×(40.6)^2= 8.03 m\]
The ball went over the crossbar
: Velocity Calculation
Each point on the trajectory has two velocity components: one on the X-axis
and one on the Y-axis
But the velocity at maximum height has only one component on the horizontal axis
The velocity on the X-axis is the projection of the initial velocity on the X-axis and is always constant:
\[𝜗_x= 𝜗_0 Cos𝜃\]
The velocity on the Y-axis is calculated from the equations of motion:
\[𝜗_𝑓𝑦= 𝜗_{0}sin 𝜃- g t\], \[𝜗_{𝑓𝑦}^2= (𝜗_{0}Sin 𝜃 )^2 -2.g.∆Y\]
Therefore, the total velocity at any position is calculated by:
\[𝜗=\sqrt{ 𝜗_X^2+𝜗_Y^2}\]
Maximum Height and Range of Projectiles
When a projectile is launched from the ground and returns to the ground:
The time to reach maximum height equals half the total time in the air
The launch velocity equals the landing velocity
At maximum height:
\[H=Y_0+\frac {v_{Y0}^2}{2g}\;\;\;\;\;\;\;\; v_{fY}=0\]
For calculating the horizontal range when:
\[Y=Y_0=0\]
\[R=X=\frac {v_0^2.}{g}.sin(2𝜃)\]
For calculating the horizontal range when:
\[Y_0≠0\]\[X=v_0.Cos{𝜃}.t\] Maximum range occurs when the launch angle is \[𝜃=45^0\]
Solved Example
A cannon on a hill 50 m above ground level fires a projectile with initial velocity \[v_0=83 \;\;m/s\] at an angle \[𝜃=40^0\] above the horizontal
Calculate the maximum height reached by the projectile
\[H=Y_0+\frac {v_{Y0}^2}{2g}\]
\[H=50+\frac {(83×sin (40))^2}{2×9.81}=195.1 m\]
Calculate the time needed for the projectile to reach the target
\[∆Y=𝜗_{0}Sin 𝜃.t -\frac{1}{2}.g.t^2\] \[0-50=83 ×Sin (40)×t -\frac{1}{2}×9.81×t^2\]
\[4.9.t^2 -53.35.t-50=0\]
\[t=11.76\;\;s\]
Calculate the horizontal range
\[X=v_0.Cos(𝜃).t\]\[X=83×Cos(40)(𝜃)11.76=747.7 \;\;m\]
Calculate the projectile's velocity upon impact
\[𝜗_X=𝜗_{0X}=𝜗_0.COS (𝜃)=83× COS (40)=63.6\;\;m/s\]
\[𝜗_𝑓Y= 𝜗_{0}sin (𝜃)- g t=83× sin (40)-9.81×11.76=-62 \;\;m/s\]
\[𝜗=\sqrt{(𝜗_X)^2+(𝜗_Y)^2}\] \[𝜗=\sqrt{(63.3)^2+(-62)^2}=88.6 \;\;m/s\]
Horizontal Projectile
This is an application of two-dimensional motion
A horizontal projectile has its initial velocity purely horizontal:
\[v_X=v_0\;\;\;\;\;\;\;\;\;v_{0Y}=0.0\]
Equations of motion on the X-axis:
\[X=v_0..t\] Equations of motion on the Y-axis:
\[𝜗_𝑓𝑦= - g t\], \[𝜗_{𝑓𝑦}^2= -2.g.∆Y\], \[∆Y= -\frac{1}{2}.g.t^2\]
Solved Example
A cannon is mounted on a building 50 m above ground level and fires a projectile horizontally at 90 m/s
Calculate the projectile's velocity upon impact
Calculate the total velocity using:
\[v= \sqrt {𝜗_X^2+𝜗_Y^2}\]
\[v_X=v_0=90 \;\;m/s\]\[v_{fY}^2=-2.g.∆𝑦 =−2×9.81×−50=981 \Rightarrow\;\; v_{fY}=31.3 m/s\] \[v= \sqrt {90^2+31.3^2 }=95.2 m/s\]
Realistic Projectile Motion
In ideal projectile motion, we neglect air resistance and the rotation of the projectile
Air resistance affects the horizontal range and maximum height of the projectile
It always opposes motion, reducing both horizontal and vertical motion components until stopping, then falling begins
The projectile's rotation causes deviation from its path, making the motion three-dimensional
Relative Velocity
Relative velocity is the velocity of a moving object relative to another object
Relative velocity changes based on direction of motion. If both objects move in the same direction
the observed object appears slower.
