Unit 3 Summary

In this unit I learned about Newton’s 3^rd Law and all the things it can be applied to. Newton’s 3^rd Law states that: every action has an equal and opposite reaction. For example:

If an apple is at rest on a table, the apple pushes down on the table, and the table pushes up on the apple.

This concept goes a long way in explaining why various things work. If you have ever gone on a horse-drawn carriage ride, I am sure you remember it being a fun, and almost magically unique experience. Well, it is nearly magical how perfectly Newton’s 3^rd Law fits into a horse and buggy system. Without this law of physics, there would be no carriage rides. You may be wondering how this works…

The horse and buggy pull on each other with equal and opposite force- the horse pulling on the buggy alone does not cause the buggy to move. They both, however, exert force on the ground. The horse pushes on the ground with more force than the buggy so the ground pushes back on the horse with more force than it pushes on the buggy; thus the horse and buggy move in the horse’s direction.

This explanation applies to tug of war, and a plethora of other real life situations.

I also learned about adding vectors. Using the horizontal and vertical velocities of an object, you can find the actual velocity.

To predict accurately the direction of the actual velocity vector, you use dotted lines that are parallel to the other vectors and the arrow of the actual velocity should be drawn through the point where these intersect. This can be used to find the direction of a boat when the velocity of a current and the velocity of a paddle are taken into account. It can also be used to explain why a box slides down a ramp.

The force of the box’s weight must be drawn straight down as that is how gravity pulls it, and the support force of the ramp is drawn perpendicular to the ramp itself. When the dotted lines are drawn parallel from these vectors, a parallel line should be sketched out to show the net force on the box, showing that the net force pulls the box down the ramp.

A similar concept is applied to tension. If a ball was hung by two ropes, the force of the ball’s weight would be drawn as a vector pointing down, then an equal and opposite vector would be drawn upwards from the ball to show its net force. After drawing lines parallel to the ropes but which intersect at the tip of the net force vector, you can draw (from the ball to the point which these dotted lines intersect the ropes) arrows depicting the force of tension on each rope. The longer arrow shows which rope has a greater tension.

Thus far, I have talked a lot about force on the Earth. However, force exists outside of our planet as well. Before considering this, it is important to know that…

 Force is directly proportional to mass: F ~ m

 Force is inversely proportional to distance squared: F ~ 1/d^2

With this in mind, we learned that there is a universal gravitational force. The formula for this is:

F = G(m1xm2)/d^2

G is equal to (6.67 x 10)^-11. This equation helps us to find the force of gravity between two object by using their masses (m1 and m2) and dividing them by the distance between the objects squared. It is very important to square the distance.

Believe it or not, the tides of the ocean are actually connected to this universal gravitational force. The Earth is so large that the distance from one end of the ocean to the opposite is great enough for each of these points to have a significant difference in their distance from the moon. This difference in distance causes the tides. The moon’s gravity pulls the water ever so slightly towards it, giving the Earth’s water an oblong shape in reality. Due to this oblong shape, when it is high tide on the end of the Earth closest to the moon, it is high tide on the opposite end farthest from the moon. Likewise, it will be low tide at the top and bottom of the Earth at this time.

If the moon pulls on point A with 25 N of force, pulls on B with 5N of force, and pulls on the center of the Earth with 15N of force, how does all this tide business work? Using simple math, we can see that 25-15= 10 N with which side A is pulled to the Earth. 5-15= -10N with which side B is pulled to the moon. Like the forces addressed in Newton’s 3^rd Law, these forces are equal and opposite. Therefore the sides are pulled with the same amount of force in opposite directions, causing the oblong shape and corresponding tides.

It takes 24 hours for the Earth to rotate once; the length of one day. There are 2 high tides and 2 low tides in every day- each tide occurs about 6 hours apart. They change as the world spins and different part of the world are closer to the moon. However, the moon moves as well. It takes the moon about a month to complete its orbit around the Earth. Each month, it is not in the exact same spot on the same days, which causes the tides to be a little different every time, just as they are a little bit different every day due to the imprecise and varying location of the moon from day to day. Since the tides change and the moon’s location changes, there are Spring and Neap tides at different times of the month. During Spring tide, the moon is either new or full, being in a line with the Earth and the sun. At this time, the high tides are higher than normal and the low tides are lower than normal. Low tide during a Spring tide is a great time to go clamming. At Neap tide, the moon is neither between the Earth and the sun nor on the other side of the planet in a line from the sun. It is on either side of the Earth; “on top” or “beneath” the planet. During Neap tides, the high tide is lower than normal and the low tide is higher than normal.

