Kinematics - In this unit, we covered topics related to motion of an object, like position, velocity, acceleration, and projectile motion. We particularly focused on graphing this motion, and learning how these graphs and models correlated with each other and represent how an object moves.
Position, Distance, and Displacement - Position is where an object is at a certain point of time. Distance is the total amount the object travels in a certain increment of time, while displacement is only the "net" amount of units traveled from the initial point to the ending point. For example, if an object's position is 2 meters and it moves 4 meters to 6, but then moves 1 meter back, the distance and displacement are 5 and 3 meters respectively.
Interpreting Position-Time Graphs: A position-graph plots the position of an object (vertical axis) as it changes due to time (horizontal axis). Our buggy lab depicted a position-time graph and its characteristics of having a y-intercept, which shows the position at zero seconds, and a slope. A p-t graph's slope shows direction and magnitude, like a vector. Its direction is shown by its sign (positive or negative), while its magnitude is essentially steepness and depicts the speed of an object. This makes sense as a p-t graph inherently plots motion over time, which is why the magnitude/steepness of a p-t graph is speed. Since the slope represents direction and speed, the slope of a p-t graph shows us the velocity of the object. Velocity as our slope makes sense, as velocity is defined as the change in position over change in time, which is precisely what the slope of a p-t graph tells us.
Determining Velocity and Speed from a Position-Time Graph: We can determine the velocity from a position-time graph by looking at its slope, which shows the velocity of an object. The steepness of this slope represents speed, so, if a graph increases quickly, then the graph's speed is increasing as time increases.
Example - For the below p-t graph, since the slope is 1 meter per second, the velocity is also 1 m/s. We also know the speed increases as at a constant rate since the steepness of the graph stays the same.
Interpreting Velocity-Time Graphs: Velocity-Time graphs depict the change in velocity (vertical axis) as time passes (horizontal axis). This, by definition, is acceleration, meaning the slope of a v-t graph is acceleration. Acceleration is defined as the how fast the speed changes. This means that, for example, a velocity time graph with a constant velocity has an acceleration of zero. In our cart on ramp lab, we depicted how the ramp affected acceleration and, thus, changed our velocity-time graph. The y-intercept of a velocity time graph represents velocity initial or when the velocity when the object moves zero seconds.
Determining Acceleration and Starting Velocity from a Position Time Graph: Since velocity is the slope of a p-t graph and acceleration is the slope of a velocity-time graph, then that means acceleration is the slope of the slope of a position-time graph. This means, for example, if a position-time graph represents a linear line, then the acceleration is zero. In this case, the velocity is constant (with the p-t graph) being a linear line, which means the acceleration is zero. You can calculate the velocity of any point/line on the position-time graph by finding the slope of that line using its minimum and maximum points. For finding the starting velocity, you can simply look at a p-t graph's initial point and use its x-coordinate as the starting velocity.
Determining Acceleration and Starting Velocity from a Position Time Graph: Since velocity is the slope of a p-t graph and acceleration is the slope of a velocity-time graph, then that means acceleration is the slope of the slope of a position-time graph. This means, for example, if a position-time graph represents a linear line, then the acceleration is zero. In this case, the velocity is constant (with the p-t graph) being a linear line, which means the acceleration is zero. You can calculate the velocity of any point/line on the position-time graph by finding the slope of that line using its minimum and maximum points. For finding the starting velocity, you can simply look at a p-t graph's initial point and use its x-coordinate as the starting velocity.
Example: You can find the starting velocity from this by looking at the starting point and seeing that, on the p-t graph) the initial point t is (0,1). This means that the initial velocity at this point (on a v-t) is 0.
Determining Position, Distance and Displacement from a Velocity Time Graph - You can find the position, distance and displacement by using the velocity equation (v = change in position/change in time). In order to find equation, you must solve for position = velocity * time by plugging in the given values from the velocity-time graph. To find distance, you calculate the area underneath a velocity-time graph by creating smaller shapes from the area, and finding the areas of the smaller shapes and them together. For displacement, the process is similar to distance, but, if there's a section/shape under the graph that's negative, then it's subtracted from the rest of the area since displacement is the "net" change in position and takes the sign (positive or negative) into account. If the graph was only positive, then the displacement and distance would be the same.
Example - I calculated the area of everything that exists under the velocity-time graph, multiplying time by velocity. This means the area is in terms of change of position, which is important to finding distance and displacement. Since part of the velocity graph is negative, we must subtract 1/2 (the area of the triangle under the x-axis) from 8 to find the displacement, but add it to find the distance.
Connecting Representations of Motion: We already covered how position, velocity, and acceleration graphs all relate; the velocity is the slope of a p-t graph and the acceleration is the slope of a v-t graph. An example showing this correlation and how the graphs relate is depicted below. There are also other models, including strobe diagrams and motion maps that depict an object's movement and the rate at which it increases.
