Physics Book - User contributions [en]

2-Dimensional Motion

2025-12-03T03:00:13Z

Ksuleimanova3:

'''Kseniia Suleimanova Fall 2025'''
==The Main Idea==
When objects move in 2-dimensional space, their motion can be described in <math>\hat{x},\ \hat{y}</math> coordinates. The motion for each of those axes can be viewed independently. Another approach is using vectors (i.e. coordinate (5, 3) can be seen as vector <math>\langle 5,\ 3 \rangle</math>).

===Displacement and distance===
Imagine we have 2 points and origin O: A = <math>\langle a,\ b \rangle</math>, B = <math>\langle c,\ d \rangle</math>

A point particle moves from origin to point A and then to point B.
The displacement can be viewed as adding those vectors:
<math>\langle a,\ b \rangle + \langle c,\ d \rangle = \langle a + c,\ b + d \rangle</math>.

The distance is the sum of magnitudes of those vectors:
<math>\sqrt{a^{2} + b^{2}} + \sqrt{c^{2} + d^{2}}</math>

===Moving with a constant velocity===
Velocity is the derivative of the position vector. In 2-dimensional space velocity looks the following:
<math>\vec{v} = \frac{\Delta \vec{r}}{\Delta t} = \left\langle \frac{\Delta x}{\Delta t},\frac{\Delta y}{\Delta t} \right\rangle</math>, so moving for <math>t</math> seconds with velocity <math>\vec{r}</math> can be represented as <math>\vec{v} \cdot t = \left\langle \frac{\Delta x}{\Delta t} \cdot t,\frac{\Delta y}{\Delta t} \cdot t \right\rangle</math>

===Moving with constant acceleration===
In 2-dimensional space acceleration is <math>\vec{a} = \frac{\Delta \vec{v}}{\Delta t}= \left\langle \frac{\Delta v_x}{\Delta t},\frac{\Delta v_y}{\Delta t} \right\rangle</math>.

A standard formula for position update is still applicable for vectors in 2-dimensional space: <math>\vec{r}(t)=\vec{r}_0+\vec{v}_0 t+\tfrac{1}{2}\vec{a}\,t^{2}=\langle x_0+v_{0x}t+\tfrac{1}{2}a_x t^{2},\ y_0+v_{0y}t+\tfrac{1}{2}a_y t^{2}\rangle</math>

==Computational models==

===Projectile motion===
https://www.glowscript.org/#/user/ksuleimanova/folder/MyPrograms/program/projectilemotiondemo - link to execute the code

[[File:p_motion.jpg]]

<pre>
Web VPython 3.2

scene.range = 10
scene.center = vector(5, 5, 0)
r = vector(0, 10, 0)
v = vector(3, 2, 0)
a = vector(0, -9.8, 0)

dt = 0.05

ball = sphere(pos=r, radius=0.2, color=color.red, make_trail=True)

gx = graph(title="Projectile Motion",
xtitle="Time, s", ytitle="Position, m")

xplot = gcurve(color=color.blue, label="x(t)")
yplot = gcurve(color=color.green, label="y(t)")

t = 0

while r.y >= 0:
rate(100)
v = v + a * dt
r = r + v * dt
ball.pos = r
xplot.plot(t, r.x)
yplot.plot(t, r.y)
t = t + dt
</pre>

In this code the position is updated every small increment of time, using the position update formula for constant acceleration, involving velocity first, and then position:

<pre>
v = v + a * dt
r = r + v * dt
</pre>

'''Circular motion'''

In the circular velocity magnitude is constant, while its direction changes.

https://www.glowscript.org/#/user/ksuleimanova/folder/MyPrograms/program/circularmotion - link to execute the code

[[File:circular_mo.jpg]]

