Question

one-dimensional - cars traveling
two-dimensional - airplanes taking off and landing
laws of motion: 3 equations -
v = v₀ + a*t
v² = v₀² + 2a*d
d = v₀t + ½a*t²

• v = final speed
• v₀ - initial speed
• t = time
• a = acceleration
• d - displacement - how far it went

acceleration (change in speed) - physics equations
lack of acceleration

you will create sample real world physics problems that represent two-dimensional motion. you will need to create two-part problems that use at least two similar and/or different two-dimensional motion equations from the equations introduced in this part of the unit. you must create at least 3 word problems using all three equations at least once. all the 3 word problems must utilize different two-dimensional motion scenarios.

components:

1. write out your problem.
2. graph your problem and solve the equations with your thinking. (you will keep this separate before the next task.)

Answer:

To fulfill the task of creating sample real - world physics problems for two - dimensional motion, we will design three problems, each using a different equation of motion at least once and set in different two - dimensional motion scenarios.

Problem 1: Projectile Motion (Throwing a Ball)

Step 1: Define the Problem

A boy throws a ball from the top of a building that is 20 m high with an initial horizontal speed \(v_{0x}=10\ m/s\) and an initial vertical speed \(v_{0y} = 0\ m/s\). The acceleration due to gravity \(a=- 9.8\ m/s^{2}\) (negative because it acts downwards). We want to find out:
a) How long does it take for the ball to hit the ground? (Use the equation \(d = v_{0}t+\frac{1}{2}at^{2}\) for vertical motion)
b) How far from the base of the building does the ball land? (Use the formula for horizontal motion \(x=v_{0x}t\) since there is no horizontal acceleration)

Step 2: Solve Part (a)

For vertical motion, the displacement \(d=- 20\ m\) (negative because it is in the downward direction), \(v_{0y} = 0\ m/s\) and \(a=-9.8\ m/s^{2}\).
Using the equation \(d=v_{0y}t+\frac{1}{2}at^{2}\)
Substitute the values: \(-20=0\times t+\frac{1}{2}\times(-9.8)\times t^{2}\)
Simplify: \(-20=- 4.9t^{2}\)
Then \(t^{2}=\frac{20}{4.9}\approx4.08\)
\(t=\sqrt{4.08}\approx2.02\ s\)

Step 3: Solve Part (b)

For horizontal motion, \(v_{0x} = 10\ m/s\) and \(t = 2.02\ s\)
Using \(x = v_{0x}t\)
\(x=10\times2.02 = 20.2\ m\)

Problem 2: Airplane Take - off (Combined Horizontal and Vertical Acceleration)

Step 1: Define the Problem

An airplane is taking off. It has an initial horizontal speed \(v_{0x}=30\ m/s\) and an initial vertical speed \(v_{0y}=0\ m/s\). The horizontal acceleration \(a_{x}=2\ m/s^{2}\) and the vertical acceleration \(a_{y}=1\ m/s^{2}\). After \(t = 10\ s\):
a) What is the final horizontal speed? (Use \(v = v_{0}+at\) for horizontal motion)
b) What is the final vertical speed? (Use \(v = v_{0}+at\) for vertical motion)
c) What is the displacement in the horizontal direction? (Use \(d=v_{0}t+\frac{1}{2}at^{2}\) for horizontal motion)
d) What is the displacement in the vertical direction? (Use \(d=v_{0}t+\frac{1}{2}at^{2}\) for vertical motion)

Step 2: Solve Part (a)

For horizontal motion, \(v_{0x}=30\ m/s\), \(a_{x}=2\ m/s^{2}\), \(t = 10\ s\)
Using \(v_{x}=v_{0x}+a_{x}t\)
\(v_{x}=30 + 2\times10=30 + 20=50\ m/s\)

Step 3: Solve Part (b)

For vertical motion, \(v_{0y}=0\ m/s\), \(a_{y}=1\ m/s^{2}\), \(t = 10\ s\)
Using \(v_{y}=v_{0y}+a_{y}t\)
\(v_{y}=0+1\times10 = 10\ m/s\)

Step 4: Solve Part (c)

For horizontal displacement, \(v_{0x}=30\ m/s\), \(a_{x}=2\ m/s^{2}\), \(t = 10\ s\)
Using \(x=v_{0x}t+\frac{1}{2}a_{x}t^{2}\)
\(x=30\times10+\frac{1}{2}\times2\times10^{2}=300 + 100=400\ m\)

Step 5: Solve Part (d)

For vertical displacement, \(v_{0y}=0\ m/s\), \(a_{y}=1\ m/s^{2}\), \(t = 10\ s\)
Using \(y=v_{0y}t+\frac{1}{2}a_{y}t^{2}\)
\(y=0\times10+\frac{1}{2}\times1\times10^{2}=50\ m\)

Problem 3: Helicopter Motion (Changing Speed and Direction)

Step 1: Define the Problem

A helicopter is moving in two - dimensions. Initially, its speed is \(v_{0}=25\ m/s\). It accelerates with an acceleration \(a = 3\ m/s^{2}\) for a distance \(d = 100\ m\). Then it changes its acceleration to \(a'=- 2\ m/s^{2}\) (deceleration) and travels a distance \(d'=50\ m\).
a) What is the speed of the helicopter after the first part of the motion? (Use \(v^{2}=v_{0}^{2}+2ad\))
b) What is the speed of the helicopter after the second part of the motion? (Use \(v^{2}=v_{1}^{2}+2a'd'\), where \(v_{1}\) is the speed after the first part)

Step 2: Solve Part (a)

Given \(v_{0}=25\ m/s\), \(a = 3\ m/s^{2}\), \(d = 100\ m\)
Using \(v^{2}=v_{0}^{2}+2ad\)
\(v_{1}^{2}=25^{2}+2\times3\times100=625 + 600 = 1225\)
\(v_{1}=\sqrt{1225}=35\ m/s\)

Step 3: Solve Part (b)

Now, \(v_{1}=35\ m/s\), \(a'=- 2\ m/s^{2}\), \(d' = 50\ m\)
Using \(v^{2}=v_{1}^{2}+2a'd'\)
\(v_{2}^{2}=35^{2}+2\times(-2)\times50=1225-200 = 1025\)
\(v_{2}=\sqrt{1025}\approx32.02\ m/s\)

These three problems cover different two - dimensional motion scenarios (projectile motion, airplane take - off, helicopter motion) and use the three equations of motion (\(v = v_{0}+at\), \(v^{2}=v_{0}^{2}+2ad\), \(d = v_{0}t+\frac{1}{2}at^{2}\)) at least once.