You can use the universal gravitational force formula to talk about tides, but you can also use it to find the force between planets. You can find the pull between the Earth and the sun, and the Earth and the moon, alike. The farther two objects are, the weaker the force is between them because of the relationship force and distance share. However, the sun does not have a weaker pull on the Earth than the moon does just because of their distances from the planet. The sun’s mass is so much greater than the moon’s that its force is greater nonetheless.

Next, I learned about momentum.

Momentum, or inertia in motion, is mass times velocity. (p=mv)

Change in momentum is simply the final momentum of an object minus its original momentum. (change in p = p final – p initial). It can also be written as change in p = mv final – mv initial.

Momentum is also equal to impulse. Impulse is force times a time interval (J= F x change in time)

All of this information may seem random when presented like this, but answering one big question can bring it all together: why do airbags keep us safe?

When a car crashes, it and you go from moving to not moving. The change in momentum here is the same no matter how you are stopped because your momentum is changed to zero since change in momentum = p final – p initial. Change in momentum is equal to impulse (change in p = J), so the impulse is also the same no matter how you are stopped. Since impulse equals force times change in time (J = F x change in t), when the time of the impulse is very long, the force is small, and vice versa. Thus, the airbag is soft and takes a long time o stop your movement, so the force is less. The smaller the force, the less you are injured. This is why airbags keep you safe.

This unit, my group made a podcast about momentum, impulse, and the relationship between the two. Here it is, hopefully it adds clarification to the concept:

According to Newton’s 3^rd Law, in any collision, all forces have an equal and opposite force. So what about momentum? If ball A and ball B are headed towards each other and collide with equal and opposite forces, the collision could be noted as F(A) = -F(B). J = F x change in t, so F(A) x t = -F(B) x t, since they both experienced the force for the same amount of time. This equation can be rewritten as J(A) = -J(B), which can then be rewritten as change in p of A = - change in p of B (because J= change in p). Change in p (A) – change in p (B) = 0, since they were equal. Therefore, this system has no leftover or extra momentum. Momentum is conserved in systems, meaning there is no net change in momentum. Collisions like this can be examined using the equation  p total before = p total after. This is rewritten as M(a)V(a) + M(b)V(b) = M(a+b) x V(ab), which presents the equation in a form to which you can plug in numbers (the mass and velocities of two different objects). It is important to remember to make one of the velocities negative since the objects are moving in opposite directions before they collide. This equation is for objects that collide and then stick together. It can be reversed to apply to objects that detach in the interaction. If the objects do not stick together, the equation M(a)V(a) + M(b)V(b) = M(a)V(a) + M(b)V(b) can be used. This is the most basic form of p total before = p total after.

If an object bounces before it stops it undergoes two changes in momentum: when it stops and when it starts moving again. Thus the object has two impulses and therefore twice the force. Bouncing objects can be more dangerous.

What I have found difficult about what I have studied is separating the equations involving force from unit 2 and the equations involving force from unit 3. F = ma seems like it should be used but it is not relevant to this unit, so it has been a struggle to keep all the equations straight in my head in correspondence with the proper concepts.

I overcame this difficulty by reviewing the material and thinking more closely about what each concept was saying. This clarified the connection between the information and the equation for me.

I put in a little less effort to the class after Thanksgiving break; this is partially due to being sick and also because it is hard to transition back into school after a longer break in the midst of the end of the semester. I still continued to try and learn everything as thoroughly as possible, though. I feel very confident about this material as it makes sense to me and I understand the connections between the concepts introduced in this unit. My confidence in my knowledge of physics this unit helped me in all aspects of group collaboration these last few weeks. Whether discussing with a partner, working together on a lab, or completing a group project, I felt that I understood what was going on and was able to help others understand a bit better as well.

My goal for next unit is to find more creative ways to apply, present, and restate the information. This would help me to deeply learn the information in a versatile way. It would also make studying more fun.