<pre>
Web VPython 3.2

scene.width = 600
scene.height = 600
R = 5
omega = 1
dt = 0.05

particle = sphere(pos=vector(R,0,0), radius=0.5, color=color.red, make_trail=True)
arrow_velocity = arrow(pos=particle.pos, axis=vector(0,1,0), color=color.blue, shaftwidth=0.1)
arrow_accel = arrow(pos=particle.pos, axis=vector(0,1,0), color=color.green, shaftwidth=0.1)

t = 0

v_mag = 3
velocity = vector(0, v_mag, 0)
path = curve(color=color.yellow)

while t < 10:
particle.pos = particle.pos + velocity*dt
accel = - (velocity.mag**2 / R) * particle.pos.norm()
velocity = velocity + accel*dt
arrow_velocity.pos = particle.pos
arrow_velocity.axis = velocity.norm()
arrow_accel.pos = particle.pos
arrow_accel.axis = accel.norm()
path.append(pos=particle.pos)
t += dt
</pre>

This code also uses position update for each time increment.

==Examples==
===Easy===
'''Problem:'''
A point particle starts at a point (0, 5)m on a flat surface an moves with a constant velocity <math>\langle 4,\ 1 \rangle m</math>. Find it's position after 5 seconds.

'''Solution:'''
Intial position vector is <math>\langle 0,\ 5 \rangle m</math>
As the velocity is constant, acceleration is 0. Insert all the data into the position update formula:
<math>\vec{r}(t)=\vec{r}_0+\vec{v}_0 t+\tfrac{1}{2}\vec{a}\,t^{2}=\vec{r}_0+\vec{v}_0 t + 0 = \langle 0,\ 5 \rangle + \langle 4,\ 1 \rangle \cdot 5 = \langle 20,\ 10 \rangle m</math>

===Medium===
'''Problem:'''
A bug on the table travels from point (1, 2)m to point (2, 1)m and then to (2, 2)m. The whole path takes 2 seconds. Find average speed and average velocity.

'''Solution:'''
Displacement is <math>\langle 2,\ 2 \rangle - \langle 1,\ 2 \rangle = \langle 1,\ 0 \rangle m</math>
Distance is the length of the path: <math>\sqrt{(2 - 1)^{2} + (1 - 2)^{2}} + \sqrt{(2 - 2)^{2} + (2 - 1)^{2}} = \sqrt{2} + \sqrt{1} = 2.41 m</math>

Average speed: <math>2.41 / 2 = 1.21 m/s</math>

Average velocity: <math>\langle 1,\ 0 \rangle / 2 = \langle 0.5,\ 0 \rangle m/s</math>

===Hard===
'''Problem:'''
The ball is thrown from the building with the height 10 meters. The initial velocity of the ball is <math>\langle 3,\ 2 \rangle \textit{m/s}</math> find the time and position when the ball reaches its maximum height.

'''Solution:'''
This is the projectile motion problem in 2-dimensional space. The most convenient approach is to view vertical and horizontal motion independently. We know that horizontal component of the velocity stays the same, while <math>\vec{v}_{y} = \vec{v}_{y_{init}} - g t</math> - the ball moves with a constant acceleration. For the <math>\hat{y}</math> we can use the position update formula for constant acceleration <math>-g</math>: <math>r_{y}(t)=r_{y_0}+\vec{v_{y_0}} t-\tfrac{1}{2}\vec{g}\,t^{2}</math>. This is a parabola that has a maximum value. Insert numerical values and solve the equation: <math>r_{y}(t)=10+2 t-\tfrac{1}{2}\cdot 9.8 \cdot t^{2}</math>. Maximum value is achieved at <math>t = 0.21 s</math>. At <math>t_{max} = 0.21 s</math> x-coordinate is <math>r_{x_{init}} = \vec{v_x} \cdot t_{max} = 0 + 3 \cdot 0.21 = 0.63 m</math> y-coordinate is <math>r_{y}(0.21)=r_{y_0}+\vec{v_{y_0}} \cdot 0.21-\tfrac{1}{2}\vec{g} \cdot 0.21^{2} = 10 + 3 \cdot 0.21 - \tfrac{1}{2} \cdot 9.8 \cdot 0.21 = 10.41 m</math>