If objects move in opposite directions
the relative velocity appears increased
When stating an object's velocity, we must specify the reference frame
For example, Ahmed drives a car east at
\[20 \;\;m/s\]
Ahmed's velocity relative to the car:
\[\vec v_{AC}=0\]
Ahmed's velocity relative to the road:
\[\vec v_{Ae}=20 m/s\] east
The ground's velocity relative to Ahmed:
\[\vec v_{eA}=20 m/s\] west
Solved Example
A boat heads north at
\[20\;\;m/s\] across a river flowing east at
\[5\;\;m/s\]
Calculate the boat's velocity relative to an observer on the bank
\[v_{BR}=20\;\;m/s\] north
\[v_{Re}=20\;\;m/s\] east
\[\vec v_{Be}=\vec v_{BR}+\vec v_{Re}\]
\[v_{Be}=\sqrt {v_{BR}^2+v_{Re}^2}=\sqrt {20^2+5^2}=20.6 m/s\]
Direction:
\[𝜃=tan^{-1}\frac {v_{BR}}{v_{Re}}=tan^{-1}\frac {20}{5}=76^0\]
Motion in Two and Three Dimensions |
Click here to view the learning outcomes
When an object moves in two or three dimensions, it has a position \[\overrightarrow r =(X,Y,Z)=( X \widehat x , Y \widehat y ,Z \widehat z )\] Therefore, the object has velocity on the three axes \[\overrightarrow v =(v_x,v_y,v_z)=(v_x \widehat x , v_y \widehat y ,v_z \widehat z )\] \[\overrightarrow v =\frac{d\vec r}{dt}=d \frac{( X \widehat x , Y \widehat y ,Z \widehat z )}{dt}=\frac{d X }{dt}\widehat x+\frac{d Y }{dt}\widehat y+\frac{d Z }{dt}\widehat z\] Cartesian components of velocity \[𝑣_x=\frac{d X }{dt}\;\;\;\;,\;\;\;\;\;𝑣_y=\frac{d Y }{dt}\;\;\;\;,\;\;\;\;\;𝑣_z=\frac{d Z }{dt}\;\;\;\;,\;\;\;\;\;\] Therefore, the object has acceleration on the three axes \[\vec a=\frac{d\vec v}{dt}=\frac{d 𝑣_x }{dt}\widehat x+\frac{d 𝑣_y }{dt}\widehat y+\frac{d 𝑣_z }{dt}\widehat z\] Cartesian components of acceleration \[a_x=\frac{d 𝑣_x }{dt}\;\;\;\;,\;\;\;\;\;a_y=\frac{d 𝑣_y }{dt}\;\;\;\;,\;\;\;\;\;a_z=\frac{d 𝑣_z }{dt}\;\;\;\;,\;\;\;\;\;\] Velocity and Acceleration in Two and Three Dimensions Motion in two or three dimensions: changing the magnitude or direction of velocity results in acceleration \[\vec a=\frac {\Delta\vec v}{\Delta t}=\frac{𝑣_2-𝑣_2}{t_2 -t_1}\]
Solved Example
A deer runs in a park in two dimensions
\[X(t)=0.5.t^2+12t+15\;\;\;\;\;\;Y(t)=-0.3.t^2+6t+20\]
where the units are \[X(m)\;\;\;Y(m)\;\;\;t(s)\]
Calculate the deer's position at time \[t=8\;\;s\]
\[X(8)=0.5× 8^2+12× 8+15=143 \;\;m \]\[Y(t)=-0.3× 8^2+6× 8+20=48.8\;\;m\] \[r=\sqrt {{r_x}^2+{r_y}^2}=\sqrt {{143}^2+{48.8}^2}=151.1 \;\;m\] Direction \[𝜃=tan^{-1}\frac {Y}{X}=tan^{-1}\frac {48.8}{143}=18.8^0\]Calculate the deer's velocity at time \[t=8\;\;s\]
\[V_x=\frac {dX}{dt} =d\frac {0.5.t^2+12t+15}{dt}=0.5×2×t+12×1=0.5×2×8+12×1=20\;\;m/s\] \[V_y=\frac {dY}{dt} =d\frac {-0.3.t^2+6t+20}{dt}=-0.3×2×t+6×1=-0.3×2×8+6×1=1.2\;\;m/s\] \[v=\sqrt {{v_x}^2+{v_y}^2}=\sqrt {{20}^2+{1.2}^2}=20.03 \;\;m/s\] Direction \[𝜃=tan^{-1}\frac {v_y}{v_x}=tan^{-1}\frac {1.2}{20}=3.4^0\]Calculate the deer's acceleration at time \[t=8\;\;s\]
\[a_x=\frac {dv_x}{dt} =d\frac {1t+12}{dt}=1×1+0=1\;\;m/s^2\] \[a_y=\frac {dv_Y}{dt} =d\frac {-0.6t+6}{dt}=1×-0.6+0=-0.6\;\;m/s^2\] \[a=\sqrt {{a_x}^2+{a_y}^2}=\sqrt {{1}^2+{-0.6}^2}=1.17 \;\;m/s^2\] Direction \[𝜃=tan^{-1}\frac {a_y}{a_x}=tan^{-1}\frac {-0.6}{1}=-31^0\] Ideal Projectile Motion An ideal projectile is any object launched with an initial velocity and moves under the influence of its weight while neglecting air resistance, wind speed, and object rotation, thus the projectile moves in two dimensionsWhen studying motion in two dimensions, it's easier to study the motion on each axis separately
The initial velocity has two components: horizontal and vertical \[v_x=v_0cos (𝜃) \] \[v_y=v_0sin (𝜃 )\]
Projectile at an angle experiment
Run the experiment and set the projectile height either from ground level or from a height within limits, set an appropriate initial velocity, specify the launch angle, and make the motion slow so you can take readings accurately. Take displacement readings on both axes every 0.5 seconds by clicking the pause button, then record the readings in the table below
Time (s)
Horizontal Distance (m)
Height (m)
Vertical Velocity (m/s)
What is the acceleration value?