I thought that the most interesting part of this unit was tides. I encounter them frequently enough, whenever I visit the beach they affect my entire time there. The connection between natural disasters and tides is especially interesting for me. Last fall, a very serious storm hit the northeast. Hurricane Sandy caused tons of destruction, injury, and chaos. One of the reasons it was such a bad storm was that it hut during Spring tide. Using the high high tides of Spring tide, Sandy caused more damage than it would have if it had hit during Neap or normal tides. This is just how nature works, though. The seemingly placid day-to-day workings of the world can just as easily turn events into drastic occurrences. I find this fascinating and I love that I now know the scientific reasoning behind this. It is also interesting to me how so much of our lives on Earth are affected by things outside of the planet itself.

Tides

If you've ever been to the beach, I'm sure you've noticed how the water is high up on the shore at some parts of the day and it is very low at other points of the day. These are called high and low tides and they are caused by the moon and the sun and their position in relation to the Earth. Force is inversely proportional to distance squared, so the closer the moon is to a point on the Earth, the stronger the force between them will be. Likewise, with different points of water on the Earth, closer blobs of water to the moon will have a greater force on them than blobs of water on the other side of the Earth. This causes a bulge in the Earth's water; it has an ovular shape which allows for high tides to occur twice a day and low tides to occur twice a day across the globe, happening on opposite ends of the world at the same time. At different times of the year, the position of the moon in relation to the sun causes what are called spring tides (very high high-tides and very low low-tides) or neap tides (tides that are not as great in difference). Although I just gave a very brief summary of this relationship, the video below explains it quite well. Adam Hart-Davis uses food to demonstrate the moon's movement, the Earth's movement, and their effect on the Earth's tides. The visuals are very helpful. He could have included more detail about the process but I think this video gives a very good, simple explanation of tides.

Unit 2 Blog Summary

Part A: What I Learned

In this unit, I learned about Newton's 2nd Law, skydiving, free fall, throwing objects straight up, throwing objects up at an angle, and falling at an angle.

Newton's 2nd law states that  acceleration is directly proportional to force and inversely proportional to mass. We completed a lab to demonstrate this relationship. In conclusion, we found that as mass increases, acceleration will decrease, and if force increases, acceleration will increase in the same ratio.
You can write Newton's 2nd law using symbols...
a~F
a~ 1/m
or as an equation....
acceleration= net force/ mass
a= f(net)/m
An object's weight can be the force acting on the object. Weight is equal to the mass of an object times gravity (w=mg). If you know the mass of an object, you can find the weight or force on the object as gravity is always 9.8 m/s^2 (or 10m/s^2 for simple calculations).

Above is an illustration of the experiment we performed in the lab, including labels of acceleration and force. If you add mass into the picture, depending on where it is placed (on the hanging weight or the cart), either the force and acceleration will increase or they will both decrease.

When you go skydiving, air resistance plays a large role in your fall through the air. As you fall, your speed increases which in turn causes your air resistance to increase. Conversely, as your speed decreases, your air resistance decreases as well. Air resistance is affected by two things: speed, and surface area.

To start at the beginning, when you first step out of a plane, your velocity is at 0m/s, which is its lowest point during the entire fall. In this same moment, acceleration and net force are at their highest points of the fall. This is because at t=0 seconds, you have not yet begun moving at speed, thus you have no velocity. However, your acceleration will decrease as you build up more air resistance. Keep in mind, you are still falling faster and faster (velocity increases), but you are gaining speed at a decreasing rate (acceleration decreases). Your net force is at its highest point at t=0 seconds as well, not only because acceleration is directly proportional to force, but also because your air resistance builds up to equal your weight (aka the force of gravity which is pulling you down). When you have no air resistance, your weight is the only force acting on your body. As the opposing force of air resistance builds, the net force decreases until it reaches 0 Newtons (the point when air resistance = your f weight). This point when the two forces are equal is also the time in your fall when your velocity has reached the fastest it can possibly be- this is called terminal velocity. When you are in terminal velocity, f net = 0, velocity is constant, and acceleration = 0 m/s^2. It is a total reversal from t=0s because velocity is at its highest point in terminal velocity while acceleration and net force are at their lowest points.

So that covers the falling part. But to skydive, there comes a point when you need a parachute. If you fell without a parachute, hitting the ground at terminal velocity would be very painful! 