==Links To Other Topics==
* 2-dimensional motion relies on vectors most of the time
* Most projectile motion problems are 2-dimensional
* Centripital motion problems are 2-dimensional as well

==See Also==
https://youtu.be/w3BhzYI6zXU?si=oivAjOifOPBi_5D4 - vectors and 2D motion

https://youtu.be/On5DpeGQ89I?si=7zqYvWx73W3HYwjG - 2D motion problems with solutions

==References==
https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/pages/week-1-kinematics/3-1-coordinate-system-and-position-vector-in-2d/

https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/pages/week-1-kinematics/3-4-projectile-motion/

2-Dimensional Motion

2025-12-03T02:59:27Z

Ksuleimanova3:

'''Kseniia Suleimanova Fall 2025'''
==The Main Idea==
When objects move in 2-dimensional space, their motion can be described in <math>\hat{x},\ \hat{y}</math> coordinates. The motion for each of those axes can be viewed independently. Another approach is using vectors (i.e. coordinate (5, 3) can be seen as vector <math>\langle 5,\ 3 \rangle</math>).

===Displacement and distance===
Imagine we have 2 points and origin O: A = <math>\langle a,\ b \rangle</math>, B = <math>\langle c,\ d \rangle</math>

A point particle moves from origin to point A and then to point B.
The displacement can be viewed as adding those vectors:
<math>\langle a,\ b \rangle + \langle c,\ d \rangle = \langle a + c,\ b + d \rangle</math>.

The distance is the sum of magnitudes of those vectors:
<math>\sqrt{a^{2} + b^{2}} + \sqrt{c^{2} + d^{2}}</math>

===Moving with a constant velocity===
Velocity is the derivative of the position vector. In 2-dimensional space velocity looks the following:
<math>\vec{v} = \frac{\Delta \vec{r}}{\Delta t} = \left\langle \frac{\Delta x}{\Delta t},\frac{\Delta y}{\Delta t} \right\rangle</math>, so moving for <math>t</math> seconds with velocity <math>\vec{r}</math> can be represented as <math>\vec{v} \cdot t = \left\langle \frac{\Delta x}{\Delta t} \cdot t,\frac{\Delta y}{\Delta t} \cdot t \right\rangle</math>

===Moving with constant acceleration===
In 2-dimensional space acceleration is <math>\vec{a} = \frac{\Delta \vec{v}}{\Delta t}= \left\langle \frac{\Delta v_x}{\Delta t},\frac{\Delta v_y}{\Delta t} \right\rangle</math>.

A standard formula for position update is still applicable for vectors in 2-dimensional space: <math>\vec{r}(t)=\vec{r}_0+\vec{v}_0 t+\tfrac{1}{2}\vec{a}\,t^{2}=\langle x_0+v_{0x}t+\tfrac{1}{2}a_x t^{2},\ y_0+v_{0y}t+\tfrac{1}{2}a_y t^{2}\rangle</math>

==Computational models==

===Projectile motion===
https://www.glowscript.org/#/user/ksuleimanova/folder/MyPrograms/program/projectilemotiondemo - link to execute the code

[[File:p_motion.jpg]]

<pre>
Web VPython 3.2

scene.range = 10
scene.center = vector(5, 5, 0)
r = vector(0, 10, 0)
v = vector(3, 2, 0)
a = vector(0, -9.8, 0)

dt = 0.05

ball = sphere(pos=r, radius=0.2, color=color.red, make_trail=True)

gx = graph(title="Projectile Motion",
xtitle="Time, s", ytitle="Position, m")

xplot = gcurve(color=color.blue, label="x(t)")
yplot = gcurve(color=color.green, label="y(t)")

t = 0

while r.y >= 0:
rate(100)
v = v + a * dt
r = r + v * dt
ball.pos = r
xplot.plot(t, r.x)
yplot.plot(t, r.y)
t = t + dt
</pre>

In this code the position is updated every small increment of time, using the position update formula for constant acceleration, involving velocity first, and then position:

