This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

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From the course by University of Houston System

Preparing for the AP Physics 1 Exam

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This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

From the lesson

Kinematics

Topics include velocity, acceleration, free fall, projectile problems and graphical analysis. You will watch 3 videos, complete 3 sets of practice problems and take 2 quizzes. Please click on the resources tab for each video to view the corresponding student handout.

- Dr. Paige K. EvansClinical Associate Professor

Math - Mariam ManuelInstructional Assistant Professor

University of Houston

This first video, Kinematics A, is called Describing Motion.

Â Let's get started and jump right in.

Â When describing motion,

Â it is important to know where an object is located at various points in time.

Â This can be more complicated than it seems.

Â Consider for a moment frames of reference.

Â A frame of reference describes the motion of an observer relative to

Â the object being observed.

Â A stationary observer like this person standing on a sidewalk is not actually at

Â rest in all reference frames.

Â While he might not seem to move relative to the Earth

Â the planet does spin at it's own axis at a rate of

Â approximately one rotation every 24 hours what we call an Earth day.

Â Even the planet itself is hurdling through space as it orbits the Sun.

Â Clearly a choice of reference frames can dramatically change our measurements and

Â must be chosen carefully.

Â Let's go back to the person on the sidewalk.

Â A car traveling at a constant 10 meters per second North passes down the street.

Â According to our observer the car was traveling 10 meters per second North,

Â like you might expect.

Â But now suppose our observer is taking a brisk jog at 4 meters per

Â second in a southward direction, when the car passes as before.

Â According to our observer on the sidewalk, the car is now traveling a total of 14

Â meters per second North, even though the car has not changed its speed relative to

Â the ground, its speed relative to the person has become larger.

Â Notice how in this example, direction was vital to the result that we obtained.

Â It would not be enough to only report the numerical speed,

Â what scientists call magnitude, of our values.

Â By including direction,

Â with magnitude, we're using a vector quantity called velocity.

Â For more information regarding vectors and

Â scalers, look for the corresponding video in the review portion of our website.

Â Looking back again, at the example from before, suppose now that our person is

Â instead jogging North at 4 meters per second, when the same car passes for

Â a third time, traveling 10 meters per second in the same direction.

Â This time the car would seem to move much more slowly to our person on the sidewalk.

Â A mere 6 meters per second.

Â In fact, if the person could run at 10 meters per second the car

Â would look like it is not moving at all.

Â For the purposes of these modules we will assume that our

Â reference frame is with respect to the Earth.

Â What we usually call the ground, unless specified otherwise.

Â >> So, let's take a step back a moment and discuss these terms.

Â Distance, displacement, vector, and scalar.

Â Keep in mind, these terms are defined in a very specific way in physics.

Â Distance is a scalar, which means it represents the numerical value, or

Â magnitude of a quantity only.

Â Another way to think of distance is as the total length traveled by an object.

Â This means that distance, much like time, continues to add on and

Â can never be negative.

Â Let's look at an example.

Â If a car travels 8 meters East, and then 6 meters West,

Â the car will have traveled a total distance of 14 meters.

Â Displacement is a vector quantity which means it includes magnitude as

Â well as direction.

Â It represents the change in the position of an object from where it

Â started its motion to where it finished.

Â In other words displacement is the difference between final and

Â initial conditions.

Â In the example from earlier, the car traveled 8 meters East and

Â then 6 meters West.

Â In this scenario the car's displacement would be 2 meters East.

Â Notice how this value is different from the distance where we had

Â simply added the two values together.

Â Mathematically, we can use the equation to take the difference between the final and

Â initial position values.

Â Notice that in physics it's customary to make North, East, up and

Â right the positive directions.

Â That's why in the example above, the positive 2 was used to represent East

Â similarly, South, West, down and left are considered negative.

Â We will stick to these conventions for the rest of these modules.

Â >> The scalar quantity speed, defined as a distance divided by time.

Â And represented with the equation seen here.