When you are falling and have reached terminal velocity, this is the point when you open your parachute. Immediately, your surface area is increased by the chute which increases the air resistance. The air resistance will be greater than the force of your weight, and since it is a force pulling you in the upward direction, your acceleration will be upward. This also causes the net force to be negative (it will no longer equal zero; at this point you are no longer in terminal velocity). As you slow down again and air resistance decreases, you reach a second terminal velocity. In this second terminal velocity, the acceleration and net force are both 0 again (equal to the first terminal velocity) and the air resistance will also be equal to the air resistance in the first terminal velocity as it has to equal the force of your weight for the net force to equal zero. The only thing that is different in the second terminal velocity is the velocity itself- it will still be constant but it will be much slower and safer in the second terminal velocity. From here you safely fall to the ground and the parachute has done its job.

When you fall straight down in free fall, another concept I learned in this unit, this means you are falling without any forces acting on you except for gravity. When you skydive, you are not in free fall. One cool thing about free fall is that mass does not matter. If you dropped a coin and a feather at the same time in a vacuum without air resistance, they would hit the bottom at the same time. This is because the only force acting on the objects is the force of gravity which is determined by an object's weight. You might be thinking, "I thought weight didn't matter?" You're wrong. Mass and weight are not the same thing. Weight is the mass of an object times gravity (w=mg). So, if you are finding the acceleration of an object, you use a=f net / mass; in free fall the f net = weight, so you can change the equation to say a= mg/m (acceleration is equal to mass times gravity divided by mass). Mass divided by itself equals one, so you can cancel out the m's. This leaves the equation as a=g (acceleration = gravity = 9.8m/s^2). So no matter what the mass of an object is, when an object is in free fall, it will always have the same acceleration. Here is a video to demonstrate:

You can use the distance equation (d=1/2gt^2) to find how far an object fell in free fall, substituting gravity for acceleration. You can also use V=gt to find how fast the object was moving when it hit the ground.

In free fall, you can (obviously) fall straight down, But you can also throw things up.

The picture above illustrates the path a ball will take. In reality, it would fall straight up and straight down, but for the sake of the drawing, the ball is shown a little displaced from this path. As the ball travels upward, it is propelled by the initial velocity with which it is thrown while gravity is acting on the ball in the opposite direction (down), causing it to slow down as it reaches the top of its path. Since the ball is in free fall, its acceleration is 10 m/s^2 during the entire time in the air. In this particular drawing, the ball is thrown up with an initial velocity of 30 m/s and its speed decreases by 10m/s each second it falls up. When the ball is at the top of its path, it has a velocity of 0m/s, but the acceleration of the ball will still be 10m/s^2. The ball then falls back toward the earth with an increasing velocity because velocity and gravity will be in the same direction again.
Just from this picture you can tell a lot about the situation. You know the balls initial velocity as well as its velocity at each point along its journey. You know the hang time of the ball (total time in the air) is 6 seconds. To find how high the ball got, you can use the distance formula. However, this formula does not account for velocity in the upward direction, so you can only plug in values from the ball's downward fall. For example:
d=(1/2)gt^2
=(1/2) (10) (3)^2
= 1/2 (90)
d= 45 m
To find how far the ball is from the ground at any given second of the fall, you would use the value found for the total distance as shown just above. Then you would count the number of seconds from the top of the path down to the second you are finding the distance for and use that number of seconds as the t value in the distance formula. Once a distance is found, subtract it from the total distance and you will have the distance from the ground at that second.

After I learned about falling in free fall and throwing things straight up in free fall, I learned about things falling at an angle in free fall and things being thrown up at an angle in free fall.

When an object falls at an angle, the vertical acceleration is constant (9.8m/s^2) and the horizontal velocity is constant (the initial velocity the object is thrown with is the velocity that the object will have in the horizontal direction for the entire fall). The vertical velocity continues to increase as the object falls, so the path it takes to the ground will be curved, or parabolic. To find the vertical distance you can use the same distance equation we are familiar with (d=(1/2)gt^2) and you can use v=gt to find the vertical velocity at a given second. You could use either of these equations to find the time of the object's fall. The time the object falls is the same vertically and horizontally; t is one variable that will remain the same in any equations for vertical or horizontal values, besides g. The time of the object's fall is determined by the vertical distance, or the height from which the object falls, which is why the distance formula is so useful and is only used for vertical values. To find the horizontal distance, velocity, or the time of the fall, we can use the equation v=d/t. Problems like this can be applied to an object being dropped from a plane or something being thrown off a cliff. 
It is also important to note that we use the vertical and horizontal velocities in our equations but we do not use the actual velocity of the object.
To find the actual velocity, we must know the vertical and horizontal velocities. When they are put together they make two sides of a box, through which a vector can be drawn diagonally to show the actual velocity of the object. In my physics class, we use either a 45 degree angle or a 3, 4, 5 triangle to find this value. The picture below is an example of how this is drawn. 

