In this challenge Hank and I paired up with another group to predict where two wind cars would intersect. First, we used a motion sensor to find the initial velocity and the acceleration of cart D. We tested this three times and then averaged our slopes to get an average acceleration of: .1203 m/s^2.
Our initial velocity was Vi: -.18.We recorded 1m away from the sensor.
Prediction: In order to make a prediction, we used the formula ∆x=1/2a(∆t)^2+(-.18)(t)
∆x=1/2(.1203)^2+(.18)(t)
We solved that equation to find time and got, 4.3s.
Next, we plugged time back into the equation to find our predicted point of intersection.
Prediction
The distance we found was 1.072 m.
Cart D: 1.072 m
Cart C: 97 cm
Results
Actual Point of intersection: 1.072 m
After we tested our prediction we discovered that it was correct. Therefor, our percent error is 0.
Friday, December 11, 2015
Sunday, December 6, 2015
Constant Acceleration Particle Model
Constant Acceleration Particle Model
This unit we learned how to calculate instantaneous velocity, how to find displacement, how to determine the acceleration of an object, create a velocity vs. time graph, an acceleration vs. time graph, and a motion map solely from a position vs. time graph, and write a mathematical model to describe the motion of an object.
Very important equations
x=1/2a•t^2+xfinal
average velocity=∆v/∆t
acceleration=∆v/t
y=∆v/∆t (t) +starting point
velocity=∆x/∆t (xfinal-xinitial/tfinal-tinitial)
∆x=1/2b•h
*∆x is also = to the displacement: final-xinitial*
*Instantanenous velocity at the midtime is = to the average velocity*
CAPM.1
There are two ways to determine the instantaneous velocity of an object.
1) You can determine the slope of the tangent to an x vs. t graph at a given point (my preferred way)
or
2) You can use the equation: instantaneous velocity=a•t+vinitial
For the first way, you can pick a point on the graph and look at the points on either side of it.
This unit we learned how to calculate instantaneous velocity, how to find displacement, how to determine the acceleration of an object, create a velocity vs. time graph, an acceleration vs. time graph, and a motion map solely from a position vs. time graph, and write a mathematical model to describe the motion of an object.
Very important equations
x=1/2a•t^2+xfinal
average velocity=∆v/∆t
acceleration=∆v/t
y=∆v/∆t (t) +starting point
velocity=∆x/∆t (xfinal-xinitial/tfinal-tinitial)
∆x=1/2b•h
*∆x is also = to the displacement: final-xinitial*
*Instantanenous velocity at the midtime is = to the average velocity*
CAPM.1
There are two ways to determine the instantaneous velocity of an object.
1) You can determine the slope of the tangent to an x vs. t graph at a given point (my preferred way)
or
2) You can use the equation: instantaneous velocity=a•t+vinitial
For the first way, you can pick a point on the graph and look at the points on either side of it.
For this graph, if you are trying to find the velocity of the object at 3 seconds, you would look at the position of the object at 2 seconds and 4 seconds. At 2s the object had traveled 10m and at 4s it had traveled 20m. Next, you set up your equation like so: 10-20m/2-4s. The distances of the positions at 4 and 2 seconds go on the numerator and the time goes on the bottom. Subtract on the top and bottom and voila, you have found the instantaneous velocity! At 3 seconds on this position vs. time graph, the instantaneous velocity at 3 seconds=5m/s.
CAPM.2
There are also two ways to find the displacement of an object. The displacement is the object's overall change in position.
1) You can find the area under a v vs. t curve
or
2) You can use the equation: x=1/2a•t^2+vinitial•t
CAPM.4
If I am given a position vs. time graph, I can describe the motion of the object, create a v vs. t graph, an a vs. t graph, a motion map, find the average and instantaneous velocity, and find the displacement of the object.
The first column contains the position vs. time graphs, the second column contains the velocity vs. time graphs, and lastly, the acceleration vs. time graphs. For the first example, the object stays at a constant velocity and does not change position. The v vs. t graph and the a vs. t graph are identical because the velocity is 0 and so is the acceleration. For the second example, the object is moving at a constant velocity. In the v vs. t graph, the line shows that the object is moving constantly and in the positive direction. The object does not accelerate so the line on the a vs. t graph stays on the x-axis at 0. For the last example, the line slopes upwards, constantly accelerating. The v vs. t graph depicts an object with a velocity that changes every second and the a vs. t graph shows the object is accelerating upwards.
v vs. t graph: line above the x-axis=moving forward & line below the x-axis=moving backwards
a vs. t graph: line above the x-axis= positive slope on v vs.t graph& line below the x-axis= negative slope on v vs. t graph.
If the object's velocity has a negative slope then the acceleration line will also be negative (under the origin).