Slope
= a = .......
From the above, we find the equations of motion on the X-axis:
\[X=v_0.Cos{𝜃}.t\]
The following graph shows the relationship between velocity and time for a projectile on the Y-axis:
What is the acceleration value?
Slope
= a = .......
From the above, we find the equations of motion on the Y-axis:
\[𝜗_𝑓𝑦= 𝜗_{0}sin 𝜃- g t\] \[𝜗_{𝑓𝑦}^2= (𝜗_{0}Sin 𝜃 )^2 -2.g.∆y\]\[∆𝑦=𝜗_{0}Sin 𝜃.t -\frac{1}{2}.g.t^2\]
Notes: Use the experiment to verify the results
The time of flight equals twice the time to reach maximum height
The projectile reaches maximum range when launched at 45 degrees
At maximum height, the vertical velocity becomes zero
Two projectiles can reach the same range when launched at angles that add up to 90 degrees
Projectile motion follows a parabolic path. To prove this:
\[X=v_0.Cos{𝜃}.t\Rightarrow\;\; t=\frac{x}{v_0.cos{𝜃}}\]
\[Y=Y_0+𝜗_{0}Sin 𝜃.t -\frac{1}{2}.g.t^2\]
\[Y=Y_0+𝜗_{0}Sin 𝜃.\frac{x}{v_0.cos{𝜃}} -\frac{1}{2}.g.(\frac{x}{v_0.cos{𝜃}})^2\]
\[Y=Y_0+Sin 𝜃.\frac {x}{cos{𝜃}} -\frac {1}{2}.g.\frac{x^2}{v_0^2.cos^2{𝜃}}\]
\[Y=Y_0+tan(𝜃).X -\frac {g}{2v_0^2.cos^2{𝜃}}.X^2\]
This equation relates the projectile's position on the horizontal axis to its position on the vertical axis
Which is the same as the equation of a parabola:
\[Y=C+bX+aX^2\]
Solved Example
A soccer player kicks a ball with velocity
\[v_0 = 25 m/s \]
at an angle
\[𝜃=33.1^0\]
above the horizontal
from a distance
from the goal
\[X=40.6 m\]
Did the ball reach the goal or go over it?
If the crossbar height is 2 meters, did the ball enter the goal or go over the crossbar?