We use this triangle system to find the actual velocity for objects thrown up at an angle as well. An example of this would be hitting a home-run in a baseball game. To find the horizontal velocity and distance, we again use the v=d/t equation. For vertical distance and velocity in this situation, we use d=1/2gt^2 and v=gt because these objects are still in free fall. We treat the vertical equations that use g in the same way we treat them in problems of objects being thrown straight up. They can only be used with values from the fall downward. The vertical height determines the hang time of the ball so these equations are really useful. If the vertical velocity of the object was 1m/s and the horizontal velocity was 1m/s, the actual velocity would be 1.41 (the square root of 2) at a 45 degree angle. If both velocities were 10m/s, the actual velocity would be 14.1 m/s, and if they were 100m/s, the actual velocity would be 141 m/s.
When an object is thrown up at an angle, vertical velocity decreases on the way up- at its highest point, the object has a vertical velocity of 0m/s. The horizontal velocity, however, is constant. If an object is thrown with a horizontal velocity of 10 m/s, it will still move horizontally at 10m/s at the top of its path when the vertical velocity is 0m/s. Since it keeps moving horizontally, it has a curved path. On the fall back to the ground, the vertical velocity will increase at a constant acceleration of 10m/s^2.


What I found difficult about this unit was the concept of skydiving with air resistance. At first, I just couldn't wrap my mind around the fact that you speed up so you have more air resistance and then once you open your parachute your air resistance will increase, but as speed decreases the air resistance decreases as well. After some practice and looking at tons of examples and diagrams, it makes a lot more sense now. The transition to falling with air resistance to falling in free fall was weird at first because I kept expecting the objects to reach a terminal velocity but I think I have gotten the hang of that concept now, too. Again, I made the lightbulb click with visual examples and just really breaking the concept down.


This unit, I think I had more effort in class but I could have put more effort into my work outside of class. I need to make more time for myself to not only get my work done but also to do it well. I do think, conversely, that the time I put into doing problems paid off. I always took the time to finish the problems and think through them which really helped me to be active and present in class. When we went over concepts and certain problems in class, I had already spent a lot of time on them and mostly understood them in my own way, so it was nice to be able to use class as a solidifier. I am getting to be very confident in physics- I feel like I really understand what is going on and I do not question everything I do. This comes from making sure I understand everything on my own time and then moving on in class, adding onto the information learned.


My goals for the next unit are to spend more time on my work outside of class and to finish my podcast early. I have found that the podcast gets really stressful when I have other work to do and my group starts finding less and less time to meet. Hopefully we can utilize the preceding weekend more efficiently int he next unit. I also want to have a very deep, holistic knowledge of the next unit. I think it would be a very bad time to start only half processing the information and I want to keep working hard all semester (all year, too, but I am thinking short-term until winter break).



Part B: Connections

I can connect this unit to soccer. I play right back, and one of my best assets in the game is my ability to clear the ball. This is similar to throwing things up at an angle; I use my foot to lift the ball and kick it across the field, in an arc through the air. I had usually just focused on my form for this, making sure I transferred my body weight over the ball and followed through my kick with my leg and foot pointed. I would love to find the actual velocity necessary to get the ball where I want it to be and then use this to find the horizontal velocity I need to swing my leg with to get this kind of power. I know that I will now start trying to kick the ball at a 45 degree angle to see if that gives it a smoother arc at all. It would also be really interesting to feel what different velocities feel like. For example, how hard will I need to swing my leg to hit a ball at 30m/s? And how will this feel on my leg? Is this an easy velocity to reach or would I need to really kick with power?

My group this unit made a podcast about things falling at an angle.


I can't wait to learn the next unit! Yay physics!

Freefall

I think we're all familiar with Tom Petty's song, "Free Fallin'," but we usually don't think of freefall as a physics concept. Free fall occurs when an object falls due to the effect of gravity only. When in free fall, an object has a constant acceleration of 10 m/s per second (which is the force of gravity). In free fall, there is no air resistance and weight does not matter. The video below explains free fall pretty thoroughly, including the example of Felix Baumgartner who set the world record for the longest free fall.