<pre>
v = v + a * dt
r = r + v * dt
</pre>

'''Circular motion'''

In the circular velocity magnitude is constant, while its direction changes.

https://www.glowscript.org/#/user/ksuleimanova/folder/MyPrograms/program/circularmotion - link to execute the code

[[File:circular_mo.jpg]]

<pre>
Web VPython 3.2

scene.width = 600
scene.height = 600
R = 5
omega = 1
dt = 0.05

particle = sphere(pos=vector(R,0,0), radius=0.5, color=color.red, make_trail=True)
arrow_velocity = arrow(pos=particle.pos, axis=vector(0,1,0), color=color.blue, shaftwidth=0.1)
arrow_accel = arrow(pos=particle.pos, axis=vector(0,1,0), color=color.green, shaftwidth=0.1)

t = 0

v_mag = 3
velocity = vector(0, v_mag, 0)
path = curve(color=color.yellow)

while t < 10:
particle.pos = particle.pos + velocity*dt
accel = - (velocity.mag**2 / R) * particle.pos.norm()
velocity = velocity + accel*dt
arrow_velocity.pos = particle.pos
arrow_velocity.axis = velocity.norm()
arrow_accel.pos = particle.pos
arrow_accel.axis = accel.norm()
path.append(pos=particle.pos)
t += dt
</pre>

==Examples==
===Easy===
'''Problem:'''
A point particle starts at a point (0, 5)m on a flat surface an moves with a constant velocity <math>\langle 4,\ 1 \rangle m</math>. Find it's position after 5 seconds.

'''Solution:'''
Intial position vector is <math>\langle 0,\ 5 \rangle m</math>
As the velocity is constant, acceleration is 0. Insert all the data into the position update formula:
<math>\vec{r}(t)=\vec{r}_0+\vec{v}_0 t+\tfrac{1}{2}\vec{a}\,t^{2}=\vec{r}_0+\vec{v}_0 t + 0 = \langle 0,\ 5 \rangle + \langle 4,\ 1 \rangle \cdot 5 = \langle 20,\ 10 \rangle m</math>

===Medium===
'''Problem:'''
A bug on the table travels from point (1, 2)m to point (2, 1)m and then to (2, 2)m. The whole path takes 2 seconds. Find average speed and average velocity.

'''Solution:'''
Displacement is <math>\langle 2,\ 2 \rangle - \langle 1,\ 2 \rangle = \langle 1,\ 0 \rangle m</math>
Distance is the length of the path: <math>\sqrt{(2 - 1)^{2} + (1 - 2)^{2}} + \sqrt{(2 - 2)^{2} + (2 - 1)^{2}} = \sqrt{2} + \sqrt{1} = 2.41 m</math>

Average speed: <math>2.41 / 2 = 1.21 m/s</math>

Average velocity: <math>\langle 1,\ 0 \rangle / 2 = \langle 0.5,\ 0 \rangle m/s</math>

===Hard===
'''Problem:'''
The ball is thrown from the building with the height 10 meters. The initial velocity of the ball is <math>\langle 3,\ 2 \rangle \textit{m/s}</math> find the time and position when the ball reaches its maximum height.

'''Solution:'''
This is the projectile motion problem in 2-dimensional space. The most convenient approach is to view vertical and horizontal motion independently. We know that horizontal component of the velocity stays the same, while <math>\vec{v}_{y} = \vec{v}_{y_{init}} - g t</math> - the ball moves with a constant acceleration. For the <math>\hat{y}</math> we can use the position update formula for constant acceleration <math>-g</math>: <math>r_{y}(t)=r_{y_0}+\vec{v_{y_0}} t-\tfrac{1}{2}\vec{g}\,t^{2}</math>. This is a parabola that has a maximum value. Insert numerical values and solve the equation: <math>r_{y}(t)=10+2 t-\tfrac{1}{2}\cdot 9.8 \cdot t^{2}</math>. Maximum value is achieved at <math>t = 0.21 s</math>. At <math>t_{max} = 0.21 s</math> x-coordinate is <math>r_{x_{init}} = \vec{v_x} \cdot t_{max} = 0 + 3 \cdot 0.21 = 0.63 m</math> y-coordinate is <math>r_{y}(0.21)=r_{y_0}+\vec{v_{y_0}} \cdot 0.21-\tfrac{1}{2}\vec{g} \cdot 0.21^{2} = 10 + 3 \cdot 0.21 - \tfrac{1}{2} \cdot 9.8 \cdot 0.21 = 10.41 m</math>