We observe that:
\[Y_0=0\;\;\;\;\;v_0 = 25\;\; m/s\;\;\;\;\; 𝜃=33.1^0 \;\;\;\;\;X=40.6\;\; m\;\;\;\;\;Y=?\]
\[Y=Y_0+tan(𝜃).X -\frac {g}{2v_0^2.cos^2{𝜃}}.X^2\]
\[Y=0+tan(33.1)×40.6 -\frac {9.81}{2×25^2×cos^2{33.1}}×(40.6)^2= 8.03 m\]
The ball went over the crossbar
Each point on the trajectory has two velocity components: one on the X-axis
and one on the Y-axis
But the velocity at maximum height has only one component on the horizontal axis
The velocity on the X-axis is the projection of the initial velocity on the X-axis and is always constant:
\[𝜗_x= 𝜗_0 Cos𝜃\]
The velocity on the Y-axis is calculated from the equations of motion:
\[𝜗_𝑓𝑦= 𝜗_{0}sin 𝜃- g t\], \[𝜗_{𝑓𝑦}^2= (𝜗_{0}Sin 𝜃 )^2 -2.g.∆Y\]
Therefore, the total velocity at any position is calculated by:
\[𝜗=\sqrt{ 𝜗_X^2+𝜗_Y^2}\]
When a projectile is launched from the ground and returns to the ground:
The time to reach maximum height equals half the total time in the air
The launch velocity equals the landing velocity
At maximum height:
\[H=Y_0+\frac {v_{Y0}^2}{2g}\;\;\;\;\;\;\;\; v_{fY}=0\]
For calculating the horizontal range when:
\[Y=Y_0=0\]
\[R=X=\frac {v_0^2.}{g}.sin(2𝜃)\]
For calculating the horizontal range when:
\[Y_0≠0\]\[X=v_0.Cos{𝜃}.t\] Maximum range occurs when the launch angle is \[𝜃=45^0\]
Solved Example
A cannon on a hill 50 m above ground level fires a projectile with initial velocity \[v_0=83 \;\;m/s\] at an angle \[𝜃=40^0\] above the horizontal
Calculate the maximum height reached by the projectile
\[H=Y_0+\frac {v_{Y0}^2}{2g}\]
\[H=50+\frac {(83×sin (40))^2}{2×9.81}=195.1 m\]
Calculate the time needed for the projectile to reach the target
\[∆Y=𝜗_{0}Sin 𝜃.t -\frac{1}{2}.g.t^2\] \[0-50=83 ×Sin (40)×t -\frac{1}{2}×9.81×t^2\]
\[4.9.t^2 -53.35.t-50=0\]
\[t=11.76\;\;s\]
Calculate the horizontal range
\[X=v_0.Cos(𝜃).t\]\[X=83×Cos(40)(𝜃)11.76=747.7 \;\;m\]
Calculate the projectile's velocity upon impact
\[𝜗_X=𝜗_{0X}=𝜗_0.COS (𝜃)=83× COS (40)=63.6\;\;m/s\]
\[𝜗_𝑓Y= 𝜗_{0}sin (𝜃)- g t=83× sin (40)-9.81×11.76=-62 \;\;m/s\]
\[𝜗=\sqrt{(𝜗_X)^2+(𝜗_Y)^2}\] \[𝜗=\sqrt{(63.3)^2+(-62)^2}=88.6 \;\;m/s\]
Horizontal Projectile
This is an application of two-dimensional motion
A horizontal projectile has its initial velocity purely horizontal:
\[v_X=v_0\;\;\;\;\;\;\;\;\;v_{0Y}=0.0\]
Equations of motion on the X-axis:
\[X=v_0..t\] Equations of motion on the Y-axis:
\[𝜗_𝑓𝑦= - g t\], \[𝜗_{𝑓𝑦}^2= -2.g.∆Y\], \[∆Y= -\frac{1}{2}.g.t^2\]
Solved Example
A cannon is mounted on a building 50 m above ground level and fires a projectile horizontally at 90 m/s
Calculate the projectile's velocity upon impact
In ideal projectile motion, we neglect air resistance and the rotation of the projectile
Air resistance affects the horizontal range and maximum height of the projectile
It always opposes motion, reducing both horizontal and vertical motion components until stopping, then falling begins
The projectile's rotation causes deviation from its path, making the motion three-dimensional
Relative velocity is the velocity of a moving object relative to another object
Relative velocity changes based on direction of motion. If both objects move in the same direction
the observed object appears slower.
If objects move in opposite directions
the relative velocity appears increased
When stating an object's velocity, we must specify the reference frame
For example, Ahmed drives a car east at
\[20 \;\;m/s\]
Ahmed's velocity relative to the car:
\[\vec v_{AC}=0\]
Ahmed's velocity relative to the road:
\[\vec v_{Ae}=20 m/s\] east
The ground's velocity relative to Ahmed:
\[\vec v_{eA}=20 m/s\] west
A boat heads north at
\[20\;\;m/s\] across a river flowing east at
\[5\;\;m/s\]
Calculate the boat's velocity relative to an observer on the bank
\[v_{BR}=20\;\;m/s\] north
\[v_{Re}=20\;\;m/s\] east
\[\vec v_{Be}=\vec v_{BR}+\vec v_{Re}\]
\[v_{Be}=\sqrt {v_{BR}^2+v_{Re}^2}=\sqrt {20^2+5^2}=20.6 m/s\]
Direction:
\[𝜃=tan^{-1}\frac {v_{BR}}{v_{Re}}=tan^{-1}\frac {20}{5}=76^0\]
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