Practice Problems
In the graph above, the object starts at 0m and ends at 50m over the course of 10 seconds. The slope of the line gets less and less steep which means the object slowed down. It is moving in a positive direction.
What is the instantaneous velocity at 3 seconds?
Well, first we should look at the two points on either side of 3. At 2 seconds the object had traveled 18m and at 4 seconds it had traveled 31m.
18-31m/2-4s
-13/-2
instantaneous velocity at 3 s=6.5 m/s
The initial velocity of the object was 5 m/s. Find the acceleration of the object.
a=∆v/∆t
6.5-0/3s
2.16m/s^2
Make a corresponding v vs. t graph and an a vs. t graph.
In this graph, the object’s velocity slowed down constantly,
from -15m/s to 0m/s after 8 seconds. The object also moved backwards since the
line is under the x-axis.
Calculate the acceleration
A=∆v/∆t
0-15/0-8
1.88ms^2
What is the velocity of the object at 1 second and 5
seconds?
V=a•t+vinitial
=(1.88)(1) + (-15)
=-13.12 m/s
V=a•t+vinitial
=(1.88)(5) + (-15)
= -5.6m/s
What is the displacement from 1 to 5 seconds?
∆x=1/2a•t^2+vinitial•t
=1/2(1.88)(5)^2+(-13.12)(5)
=23.5+ (-65.6)
=-42.6m
Physics in every day life
We may not take the time to find the average velocity or
acceleration of an object on a daily basis but we do have devices that do it
for us instantaneously. After learning about this constant acceleration
particle model, I can calculate my average velocity on my run if I were to
measure the number of steps I take or my displacement if I kept track of my
acceleration & velocity. Although unlikely to use while running, these
skills are useful in other areas of life. These concepts were seemingly
abstract until we applied them to our everyday lives with the extra credit
opportunity of finding the height of 3rd Mitchell or the displacement
of a driveway after a dog ran from one end to the other.
Helpful Reminders
Thursday, November 19, 2015
CAPM Challenge 1
For this lab, Hank, Ally & I worked on a slanted table with one side propped up by books. Two meter sticks were placed along the side of the table and chalk was used to mark the ball's distance. A metronome clicked every half second and that signaled a chalk mark on the desk. In total we collected 2 trials.
When we created our graph we quickly realized that it was not linear. We had to linearize it in order to create a prediction and in order to do that, the x values were squared (time). To predict the position of the car at 4 seconds, we squared 4 and plugged that in for x in the equation. Slope=1/2 the acceleration. To find the acceleration, multiply the slope by 2.
Position=7.056 (16)+6.0772
Prediction: 118.9733 cm
Position=1/2a(t^2)
A=7.056m/s^2=14.112m/s^2
A=14.112m/s^2
When we created our graph we quickly realized that it was not linear. We had to linearize it in order to create a prediction and in order to do that, the x values were squared (time). To predict the position of the car at 4 seconds, we squared 4 and plugged that in for x in the equation. Slope=1/2 the acceleration. To find the acceleration, multiply the slope by 2.
Position=7.056 (16)+6.0772
Prediction: 118.9733 cm
Position=1/2a(t^2)
A=7.056m/s^2=14.112m/s^2
A=14.112m/s^2
| Time | Trial 1 | Trial 3 |
| 0.5 | 6 | 5 |
| 1 | 13 | 14.5 |
| 1.5 | 22 | 22.5 |
| 2 | 34.5 | 37 |
| 2.5 | 51 | 55 |
| 3 | 69 | 77 |
| 3.5 | 92 | 98 |
Friday, November 6, 2015
BFPM Practicum
cos (21) x 2.2
2.0538 (FtyA)
cos (53) x .85
.5115 (FtyB)
2.0538+0.5115= 2.653 N
We did the cosine of both 2.2 N and .85 N and then added the two numbers together to find our predicted weight. It is predicted that the weight of the mystery bag is 2.653 N.
Actual Weight:
Percent Error:
Thursday, November 5, 2015
Balanced Force Particle Model
Introduction
This unit our class learned about Newton's first and third law, force vectors, free body diagrams, and mass and weight. We practiced creating FBDs that depicted an object at rest, accelerating, and in motion. A force is an interaction between 2 objects, either pushed or pulled.
Newton's First Law
Newton's First Law of Motion states that "an object at rest and an object in motion will remain in motion with the same speed and and in the same direction unless acted upon by an unbalanced force." An example of this law is when Ms. Lawrence threw a ball straight up in the air and it came back down, with no forces acting on it.
Newton's Third Law
Newton's Third Law of Motion states that "for every action, there is an equal and opposite reaction." For every interaction between two objects there is a pair of forces acting on both. The size of the force acting on the first object is equivalent to the size of the force acting on the second object. When you press your shoe into the grass, there is a force that the sole of your shoe is pushing onto the ground and a force coming back from the grass, to your shoe. (Every action has an equal & opposite reaction.)