The position-time graph here is quadratic and decreases as it approaches 2 seconds. Since the graph is quadratic, the velocity-time graph is linear and the velocity decreases at about 1 m/s as it approaches 2 seconds. The linear line also starts at (0,0) since the vertex position-time graph is at t = 0, meaning a horizontal zero has to exist for the velocity-time graph. Because the v-t graph is linear, the acceleration is constant at 0, and does not affect the velocity.
This strobe diagram and the arrows increasing in size depicts how the object's position is increasing as time increases.
This motion map represents how the object is moving forward and getting faster.
Solving Problems Using Constant Velocity and Uniform Accelerated Models:
x = Xi + Vi * t
Xf = Xi + Vt + 1/2 * a * t ^2 ; change in x = vt + 1/2at^2
When solving for problems using these equations, you have to figure out at least three variables (given directly or indirectly).
Example 1: A bus leaving an intersection accelerates at 2 m/s^2. Where is the bus after 4.4 seconds? (From Mastering Physics 1.2 Number 4)
Initial Velocity = 0 m/s
a = 2 m/s^2
t = 4.4 seconds
Use UAM -> 0 * 4.4 seconds + 1/2 * 2 sec * (4.4 m/s^2)^2 = 19.36 m
Example 2: You throw a tennis ball straight upward. The initial speed is 12 m/s. How long will it take for the ball to reach it maximum height? (From Mastering Physics 1.2 Number 9)
Vi = 12 m/s
Vf = 0 m/s
a = -9.81 m/s^2
Use constant velocity -> 0 = 12 - 9.81t -> -12 = -9.81t = t = 1.22 seconds
Projectile Motion - A projectile, an object in motion, experiences projectile motion when it moves in a curved path that is only affected by gravity. The x-motion/velocity determines how far the object moves, while the y-motion/velocity determines how long the object is in the air for. This is represented by the penny activity we did in class, as, even though the pennies were dropped in different spots on the counter, they landed at the same time since they were at the same height in the air. The horizontal motion is show by Vx * t and the vertical motion is show by 1/2at^2 + vi * t. The maximum height is when Vy (the y-motion) is equal to zero, so you can solve for it by using the equation for vertical motion, trigonometry, and plugging in zero and 9.8 m/s^2 (gravity) for a. You use the horizontal motion equation to find the range since range is the displacement in the horizontal direction. It's important to note how both components, horizontal and vertical, are important for projectile motion.
x = Xi + Vi * t
Xf = Xi + Vt + 1/2 * a * t ^2 ; change in x = vt + 1/2at^2
When solving for problems using these equations, you have to figure out at least three variables (given directly or indirectly).
Example 1: A bus leaving an intersection accelerates at 2 m/s^2. Where is the bus after 4.4 seconds? (From Mastering Physics 1.2 Number 4)
Initial Velocity = 0 m/s
a = 2 m/s^2
t = 4.4 seconds
Use UAM -> 0 * 4.4 seconds + 1/2 * 2 sec * (4.4 m/s^2)^2 = 19.36 m
Example 2: You throw a tennis ball straight upward. The initial speed is 12 m/s. How long will it take for the ball to reach it maximum height? (From Mastering Physics 1.2 Number 9)
Vi = 12 m/s
Vf = 0 m/s
a = -9.81 m/s^2
Use constant velocity -> 0 = 12 - 9.81t -> -12 = -9.81t = t = 1.22 seconds
Projectile Motion - A projectile, an object in motion, experiences projectile motion when it moves in a curved path that is only affected by gravity. The x-motion/velocity determines how far the object moves, while the y-motion/velocity determines how long the object is in the air for. This is represented by the penny activity we did in class, as, even though the pennies were dropped in different spots on the counter, they landed at the same time since they were at the same height in the air. The horizontal motion is show by Vx * t and the vertical motion is show by 1/2at^2 + vi * t. The maximum height is when Vy (the y-motion) is equal to zero, so you can solve for it by using the equation for vertical motion, trigonometry, and plugging in zero and 9.8 m/s^2 (gravity) for a. You use the horizontal motion equation to find the range since range is the displacement in the horizontal direction. It's important to note how both components, horizontal and vertical, are important for projectile motion.
Dynamics - For this unit, we covered topics related to force. Specifically, we focused on the laws of how forces interact on an object together, learning how to use models and diagrams to represent different types of scenarios of forces interacting on an object. We also dedicated significant time to learning how to use our understanding of forces and kinematic equations to problem solve and find solutions to application questions.
Newton's First Law:
Newton's First Law of inertia tells us that an object in motion will have a constant velocity and direction unless acted on by an unbalanced force. So, for example, a soccer ball will have a constantly velocity after it is kicked, but, if another player stops it, then the velocity and direction changes.
