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

In the last video, we introduced to the three kinematic equations seen here.

Â We made a point to emphasize that these relationships are only accurate in

Â scenarios that experience constant acceleration.

Â Lucky for us, there are many situations which we encounter in our

Â daily lives in which the acceleration of an object is constant.

Â A classic example is that of a freely falling object near the Earth's surface.

Â Let's be clear here about what we mean by an object in free-fall.

Â First, let's ignore air resistance in all of our problems,

Â as this is beyond the scope of AP Physics 1.

Â A good example of free fall is a baseball flying through the air.

Â The only force acting on the ball is the gravitational attraction between it and

Â the Earth causing the ball to accelerate downward.

Â An airplane however is not in free fall.

Â Air molecules collide with the wings of the plane and

Â provide substantial lift to keep the vehicle aloft.

Â Any object on which gravity is the only force acting is said to be in free fall.

Â You have probably heard the statement that all objects near the Earth's surface

Â accelerate toward the center of the planet at the same rate in the absence of

Â air friction.

Â Yes, it is true, as an object gets further away from the Earth,

Â the gravitational attraction between the object and the planet becomes less.

Â This, in turn,

Â means that the acceleration of that object in free fall would also be less.

Â But, in physics, we like to use approximations to

Â simplify our calculations.

Â For all scenarios near the surface of the Earth the change in

Â acceleration as an object falls is so small that it can be ignored.

Â A good average for this constant acceleration across the surface of

Â the Earth is 9.8 meters per second squared.

Â Since it is perfectly acceptable on the AP exam to simplify even further to 10 meters

Â per second squared, you will often see us also using this value in our calculations.

Â When analyzing objects in free fall there are a few common misconceptions that we

Â should also address.

Â In our daily lives, objects may not always fall with the same acceleration.

Â A feather and a bowling bowl will not hit the ground at the same time if

Â dropped from the same height.

Â This discrepancy is due to air friction.

Â Place these two objects in a vacuum tube without air and

Â you will see both hit the ground at the same time.

Â Similarly, heavy objects, once with larger mass, do not accelerate any

Â faster towards the center of the Earth than a lighter object, one with less mass.

Â The same demonstration with the vacuum tubes, feather and

Â bowling ball show us that this is the case.

Â This result may seem odd to you, since a bowling ball does indeed experience more

Â gravitational force from the Earth.

Â This is why it weighs more.

Â But, the ball still accelerates at the same rate as the feather.

Â This behavior will be discussed in more detail when we talk about

Â gravitation in a later module.

Â >> Now let's look at one of the examples from your

Â Kinematics A practice problem set.

Â We're going to work through problems nine and ten because they serve as an excellent

Â example of an object experiencing constant acceleration while in free fall.

Â So this problem says that a soccer ball is kicked vertically upward into the air with

Â an initial velocity of five meters per second.

Â It takes 0.5 seconds for

Â the ball to reach its highest point before beginning its descent.

Â It then asks us two questions about the ball's motion while it is in the air.

Â So the first part of this question asks us to describe the ball's motion as it

Â travels through the air.

Â In particular, note any points at which displacement, velocity, or

Â acceleration of the ball are zero.

Â Notice over here we have plotted the y versus t for this soccer ball's motion.

Â So, let's look at the very start.

Â Right over here where I will label this point as x

Â equals 0, you notice that when it's launched vertically it has it's greatest,

Â it has it's maximum velocity.

Â Well, this velocity begins to change as we travel up

Â towards the very top of the flight.

Â Change in velocity over time.

Â Well that's acceleration, so something is causing this object to accelerate.

Â Well that something in this case, this object's in free fall and

Â that force that's causing it to accelerate is gravity.

Â So, it's accelerating at a rate of 10, -10 that is,

Â meters per second squared due to gravity.

Â Remember, that this acceleration is directed towards the center of

Â the Earth each time.

Â So at each point, you have that constant acceleration acting on this object.

Â When it reaches its very highest point.

Â I know that my vertical velocity is going to equal 0 at this point.

Â But people often think that acceleration might be 0 here as well.

Â This isn't true.

Â Remember that you can have an acceleration without having a velocity.

Â My acceleration at this point is still the -10 meters per

Â second squared due to gravity.

Â Now let's look again at something over here.

Â You notice where, right over here, I drew out,

Â rough sketched that is, of the vector for acceleration.

Â Acceleration velocity are in opposite directions.

Â We talked about this in the previous module, where when acceleration and

Â velocity are in opposite directions, the object is slowing down.

Â Well if that's true then let's look at what's happening at this point

Â over here now.

Â Now I have velocity directed downward, so I have a negative velocity.

Â But I also have acceleration acting, as well, downwards.

Â If acceleration and velocity are in the same direction,

Â the object is going to increase in its magnitude of velocity.

Â So, it reaches its maximum velocity, look at that, once again, at x equals 0.

Â So you no, you start noticing that there's symmetry here, right?

Â If I look at velocity say, at point over here, and

Â I look at the velocity over here, while they're in

Â opposite directions the magnitude of these vectors is still the same.

Â And that continues.

Â What happens right over here at the point where my vertical velocity is 0?

Â Well, this is half of the time of the flight.

Â They tell me that it reached its highest point at t equals 0.5 seconds.

Â Well, according to the symmetry, if it took 0.5 seconds for this object to

Â move up, then it's going to take 0.5 seconds for this object to come back down.

Â Which means right over here at the very end when the object comes back to

Â the ground, the time should be 0.5 plus 0.5.

Â One second.

Â Let's go ahead and look at a little bit more detail about these, vectors.