==Links To Other Topics==
* 2-dimensional motion relies on vectors most of the time
* Most projectile motion problems are 2-dimensional
* Centripital motion problems are 2-dimensional as well

==See Also==
https://youtu.be/w3BhzYI6zXU?si=oivAjOifOPBi_5D4 - vectors and 2D motion

https://youtu.be/On5DpeGQ89I?si=7zqYvWx73W3HYwjG - 2D motion problems with solutions

==References==
https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/pages/week-1-kinematics/3-1-coordinate-system-and-position-vector-in-2d/

https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/pages/week-1-kinematics/3-4-projectile-motion/

File:Circular mo.jpg

2025-12-03T02:57:20Z

Ksuleimanova3:

File:P motion.jpg

2025-12-03T02:45:45Z

Ksuleimanova3:

2-Dimensional Motion

2025-12-03T02:44:40Z

Ksuleimanova3: /* Computational models */

'''Kseniia Suleimanova Fall 2025'''
==The Main Idea==
When objects move in 2-dimensional space, their motion can be described in <math>\hat{x},\ \hat{y}</math> coordinates. The motion for each of those axes can be viewed independently. Another approach is using vectors (i.e. coordinate (5, 3) can be seen as vector <math>\langle 5,\ 3 \rangle</math>).

===Displacement and distance===
Imagine we have 2 points and origin O: A = <math>\langle a,\ b \rangle</math>, B = <math>\langle c,\ d \rangle</math>

A point particle moves from origin to point A and then to point B.
The displacement can be viewed as adding those vectors:
<math>\langle a,\ b \rangle + \langle c,\ d \rangle = \langle a + c,\ b + d \rangle</math>.

The distance is the sum of magnitudes of those vectors:
<math>\sqrt{a^{2} + b^{2}} + \sqrt{c^{2} + d^{2}}</math>

===Moving with a constant velocity===
Velocity is the derivative of the position vector. In 2-dimensional space velocity looks the following:
<math>\vec{v} = \frac{\Delta \vec{r}}{\Delta t} = \left\langle \frac{\Delta x}{\Delta t},\frac{\Delta y}{\Delta t} \right\rangle</math>, so moving for <math>t</math> seconds with velocity <math>\vec{r}</math> can be represented as <math>\vec{v} \cdot t = \left\langle \frac{\Delta x}{\Delta t} \cdot t,\frac{\Delta y}{\Delta t} \cdot t \right\rangle</math>

===Moving with constant acceleration===
In 2-dimensional space acceleration is <math>\vec{a} = \frac{\Delta \vec{v}}{\Delta t}= \left\langle \frac{\Delta v_x}{\Delta t},\frac{\Delta v_y}{\Delta t} \right\rangle</math>.

A standard formula for position update is still applicable for vectors in 2-dimensional space: <math>\vec{r}(t)=\vec{r}_0+\vec{v}_0 t+\tfrac{1}{2}\vec{a}\,t^{2}=\langle x_0+v_{0x}t+\tfrac{1}{2}a_x t^{2},\ y_0+v_{0y}t+\tfrac{1}{2}a_y t^{2}\rangle</math>

==Computational models==

===Projectile motion===
https://www.glowscript.org/#/user/ksuleimanova/folder/MyPrograms/program/projectilemotiondemo - link to execute the code