It also explains inertia, which is essentially an object's resistance to acceleration or deacceleration. If a change in motion occurs, a object with a high mass (and therefore high inertia) will be more likely to resist this change. This explains why when a car comes to a short stop, passengers keep moving forward since their larger mass makes them more resistance to the sudden stop of the car.
Newton's Second Law:
Newton's Second Law states that objects accelerate more when they feel a net force with a larger magnitude, meaning acceleration are net force are proportional. It also reinforces the law of inertia, as there is an inverse relationship between mass and acceleration, because a larger mass means more inertia, which equates to a lesser acceleration. This was confirmed in our unbalanced forces lab, as we found that applying more force to the cart by adding mass to the hangar resulted in an acceleration increase, while adding more mass to the cart itself resulted in an acceleration decrease. The graphs are pictured below, and show the relationships between net force and mass with acceleration. From the law and these experiments, you can also find the ever-important Zf (net force) = m * a equation.
Newton's First Law:
Newton's First Law of inertia tells us that an object in motion will have a constant velocity and direction unless acted on by an unbalanced force. So, for example, a soccer ball will have a constantly velocity after it is kicked, but, if another player stops it, then the velocity and direction changes.
It also explains inertia, which is essentially an object's resistance to acceleration or deacceleration. If a change in motion occurs, a object with a high mass (and therefore high inertia) will be more likely to resist this change. This explains why when a car comes to a short stop, passengers keep moving forward since their larger mass makes them more resistance to the sudden stop of the car.
Newton's Second Law:
Newton's Second Law states that objects accelerate more when they feel a net force with a larger magnitude, meaning acceleration are net force are proportional. It also reinforces the law of inertia, as there is an inverse relationship between mass and acceleration, because a larger mass means more inertia, which equates to a lesser acceleration. This was confirmed in our unbalanced forces lab, as we found that applying more force to the cart by adding mass to the hangar resulted in an acceleration increase, while adding more mass to the cart itself resulted in an acceleration decrease. The graphs are pictured below, and show the relationships between net force and mass with acceleration. From the law and these experiments, you can also find the ever-important Zf (net force) = m * a equation.
Newton's Third Law -
It states that there there is a force exerted upon each of the objects. They are balanced if they feel the same force but in opposite directions. A good example of this would be a sliding puck that is balanced in the y-direction since normal and gravitational force cancel out in this case.
Identifying Interactions (System Schema + Force Diagram) -
We can use different kinds of models and diagrams to represent how forces interact with each. For system schema, we can depict the forces between objects by drawing bubbles that represent the objects and connect them with lines that represent the forces. They are especially useful since they can provide a clearer representation of the forces at play and how certain objects interact with each other.
It states that there there is a force exerted upon each of the objects. They are balanced if they feel the same force but in opposite directions. A good example of this would be a sliding puck that is balanced in the y-direction since normal and gravitational force cancel out in this case.
Identifying Interactions (System Schema + Force Diagram) -
We can use different kinds of models and diagrams to represent how forces interact with each. For system schema, we can depict the forces between objects by drawing bubbles that represent the objects and connect them with lines that represent the forces. They are especially useful since they can provide a clearer representation of the forces at play and how certain objects interact with each other.
However, a system schema can be flawed since it does not depict direction or magnitude in anyway, which is why a force diagram is even better. It depicts the direction of the force, allowing us to see whether it's in the x or y direction. It also lets us see the relative ratio between different forces and tell whether forces are balanced or unbalanced.
For the same scenario as the system schema example, I create a practice force diagram. Now, I was able to depict the net force of 65N in addition to the other forces at play. I was able to show how the applied force propels the ball upwards in the y-direction, while gravity drags it down in the y-direction but to a lesser extent.
Force Calculations -
When calculating forces, it is important to keep in mind a few important equations, which can be found from our knowledge on different kinds of forces.
- We know that Net force = mass * acceleration from our unbalanced force experiments when learning about Newton's second law. We can use this to find the total net force of a certain direction. We also know that Fg (force of gravity) = mass * gravity (and gravity is always +/- 9.8), meaning we can always find the mass or force of gravity if given the other.
- Fs = - (k * deltaX). Spring force, as proven by our findings in our spring demo, is the negative of the product of the spring's constant of k by the distance the spring is stretched/compressed. The spring's constant is essentially the sturdiness or stretchiness of the spring.
- The force of friction is when two objects rub against each other, creating a force in the opposite direction of an object's motion.
f = mew * Fn -> Force of friction = coefficient of friction * normal force. This in accordance with Hooke's law that states the coefficient of friction is equal to the force of friction divided b the normal force. The coefficient of friction, or mew, is essentially the surface's roughness that plays a huge role in the magnitude of the friction force. Oftentimes, we can use our knowledge of the force of gravity, which can help us find normal force and, other given variables to find the force of friction. This can help us determine the net force of whatever direction the friction force is in.