Â The vector quantity velocity is defined as a displacement divided by time,

Â and represented with a similar equation, as seen here.

Â Let's look at the example of Paige on her bicycle.

Â Paige rides her bike 120 meters North in a period of 40 seconds,

Â with a constant velocity.

Â Realizing that she's lost, she stops for 30 seconds to get her bearings.

Â Finding that she has gone too far, Page turns around and

Â travels back 30 meters in a period of 20 seconds with a constant velocity.

Â Calculate the average speed and average velocity of Page's trip.

Â To solve this problem, I'm going to have to find Paige's average speed, and

Â average velocity, and those two numbers will not necessarily be the same.

Â So I'm going to start with speed in this problem, and

Â I'm going to be using my equation, that speed equals distance divided by time.

Â Since we're talking about distance, that's a scalar value.

Â Which means that we don't want to include positive and negative signs for direction.

Â Instead I'll be adding together all the different portions of

Â motion here together.

Â In this case, 2, over the total time of the trip, in this case, 3.

Â Yes we do want to include the time where she's at rest because that is a portion of

Â the trip and will affect the average speed at the end.

Â Like I said we're ignoring negative signs so

Â I'm just including the magnitude, the positive value of these distances.

Â 20 or 120 meters and 30 meters.

Â And then the time.

Â The first portion was 40 seconds.

Â Then there was a 30 second portion.

Â And then there was a 20 second portion.

Â And so I would add the numerator together.

Â Add the new denominator together and divide to find my average speed.

Â Which in this case is about 1.67 meters per second.

Â And I am not going to include the direction with this because again we're

Â talking about speed not velocity.

Â But solving for the velocity now, my equation looks very similar.

Â These arrowheads on top indicate the vector quantity reminding me that I

Â am talking about a v, a vector like velocity and displacement.

Â Well, let's talk about this displacement for a moment.

Â Because, Paige goes North 120 meters at the beginning of the problem.

Â Then waits for a moment.

Â Then heads backwards 30 meters.

Â If I want to know the displacement,

Â it's how far from where I started to where I finished the motion.

Â In this case, Paige has gone a total displacement of 90 meters

Â North, because I went 120 North, then South 30 meters, I'm now 90 meters North.

Â That means here from my displacement, that's the value I will include for

Â this displacement.

Â The time again will be the total time of the trip.

Â So I'm going to need all three times added together for the denominator.

Â And so this is what that might look like.

Â A positive 90 for my displacement.

Â That will help me keep track of my direction.

Â Again, I'm going to add together my times, which looks

Â to me that it's going to give me 90 divided by 90.

Â I get a positive 1 meters per second for my average velocity.

Â And that would give me my direction in this case which is North.

Â Typically you want to only report the direction once because you have a risk for

Â double negatives.

Â So either put the positive sign or the North, I wouldn't keep both.

Â I'm just going to scratch this out from here.

Â Two other things to keep in mind.

Â Notice that the velocity and the speed were very different.

Â They don't have to be the same value.

Â And I made sure to include that nega, or that direction.

Â The other thing that I want to point out is that sometimes what people will try to

Â do is average the two different speeds for

Â each section together to get an average speed, and that won't work.

Â Because they didn't cover each portion of the motion, in the same amount of time and

Â the same distance.

Â So just averaging the two won't give you the appropriate average speed, or

Â average velocity.

Â >> Acceleration is much like velocity and

Â displacement, in that it is a vector quantity.

Â And it is defined as the change in velocity divided by time.

Â It's represented by the following equation.

Â Note for most problems the initial time equals 0 as the start of the problem.

Â Rearranging this equation is also very common as shown here.

Â This is the first of our kinematic equations.

Â The others are also listed here.

Â Please keep in mind that acceleration must remain constant when using these.

Â The equations will be derived at the end of this video, but for

Â now, let's see them in practice.

Â Here's an example.

Â Paige is driving a car with a velocity of 20 meters per second West.

