2-Dimensional Motion - Revision history

Ksuleimanova3 at 03:00, 3 December 2025

2025-12-03T03:00:13Z

← Older revision		Revision as of 23:00, 2 December 2025
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	t += dt		t += dt
	</pre>		</pre>

			This code also uses position update for each time increment.

Ksuleimanova3 at 02:59, 3 December 2025

2025-12-03T02:59:27Z

@@ Line 27: / Line 27: @@
 ===Projectile motion===
 https://www.glowscript.org/#/user/ksuleimanova/folder/MyPrograms/program/projectilemotiondemo - link to execute the code
 <pre>
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      t = t + dt
 </pre>
 ==Examples==
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 '''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:'''

Ksuleimanova3: /* Computational models */

2025-12-03T02:44:40Z

Computational models

← Older revision		Revision as of 22:44, 2 December 2025
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	==Computational models==		==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==		==Examples==

Ksuleimanova3 at 01:02, 3 December 2025

2025-12-03T01:02:20Z

← Older revision		Revision as of 21:02, 2 December 2025
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	==Examples==		==Examples==
	===Easy===		===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===		===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===		===Hard===
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	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>		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>

	==~~History~~==		==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==		==See Also==
	https://youtu.be/w3BhzYI6zXU?si=oivAjOifOPBi_5D4 - vectors and 2D motion		https://youtu.be/w3BhzYI6zXU?si=oivAjOifOPBi_5D4 - vectors and 2D motion
	https://youtu.be/On5DpeGQ89I?si=7zqYvWx73W3HYwjG - 2D motion problems
			https://youtu.be/On5DpeGQ89I?si=7zqYvWx73W3HYwjG - 2D motion problems with solutions


	==References==		==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/

Ksuleimanova3 at 00:17, 3 December 2025

2025-12-03T00:17:31Z

← Older revision		Revision as of 20:17, 2 December 2025
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	==Examples==		==Examples==
			===Easy===

			===Medium===

			===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>

	==History==		==History==

	==See Also==		==See Also==
	=~~==Videos===~~		https://youtu.be/w3BhzYI6zXU?si=oivAjOifOPBi_5D4 - vectors and 2D motion
	https://youtu.be/~~V1I-vrXGl3A~~?si=~~G30q3yECGvbwonP7~~		https://youtu.be/On5DpeGQ89I?si=7zqYvWx73W3HYwjG - 2D motion problems


	==References==		==References==

Ksuleimanova3 at 23:50, 2 December 2025

2025-12-02T23:50:59Z

← Older revision		Revision as of 19:50, 2 December 2025
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	<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>		<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==

Ksuleimanova3 at 23:46, 2 December 2025

2025-12-02T23:46:32Z

← Older revision		Revision as of 19:46, 2 December 2025
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	'''Kseniia Suleimanova Fall 2025'''		'''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>




			==Examples==

			==History==

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

			==References==

Ksuleimanova3: Created page with "'''Kseniia Suleimanova Fall 2025'''"

2025-12-02T23:09:57Z

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