<pre>
Web VPython 3.2

scene.range = 10
scene.center = vector(5, 5, 0)
r = vector(0, 10, 0)
v = vector(3, 2, 0)
a = vector(0, -9.8, 0)

dt = 0.05

ball = sphere(pos=r, radius=0.2, color=color.red, make_trail=True)

gx = graph(title="Projectile Motion",
xtitle="Time, s", ytitle="Position, m")

xplot = gcurve(color=color.blue, label="x(t)")
yplot = gcurve(color=color.green, label="y(t)")

t = 0

while r.y >= 0:
rate(100)
v = v + a * dt
r = r + v * dt
ball.pos = r
xplot.plot(t, r.x)
yplot.plot(t, r.y)
t = t + dt
</pre>

==Examples==
===Easy===
'''Problem:'''
A point particle starts at a point (0, 5)m on a flat surface an moves with a constant velocity <math>\langle 4,\ 1 \rangle m</math>. Find it's position after 5 seconds.

'''Solution:'''
Intial position vector is <math>\langle 0,\ 5 \rangle m</math>
As the velocity is constant, acceleration is 0. Insert all the data into the position update formula:
<math>\vec{r}(t)=\vec{r}_0+\vec{v}_0 t+\tfrac{1}{2}\vec{a}\,t^{2}=\vec{r}_0+\vec{v}_0 t + 0 = \langle 0,\ 5 \rangle + \langle 4,\ 1 \rangle \cdot 5 = \langle 20,\ 10 \rangle m</math>

===Medium===
'''Problem:'''
A bug on the table travels from point (1, 2)m to point (2, 1)m and then to (2, 2)m. The whole path takes 2 seconds. Find average speed and average velocity.

'''Solution:'''
Displacement is <math>\langle 2,\ 2 \rangle - \langle 1,\ 2 \rangle = \langle 1,\ 0 \rangle m</math>
Distance is the length of the path: <math>\sqrt{(2 - 1)^{2} + (1 - 2)^{2}} + \sqrt{(2 - 2)^{2} + (2 - 1)^{2}} = \sqrt{2} + \sqrt{1} = 2.41 m</math>

Average speed: <math>2.41 / 2 = 1.21 m/s</math>

Average velocity: <math>\langle 1,\ 0 \rangle / 2 = \langle 0.5,\ 0 \rangle m/s</math>

===Hard===
'''Problem:'''
The ball is thrown from the building with the height 10 meters. The initial velocity of the ball is <math>\langle 3,\ 2 \rangle \textit{m/s}</math> find the time and position when the ball reaches its maximum height.

'''Solution:'''
This is the projectile motion problem in 2-dimensional space. The most convenient approach is to view vertical and horizontal motion independently. We know that horizontal component of the velocity stays the same, while <math>\vec{v}_{y} = \vec{v}_{y_{init}} - g t</math> - the ball moves with a constant acceleration. For the <math>\hat{y}</math> we can use the position update formula for constant acceleration <math>-g</math>: <math>r_{y}(t)=r_{y_0}+\vec{v_{y_0}} t-\tfrac{1}{2}\vec{g}\,t^{2}</math>. This is a parabola that has a maximum value. Insert numerical values and solve the equation: <math>r_{y}(t)=10+2 t-\tfrac{1}{2}\cdot 9.8 \cdot t^{2}</math>. Maximum value is achieved at <math>t = 0.21 s</math>. At <math>t_{max} = 0.21 s</math> x-coordinate is <math>r_{x_{init}} = \vec{v_x} \cdot t_{max} = 0 + 3 \cdot 0.21 = 0.63 m</math> y-coordinate is <math>r_{y}(0.21)=r_{y_0}+\vec{v_{y_0}} \cdot 0.21-\tfrac{1}{2}\vec{g} \cdot 0.21^{2} = 10 + 3 \cdot 0.21 - \tfrac{1}{2} \cdot 9.8 \cdot 0.21 = 10.41 m</math>

==Links To Other Topics==
* 2-dimensional motion relies on vectors most of the time
* Most projectile motion problems are 2-dimensional
* Centripital motion problems are 2-dimensional as well