Â When she applies the brakes, and

Â accelerates down to 8 meters per second in 2 seconds.

Â Calculate the average acceleration of the car during this period.

Â Okay, so as they tell us over here, Paige is driving at 20 meters per second West.

Â So I draw out my vector over here.

Â And then they tell me that she applies the brake and

Â accelerates down to 8 meters per second in 2 seconds.

Â So what I know is that she's still traveling in the same direction, and

Â that this transition took her 2 seconds.

Â So they want me to calculate the average acceleration.

Â Well, I look at my equation for acceleration.

Â I have the change in velocity, final minus initial.

Â They're just the vector symbols that I'm adding on here, over the change in time.

Â Again remember, my time, my initial time would just be 0.

Â So my total time here is 2 seconds.

Â So now what I can go ahead and do, is plug in the values that I have.

Â Well, we had talked earlier about directions in physics, ones that

Â are considered positive, and ones that are generally considered negative.

Â Remember South, down, left, and West is considered negative, and

Â since this is a vector, I have to implement that direction.

Â Well, my final velocity is 8 meters per second West, but

Â I'm going to write that down over here as -8.

Â My initial velocity is 20 meters per second West.

Â I'm going to write that out as -20.

Â So I have -8 minus -20.

Â All of that over 2.

Â So if I have -8 and I have minus the -20,

Â well that's just -8 plus 20.

Â So then what I end up with is 12 over 2,

Â I end up with 6 meters per second squared.

Â So now note that this was positive.

Â Positive 6 meters per second squared.

Â Another way to write that out.

Â Would be by saying that the acceleration, the average acceleration,

Â is 6 meters per second squared, going East.

Â You should use one or the other, either the positive or

Â negative sign, or write out the direction.

Â This problem uncovers a common misconception.

Â That is, that students often believe an object's velocity and

Â acceleration need to be in the same direction.

Â But in the example we just saw, you'll notice that while the car is moving West,

Â it is accelerating in the East direction because it's slowing down.

Â So the velocity and

Â acceleration vectors point in the opposite direction lowering the car's speed.

Â This also tells us that if the velocity and acceleration vectors are in

Â the same direction that means that the car is speeding up.

Â Let's use the information we have covered so far to answer a few questions.

Â One.

Â Can an object have a 0 velocity while experiencing an acceleration?

Â Absolutely.

Â Consider what happens when a car starts from rest and accelerates.

Â At the very initial state the car has a velocity of 0.

Â But there must be change in velocity occurring to

Â cause movement in the vehicle.

Â And we know that change in velocity over time is acceleration.

Â Two, can an object have 0 acceleration while having a velocity?

Â Yes!

Â Think of a car traveling at a constant speed by using cruise control.

Â Because there is no change in the magnitude of the velocity or

Â direction there is 0 acceleration.

Â Can an object have a negative acceleration but positive velocity?

Â Sure it can.

Â Remember the example of Paige driving 20 meters per second West and

Â accelerating down to 8 meters per second?

Â In that example we discussed how acceleration in the opposite direction of

Â the velocity caused a decrease in the magnitude of the overall velocity.

Â In other words, she slowed down.

Â >> You have probably noticed that some of the problems we have

Â discussed have included a graph of that object's motion over time.

Â Let's look at some of the information you can learn from these graphs.

Â Recall the example from earlier of Paige riding her bicycle.

Â She had traveled 120 meters North in 40 seconds.

Â Paused for 30 seconds.

Â And traveled South 30 meters in 20 seconds.

Â We can already tell a great deal from this graph.

Â There are three portions of uniform motion during her trip.

Â Based upon the information above,

Â we can tell where Paige is located at any point in time.

Â For instance, she is located at a position x of 120 meters,

Â at time 40 seconds, and remains there until a time of 70 seconds.

Â At the end of her trip we can tell she is located 90 meters north of her

Â starting location after 90 seconds have passed.

Â In physics, the slope of a graph usually has some physical significance.

Â Recall that slope is rise divided by run.