I think that video explains the concept of free fall really well, but I am including this second video which is a demonstration of free fall- the experiment is conducted with a feather and a coin in a vacuum. My physics teacher conducted this experiment in class and I was really excited to find an easy-to-follow video of the same experiment.



I hope this clears up free fall for you!

Newton's 2nd Law of Motion

In this video, Newton's 2nd Law of Motion is demonstrated in several different ways; first the video shows the effects of mass on acceleration in space as compared to on Earth. Several different examples follow this as the video moves on to explain the effects of force on acceleration, specifically the force of friction.
I think it is super cool how Newton's 2nd Law is shown in space and it is helpful as we can take gravity out of the equation and just focus on the relationship between force, acceleration, and mass. According to Newton's 2nd Law, acceleration is directly proportional to force and inversely proportional to mass. This means that acceleration will increase as force increase, while it will decrease when the mass of the object increases.
The relationship between mass and acceleration is very clearly demonstrated in the first part of the video. I think the later parts of this video are less clear in their explanation of force in relation to acceleration. However, this is still a helpful and unique resource. I hope this broadens your understanding of Newton's 2nd Law! Enjoy!

Unit 1 Reflection

In this unit, I learned about many concepts of physics. 

I learned about Newton's 1st Law, which states that an object in motion tends to stay in motion, while an object at rest will stay at rest, unless acted upon by an outside force. Inertia, or the "laziness" of an object, is the property of an object to follow Newton's 1st Law. The more mass an object has, the more inertia it has.(You can even describe mass as a measure of inertia). Applying the concept of inertia and Newton's Law to everyday life: if a table is set with dishes and such, and the tablecloth is pulled out from underneath the settings, the dishes will remain on the table. This is because both the tablecloth and the place settings are at rest. When you pull the tablecloth, you are exerting force on it which causes it to no longer be at rest. The dishes, however, have not been acted upon by an outside force. Thus, in accordance with Newton's 1st Law, the table settings will remain at rest; the tablecloth is pulled cleanly from the table while the dishes remain in their original spots.

I just mentioned force and how they can change an object's motion. Force, measured in Newtons, is defined as a push or a pull- a pretty broad definition if you ask me. To make things simpler for discussion's sake, we measure the net force of objects. Net Force is the total force acting on an object. 
For example, in the drawing below, a box is being pushed with a force of 50 Newtons. Assuming there are no other forces acting on the box at this moment, the box would have a net force of 50N.

However, in the next image, two opposing forces are pushing on the box. The force from the left is pushing with 100N, while the force from the right is pushing with 50N. As these are opposing forces, you must subtract them to find the net force. 100-50=50, so the net force= 50N.

If these opposing forces were equal to each other, the box would be at equilibrium. Equilibrium occurs anytime the net force adds up to 0 Newtons. It occurs when an object is 1) moving at constant velocity or 2) at rest.

This brings me to the next lesson: velocity. Many people think of velocity as the speed of an object. This is only partially accurate. While "speed" describes how fast an object is moving, "velocity" describes the speed of an object, as well as the direction in which the object is moving. The equation of velocity is: V = d/t (Velocity is equal to distance over time). Velocity is measure in meters per second (m/s) and measures the distance covered in a certain amount of time, as shown by the equation.
If the direction of movement changes, the velocity changes. Thus to be moving at constant velocity, an object must maintain both constant speed and direction. An object is moving at constant velocity when it is at equilibrium (or at rest).

A common mistake is confusing velocity to be the same thing as acceleration. Acceleration describes the rate at which an object is changing speed. Its equation is: A= change in velocity/time interval. Acceleration is measured in meters per seconds squared (m/s^2). For an object to be accelerating, it needs to experience a change in velocity, which can occur one of three ways: 1) changing direction, 2) speeding up, or 3) slowing down. Acceleration can be increasing, decreasing, or constant, depending on the surface upon which the object is moving. If an object has constant acceleration, it cannot simultaneously have constant velocity. An object falling straight down will always have an acceleration of 10 m/s^2, meaning the object increases its speed by 10 m/s per each second.


To calculate how fast an object is moving, you can use the equation: V= at, which shows that velocity is equal to acceleration times time. 