Solving Force Problems -
A first pivotal step is writing down as many given variables as possible and also creating a force diagram to gain a better understanding of what forces are at play. By writing down given variables, we can use the equations listed above to create a system of equations where we can ultimately find the x and y-component forces. Then, after filling out a force table, we can solve for the net force in all directions.
When calculating forces, it is important to keep in mind a few important equations, which can be found from our knowledge on different kinds of forces.
- We know that Net force = mass * acceleration from our unbalanced force experiments when learning about Newton's second law. We can use this to find the total net force of a certain direction. We also know that Fg (force of gravity) = mass * gravity (and gravity is always +/- 9.8), meaning we can always find the mass or force of gravity if given the other.
- Fs = - (k * deltaX). Spring force, as proven by our findings in our spring demo, is the negative of the product of the spring's constant of k by the distance the spring is stretched/compressed. The spring's constant is essentially the sturdiness or stretchiness of the spring.
- The force of friction is when two objects rub against each other, creating a force in the opposite direction of an object's motion.
f = mew * Fn -> Force of friction = coefficient of friction * normal force. This in accordance with Hooke's law that states the coefficient of friction is equal to the force of friction divided b the normal force. The coefficient of friction, or mew, is essentially the surface's roughness that plays a huge role in the magnitude of the friction force. Oftentimes, we can use our knowledge of the force of gravity, which can help us find normal force and, other given variables to find the force of friction. This can help us determine the net force of whatever direction the friction force is in.
Solving Force Problems -
A first pivotal step is writing down as many given variables as possible and also creating a force diagram to gain a better understanding of what forces are at play. By writing down given variables, we can use the equations listed above to create a system of equations where we can ultimately find the x and y-component forces. Then, after filling out a force table, we can solve for the net force in all directions.
(Problem From OneNote Slides) - First, I wrote out all my givens, applied force, friction force, and the mass. I then created a force diagram depicting the motion of the applied force to the left being stronger than the friction force. I knew gravity and normal force had to cancel out here, because there was no vertical motion. Then, I used the equation for gravity force (37kg * -10 N/kg) to find that the gravity force is -370 N, and, from there, I knew the normal force was 370 since they had to cancel out. I showed this in a force table, and found there is a net force of -26 N (to the left) in the x-component.
Relating Representations of Motion and Force Models:
The major representations of motion, as we learned in the kinematics unit, are position-time, velocity-time, and acceleration-time graphs. To recap, a velocity-time graph represents velocity over time (the change in position), the slope of the position time-graph. In turn, an acceleration-time graph represents acceleration over time, which is the slope and derivative of a v-t graph. This is why an a-t graph is the derivative of a v-t graph that represents its slope. A net force-time graph is the same as an acceleration-time graph, as they both depict the change in velocity over time. This means that, like an a-t graph, a force-time graph also is the derivative and slope of a v-t graph and the second derivative of a p-t graph.
The major representations of motion, as we learned in the kinematics unit, are position-time, velocity-time, and acceleration-time graphs. To recap, a velocity-time graph represents velocity over time (the change in position), the slope of the position time-graph. In turn, an acceleration-time graph represents acceleration over time, which is the slope and derivative of a v-t graph. This is why an a-t graph is the derivative of a v-t graph that represents its slope. A net force-time graph is the same as an acceleration-time graph, as they both depict the change in velocity over time. This means that, like an a-t graph, a force-time graph also is the derivative and slope of a v-t graph and the second derivative of a p-t graph.
The graphs above depict this relationship, as the f-t graph is 0, and, as a result, the derivative of the v-t graph where y = a certain nonzero number. The slope of the v-t graph is 0, which is why the f-t graph's acceleration is constant at zero. The v-t graph represents the p-t graph's slope and is its derivative. This problem is inspired by question 3a on the 2.2 Mastering Physics homework.
Solving Problems of Forces and Motion -
These problems take the previously-mentioned force calculations to another level, as they require us to use the kinematic equations in addition to the force equations found above. In doing so, we employ systems of equations because there are more variables to deal with. This is why writing down all the variables beforehand and using force diagrams and tables are even more important, as they help us better organize problems with a lot of variables. Acceleration is key, as it is involved in both the equation for newton's second law ( F = ma) and also appears in several kinematic equations.
These problems take the previously-mentioned force calculations to another level, as they require us to use the kinematic equations in addition to the force equations found above. In doing so, we employ systems of equations because there are more variables to deal with. This is why writing down all the variables beforehand and using force diagrams and tables are even more important, as they help us better organize problems with a lot of variables. Acceleration is key, as it is involved in both the equation for newton's second law ( F = ma) and also appears in several kinematic equations.