Free Body Diagrams
Tension Force
This is a force that is transmitted through a rope or wire and is represented by Ft. The tension runs along the direction the rope is pulled.
Tilting an Axis
When an object is being pulled or resting on a surface that is at an angle, we can rotate our free body diagrams. The x-axis is tilted to the angle of the surface.
Vectors
Vectors are the arrows we use in free body diagrams. In the example above, both Fnorm and Fgrav have arrows in equal length. This signifies that the forces are equal, but if Fgrav was longer that would mean it has a larger force than the smaller one (Fnorm).
Balanced Diagrams
This unit our class learned about Newton's first and third law, force vectors, free body diagrams, and mass and weight. We practiced creating FBDs that depicted an object at rest, accelerating, and in motion. A force is an interaction between 2 objects, either pushed or pulled.
Newton's First Law
Newton's First Law of Motion states that "an object at rest and an object in motion will remain in motion with the same speed and and in the same direction unless acted upon by an unbalanced force." An example of this law is when Ms. Lawrence threw a ball straight up in the air and it came back down, with no forces acting on it.
Newton's Third Law
Newton's Third Law of Motion states that "for every action, there is an equal and opposite reaction." For every interaction between two objects there is a pair of forces acting on both. The size of the force acting on the first object is equivalent to the size of the force acting on the second object. When you press your shoe into the grass, there is a force that the sole of your shoe is pushing onto the ground and a force coming back from the grass, to your shoe. (Every action has an equal & opposite reaction.)
Free body diagrams were used throughout this unit to help illustrate forces acted upon an object. We only include forces acting upon the object and not forces that the object is creating.
Normal Force
The normal force acts in the opposite direction of gravity. It is the support force exerted upon an object that is touching another stable object. It is represented with Fn. This vector always goes perpendicular to the surface the object is resting upon.
Gravitational Force
A gravitational force is the pull of gravity towards the center of the earth, therefor the vector always points downward. It is a force that attracts any object with mass and is represented with Fg.
Frictional Force
A frictional force is opposite the direction the object is moving. This force is exerted by a surface as an object moves across it or tries to. It is always parallel to the surface the object is on and is represented with Ff.Tension Force
This is a force that is transmitted through a rope or wire and is represented by Ft. The tension runs along the direction the rope is pulled.
Tilting an Axis
When an object is being pulled or resting on a surface that is at an angle, we can rotate our free body diagrams. The x-axis is tilted to the angle of the surface.
Vectors
Vectors are the arrows we use in free body diagrams. In the example above, both Fnorm and Fgrav have arrows in equal length. This signifies that the forces are equal, but if Fgrav was longer that would mean it has a larger force than the smaller one (Fnorm).
Balanced Diagrams
In the example above, the free body diagram is balanced despite the fact that the object is in motion. There is a normal force pushing up, a gravitational force pulling it down towards the earth, a frictional force going against the direction the object is moving, and a push force directing the object to the right. We can conclude that the object is moving at a constant velocity since it is balanced. As shown at the bottom of the picture, F(net)=0N.
Unbalanced Diagrams
As discussed in class, tug of war is a great example of balanced and unbalanced forces. If the two teams are exerting an equal amount of force, 300 N vs. 300 N, then neither team would win. However, in the picture above, the team to the left is exerting more force into the ground and thus has 100 more newtons of force than the team on the right. Whoever has more friction between the ground and their shoes wins the game.
Part B
This unit was more applicable to everyday life in comparison to the last unit, as the class used more real world examples to demonstrate our knowledge. Initially we used the hovercraft, then simply throwing the ball in the air, tug of war, horse pulling a cart, and wearing a seatbelt. Forces are used all the time, everyday. When we place objects down on the table or participate in a game of tug of war, forces are acting unknowingly to us. Newton's Laws also helped prove some counterintuitive thinking wrong.
Monday, October 5, 2015
Texting While Driving
Speed:
35 mph=56.327 kmh
Constant
variables: speed of car, the car, the phone, person driving
Amount
of time to text LOL (in seconds)
Trial
1: 3.99 s
Trial
2: 4.18 s
Trial
3: 3.8 s
Average:
3.99 s
56.327/60=
93.8/60=
1.563 m/s
1.563
m/s*3.99s = 6.2377 average meters are covered
Conclusion
The
average amount of time it takes to text LOL while driving is 3.99 seconds. The
car travels at a speed of 35 mph or 56.327 km per hour when converted.
Kilometers is converted to meters per hour and then divided by 60.938, which is
then divided by 60 and then we find the number of meters covered per second,
1.563. That number is then multiplied by the average time, 3.99s. This comes
out to equal 6.2377 meters are covered texting LOL while driving.
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