==See Also==
https://youtu.be/w3BhzYI6zXU?si=oivAjOifOPBi_5D4 - vectors and 2D motion

https://youtu.be/On5DpeGQ89I?si=7zqYvWx73W3HYwjG - 2D motion problems with solutions

==References==
https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/pages/week-1-kinematics/3-1-coordinate-system-and-position-vector-in-2d/

https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/pages/week-1-kinematics/3-4-projectile-motion/

2-Dimensional Motion

2025-12-03T01:02:20Z

Ksuleimanova3:

'''Kseniia Suleimanova Fall 2025'''
==The Main Idea==
When objects move in 2-dimensional space, their motion can be described in <math>\hat{x},\ \hat{y}</math> coordinates. The motion for each of those axes can be viewed independently. Another approach is using vectors (i.e. coordinate (5, 3) can be seen as vector <math>\langle 5,\ 3 \rangle</math>).

===Displacement and distance===
Imagine we have 2 points and origin O: A = <math>\langle a,\ b \rangle</math>, B = <math>\langle c,\ d \rangle</math>

A point particle moves from origin to point A and then to point B.
The displacement can be viewed as adding those vectors:
<math>\langle a,\ b \rangle + \langle c,\ d \rangle = \langle a + c,\ b + d \rangle</math>.

The distance is the sum of magnitudes of those vectors:
<math>\sqrt{a^{2} + b^{2}} + \sqrt{c^{2} + d^{2}}</math>

===Moving with a constant velocity===
Velocity is the derivative of the position vector. In 2-dimensional space velocity looks the following:
<math>\vec{v} = \frac{\Delta \vec{r}}{\Delta t} = \left\langle \frac{\Delta x}{\Delta t},\frac{\Delta y}{\Delta t} \right\rangle</math>, so moving for <math>t</math> seconds with velocity <math>\vec{r}</math> can be represented as <math>\vec{v} \cdot t = \left\langle \frac{\Delta x}{\Delta t} \cdot t,\frac{\Delta y}{\Delta t} \cdot t \right\rangle</math>

===Moving with constant acceleration===
In 2-dimensional space acceleration is <math>\vec{a} = \frac{\Delta \vec{v}}{\Delta t}= \left\langle \frac{\Delta v_x}{\Delta t},\frac{\Delta v_y}{\Delta t} \right\rangle</math>.

A standard formula for position update is still applicable for vectors in 2-dimensional space: <math>\vec{r}(t)=\vec{r}_0+\vec{v}_0 t+\tfrac{1}{2}\vec{a}\,t^{2}=\langle x_0+v_{0x}t+\tfrac{1}{2}a_x t^{2},\ y_0+v_{0y}t+\tfrac{1}{2}a_y t^{2}\rangle</math>

==Computational models==

==Examples==
===Easy===
'''Problem:'''
A point particle starts at a point (0, 5)m on a flat surface an moves with a constant velocity <math>\langle 4,\ 1 \rangle m</math>. Find it's position after 5 seconds.

'''Solution:'''
Intial position vector is <math>\langle 0,\ 5 \rangle m</math>
As the velocity is constant, acceleration is 0. Insert all the data into the position update formula:
<math>\vec{r}(t)=\vec{r}_0+\vec{v}_0 t+\tfrac{1}{2}\vec{a}\,t^{2}=\vec{r}_0+\vec{v}_0 t + 0 = \langle 0,\ 5 \rangle + \langle 4,\ 1 \rangle \cdot 5 = \langle 20,\ 10 \rangle m</math>

===Medium===
'''Problem:'''
A bug on the table travels from point (1, 2)m to point (2, 1)m and then to (2, 2)m. The whole path takes 2 seconds. Find average speed and average velocity.