Â Represented by the equation m equals y2 minus y1 divided by x2 minus x1.

Â By using our variables in this graph,

Â we can re-write this equation to look something more like this.

Â m equals x final, minus x initial, divided by t final, minus t initial.

Â This equation looks a lot like our equation for velocity.

Â This is a big idea for us.

Â The slope of a displacement versus time graph

Â represents the velocity of an object.

Â A similar argument can be made regarding the slope of a velocity versus time graph.

Â Dividing the change in the y-axis by the change in the x-axis.

Â Or in this case the change in velocity over time,

Â represents the acceleration of an object.

Â Looking at units, we're actually dividing meters per second, by seconds.

Â Building a quantity that is meters per second squared at acceleration.

Â What else can we tell from a velocity versus time graph?

Â Notice in this example, that the object did not start at rest.

Â The object is moving with a positive velocity, and

Â therefore in a positive direction at the start of the problem when t equals 0.

Â The area under a velocity versus time graph also has significance.

Â Look at the first portion of this object's motion.

Â The area underneath this graph from the function to the x-axis is a triangle.

Â To find the area of this triangle,

Â we will use our equation A equals one half base times height.

Â The base of our triangle would represent the time over which this portion of

Â motion occurred.

Â And the height would represent the change in velocity of the object.

Â Subbing in, we can get an equation that looks like A equals one half t times v.

Â looking at units again, we have seconds multiplied by meters per second.

Â After the seconds cancel out we are left with only meters.

Â We have multiplied it velocity by time and arrived at a displacement.

Â This is one last big idea for graphs.

Â The area can also have significance in physics.

Â When analyzing a velocity versus time graph like the one here,

Â the area under the curve represents the displacement of the object over time.

Â We will leave it up to the student to show that the area under an acceleration versus

Â time graph represents the change in velocity of an object over time.

Â >> To wrap up, let's look at a few scenarios using graphs.

Â Assume that all of the following cases have constant acceleration.

Â In the first example, the car starts from rest,

Â begins to move in the positive x direction, and speeds up.

Â The fact that it is speeding up tells me that the acceleration is in

Â the same direction as the velocity, which is positive.

Â This also tells me that there is an increase in velocity.

Â In the next case we have the same car, now slowing down.

Â This means my acceleration and velocity vectors are in opposite directions.

Â Acceleration has to move against velocity to decrease it.

Â So acceleration is negative, and velocity is decreasing.

Â The third case involves a car moving in the negative direction at

Â a constant speed.

Â Notice that the car did not start from rest.

Â It is already in motion and what I know about that motion is that it is constant.

Â Well if there is no change occurring in the velocity than the acceleration

Â must be 0.

Â Now we have a car moving in the positive direction and

Â speeding up, much like the first example,

Â the car is accelerating in the positive direction, and increasing in its velocity.

Â But, notice the graphs are different between the first and fourth example.

Â This is because, in number four, the car does not start from rest.

Â Lastly we have a car moving in the negative direction and slowing down.

Â So slowing down tells me that the car's velocity and

Â acceleration are in opposite directions.

Â Since the car is moving in the negative direction,

Â acceleration again being in the opposite direction would be positive.

Â As promised earlier, it is now time to derive the equations.

Â Suppose we have a v verses t graph like the one seen here.

Â Notice the constant slope of the graph.

Â Remember the slope of the v verses t graph provides acceleration.

Â Therefore this slope showcases a constant acceleration.

Â Displacement can be found by solving for

Â the area under the curve of a v verses t graph.

Â I'm shading in the triangle and rectangle.

Â I know that the area of the triangle is one-half base times height, and

Â the area of the rectangle is base times height.

Â By combining these, I end up with the following.

Â Well now I can sub in our variables.

Â Note that the base is time, while the height is velocity.

Â I end up with the following.

Â By combining this with my equation for

Â acceleration, I derived the equation seen here.

Â You can also derive the very last equation, shown here,

Â by combining the following two.

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