To calculate how far an object has moved, use the equation: d= 1/2at^2, which indicates that the distance an object has traveled is equal to one half of the object's acceleration multiplied by the time it has been moving, squared. Both of these equations describe a relationship between acceleration and velocity, as do the three charts shown below.

The final lesson of the unit was about graphs and applying their equations to physics. The equation of a straight line is y= mx + b. This line, when graphed with time (squared) on the x-axis and distance on the y-axis, can be translated into the equation for distance (d= 1/2at^2). The slope (m) would stand for ½ acceleration, x would correlate with time, and the b is irrelevant from here on out. Using that information, you can visually represent data of an object’s distance traveled by graphing it. Also, you can translate this equation into the “how fast” equation. Since m= ½ a, to find a (in V = at), you would multiply whatever value you’ve found for m by 2. You already would know the value for time, so you could then use the information from the graph to find not only the distance traveled by an object in a specified amount of time, but also how fast the object traveled this distance.

What I found difficult about what I have studied is connecting the topics of acceleration and velocity and then translating this connection into graph-able data. At first I was so determined to understand the difference between acceleration and velocity that I compartmentalized the two concepts in my mind. Really, they cannot be completely separated if you want to fully understand each concept. I overcame this by working out different problems using the equations involving velocity and acceleration and studying over the explanations of each. The light bulb really clicked when I could put into words that “velocity measures the direction and speed of an object and acceleration measures how fast the speed of said object is changing; acceleration is defined as the change in velocity over a given time.” This really makes sense to me, as both concepts involve speed and can be linked in that way. Also, the graphs we worked with this unit were very intimidating to me at first glance. I am not the most computer savvy person and often find myself completely overwhelmed by tasks involving Excel. These graphs, however, made more sense as our class got more comfortable working with them. It helps that I have been working with plotting data on a graph in my math classes for several years now. Going over the correlation between the equations for the graph and the corresponding physics equations clarified the entire process for me; now I understand why I am putting which value on which axis.

I started this unit off very strong, in my opinion. I took detailed notes and gave myself ample time to complete each assignment with full understanding of the material. As my workload increased, this diligence lost some of its initial luster. I continued to work hard though; my homework has consistently been completed on time (save for one instance in which I had only 30 minutes of study hall along with a packed Thursday night schedule), my lab work done with care and elaboration, and blog postings finished promptly and with extensive explanation. Over the course of the year I hope to become more creative in my work and less “cut and dry.” This is definitely an area with room for improvement. My confidence in physics has grown so much already and I enjoy being able to discuss a concept with self-assurance. Not only do I understand the concepts but I am able to apply the concepts to examples we are given in class, etc. When it comes to solving different problems, my skills are varied. I consider writing to be my strongest asset in school. Thus, short answer problems are very easy for me as I find them logical; when I can write out the process it makes more sense. However, the shorter math problems that involve a conceptual formula make sense to me as well and I think I am generally rather good at these. I have a lot of trouble when it comes to “problem solving” questions in which harder math is needed as well as the application of several concepts at once. This is something I need to work on and would like to get better at as these types of questions will show up in my future math and science courses.

My goal for the next unit is to study more frequently by redoing problems I do not understand or have gotten wrong in the past, as well as doing harder level problems of concepts I do understand just to challenge myself and make sure I have taken my knowledge to the next level. I also have a goal to come to conference period at least once a week in the future as a way to make sure I have all my questions answered and every point is clarified. I think these extra steps will boost my grade as well as my confidence in the material.

In the fall, I run cross-country. I love running and spend a lot of my time outside of the season running and learning about the sport. Physics relates to running; I have thought about my acceleration time and time again as I tackle large hills or sprint down the final stretch of track. I enjoy playing soccer just as much, if not more, than I enjoy running. It is not soccer season yet, but 90% of the time soccer is on my mind in one way or another. Going over physics concepts has made me reflect on my game a lot in the past few weeks. For example, the force with which I kick a ball will need to be greater than the force of friction from the field or the air resistance the ball will meet if I were to chip it through the air. Additionally, the velocity with which I kick the ball will influence how far the ball travels and how fast it does so. I cannot wait until soccer season so I can apply these new concepts out on the field. I am also very excited to learn more about the physics involved sports (mostly soccer, but other sports as well, like running, football, and tennis) and why they work the way they do. Who knew athletics and science worked together so well?!

Below is a podcast I made with  group about making graphs and how to use them efficiently. Enjoy!