'''Solution:'''
Displacement is <math>\langle 2,\ 2 \rangle - \langle 1,\ 2 \rangle = \langle 1,\ 0 \rangle m</math>
Distance is the length of the path: <math>\sqrt{(2 - 1)^{2} + (1 - 2)^{2}} + \sqrt{(2 - 2)^{2} + (2 - 1)^{2}} = \sqrt{2} + \sqrt{1} = 2.41 m</math>

Average speed: <math>2.41 / 2 = 1.21 m/s</math>

Average velocity: <math>\langle 1,\ 0 \rangle / 2 = \langle 0.5,\ 0 \rangle m/s</math>

===Hard===
'''Problem:'''
The ball is thrown from the building with the height 10 meters. The initial velocity of the ball is <math>\langle 3,\ 2 \rangle \textit{m/s}</math> find the time and position when the ball reaches its maximum height.

'''Solution:'''
This is the projectile motion problem in 2-dimensional space. The most convenient approach is to view vertical and horizontal motion independently. We know that horizontal component of the velocity stays the same, while <math>\vec{v}_{y} = \vec{v}_{y_{init}} - g t</math> - the ball moves with a constant acceleration. For the <math>\hat{y}</math> we can use the position update formula for constant acceleration <math>-g</math>: <math>r_{y}(t)=r_{y_0}+\vec{v_{y_0}} t-\tfrac{1}{2}\vec{g}\,t^{2}</math>. This is a parabola that has a maximum value. Insert numerical values and solve the equation: <math>r_{y}(t)=10+2 t-\tfrac{1}{2}\cdot 9.8 \cdot t^{2}</math>. Maximum value is achieved at <math>t = 0.21 s</math>. At <math>t_{max} = 0.21 s</math> x-coordinate is <math>r_{x_{init}} = \vec{v_x} \cdot t_{max} = 0 + 3 \cdot 0.21 = 0.63 m</math> y-coordinate is <math>r_{y}(0.21)=r_{y_0}+\vec{v_{y_0}} \cdot 0.21-\tfrac{1}{2}\vec{g} \cdot 0.21^{2} = 10 + 3 \cdot 0.21 - \tfrac{1}{2} \cdot 9.8 \cdot 0.21 = 10.41 m</math>

==Links To Other Topics==
* 2-dimensional motion relies on vectors most of the time
* Most projectile motion problems are 2-dimensional
* Centripital motion problems are 2-dimensional as well

==See Also==
https://youtu.be/w3BhzYI6zXU?si=oivAjOifOPBi_5D4 - vectors and 2D motion

https://youtu.be/On5DpeGQ89I?si=7zqYvWx73W3HYwjG - 2D motion problems with solutions

==References==
https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/pages/week-1-kinematics/3-1-coordinate-system-and-position-vector-in-2d/

https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/pages/week-1-kinematics/3-4-projectile-motion/

2-Dimensional Motion

2025-12-03T00:17:31Z

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2-Dimensional Motion

2025-12-02T23:50:59Z

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2-Dimensional Motion

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Translational Angular Momentum

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2-Dimensional Motion

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Translational Angular Momentum

2025-12-02T23:07:56Z

Ksuleimanova3:

'''Kseniia Suleimanova Fall 2025'''

==The Main Idea==
Translational angular momentum is used when compution the motion of rotating systems. Translational angular momentum is the part of total angular momentum responsible for the motion of the center of mass with respect to the some point.

==Examples==

==History==

==See Also==
===Videos===
https://youtu.be/V1I-vrXGl3A?si=G30q3yECGvbwonP7

==References==

Translational Angular Momentum

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Ksuleimanova3:

'''Kseniia Suleimanova Fall 2025'''

==The Main Idea==

==Examples==

==History==

==See Also==
===Videos===
https://youtu.be/V1I-vrXGl3A?si=G30q3yECGvbwonP7

==References==

Translational Angular Momentum

2025-12-02T16:24:23Z

Ksuleimanova3:

'''Kseniia Suleimanova Fall 2025'''

==The Main Idea==

Translational Angular Momentum

2025-12-02T16:17:57Z

Ksuleimanova3:

'''Kseniia Suleimanova Fall 2025'''

Translational Angular Momentum

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Kseniia Suleimanova Fall 2025