Â So, at the point where my time is 0, my x was 0, velocity was maximum and upwards.

Â Time 0.5, my x value was at its maximum,

Â that's when it had reached its highest point.

Â But my velocity was at its lowest, my vertical velocity equaled 0.

Â At the full length of its travel, time equals 1 second.

Â Position, back again, same place, so my displacement is 0.

Â Velocity is maximum at this time, downwards, or negative.

Â And notice that the acceleration, the whole way through, was constant.

Â Downwards, pointing towards the center of the Earth,

Â the object was accelerating due to gravity.

Â Gravity being the only force acting on it.

Â Okay, so now it wants us to calculate the acceleration of the ball as it

Â travels through air.

Â In what direction is this acceleration vector oriented?

Â So, if you've watched the module and worked through the previous question,

Â you probably already know what this answer's going to be.

Â But, let's go ahead and prove it.

Â So, my acceleration equation tells me that acceleration

Â equals final velocity minus initial velocity over the change in time.

Â Note that in most cases, such as this one, the initial time is 0.

Â So, let's go ahead and solve for that acceleration.

Â Well, based on this previous question,

Â we knew that we had time of 0.5 seconds at the very top of the flight.

Â Due to the symmetry that we've been discussing,

Â we know that time up equals time down, so I know that my total time is 1 second.

Â Also, same symmetry.

Â I know that my initial velocity here was 5 meters per second.

Â I note that the final velocity is going to be the same in magnitude.

Â But notice there's a difference in the direction.

Â Now my final velocity vector is pointing downwards.

Â So this final velocity.

Â Is -5 meters per second.

Â Let's plug these values in and see what we get.

Â So, my acceleration is going to equal -5.

Â Minus that positive 5 initial velocity over time, which is 1.

Â This gives me an acceleration of -10 meters per second squared.

Â Perfect.

Â That's what we expected, right?

Â That is the acceleration due to gravity.

Â Well, another thing we talked about that was that when we're using these

Â kinematics equations it's under the assumption that acceleration is constant.

Â Well, let's check to make sure that that, that this acceleration is constant.

Â Say I look at the very top of the flight at this point right here.

Â Where I had T equals 0.5 seconds.

Â At this point I knew that my vertical velocity was equaling 0.

Â So let's go ahead and

Â solve for the acceleration at that point just up to there.

Â So, going to scroll down giving us some room.

Â And now I once again have my equation.

Â Minus v initial over the change in time.

Â This time, my time is different.

Â Final velocity vertically is 0.

Â We're looking at the half way point.

Â Initial velocity is still the positive 5.

Â Time is 0.5.

Â I once again end up with -10 meters per second squared.

Â So, there you go.

Â Acceleration due to gravity is -10 meters per second squared.

Â It had asked us which direction the vector was oriented in.

Â The negative indicates that it's oriented downwards.

Â Now that we have broken this object's motion down into its parts, we know a lot

Â more information about its displacement, velocity and acceleration over time.

Â So now we can grasp the object's motion, and

Â this is going to provide us with even more insight.

Â Let's start with the acceleration graph for the soccer ball.

Â The object is in free fall, just like the mat in the earlier problem, we know that

Â the object's acceleration is at a constant of -10 meters per second squared.

Â Since this is constant, I'm going to draw a horizontal line on the graph,

Â showcasing this.

Â Now let's look at the velocity versus time graph.

Â We know that the object starts with a positive upward velocity of

Â 5 meters per second.

Â At the height of its flight, which is at 0.5 seconds,

Â the ball now has a velocity of 0 in the y direction.

Â This is important because remember, once the ball reaches its highest point,

Â it's going to start coming back down.

Â In order for it to do that, it has to stop moving vertically up.

Â So, we have a 0 velocity in that vertical direction.

Â Because of the symmetry of the situation the ball takes the same amount of time to

Â return to its original location.

Â We often refer to this as time up equaling time down.

Â Now, it has a negative, or in this case, downwards velocity.

Â We can see that as -5 meters per second.

Â We must now connect three points using the information from our

Â acceleration versus time graph.

Â Recall that the slope of the velocity versus time graph is the acceleration of

Â the object.

Â In this case, our graph needs to have a constant slope,

Â -10 meters per second squared.

Â Which is a line of negative slope.

Â Now let's look at the displacement versus time graph.

Â The soccer ball starts at what we call a height of 0.

Â This is the point in which the ball leaves the kicker's foot and is in free fall.

Â From my calculations, I know that the ball would reach a maximum height of

Â 1.25 meters at time 0.5 seconds.

Â So we can plot this point as well.

Â Since it took the ball 0.5 seconds to reach the height,

Â it will take the ball equally as long to reach back down to the ground.

Â This means that at t equals 1 second, the ball will have returned to

Â the position x equals 0, in other words, the ground.

Â To connect these points, we have to look again at the values of the previous graph.

Â The slope of a Displacement verses time graph is the velocity of the object.

Â The velocity of the ball was positive, became 0, and then became negative.

Â This means the slope of our Position verses time graph should be positive,

Â then 0, and then negative.

Â Connecting our three points yields a parabola to

Â describe the ball's displacement over time.

Â Earlier we solved for the height, mathematically.

Â But couldn't we do this graphically, as well?

Â Sure.

Â Let's look at the area underneath the velocity versus time graph, and

Â compare that result to the displacement calculated earlier.

Â Calculating the area underneath this graph for the first portion of motion,

Â will tell us how far the ball has traveled on the upward portion of its motion.

Â That is during time up.

Â By calculating the area of the triangle as seen here,

Â we can see that the ball should have moved upwards a total of 1.25 meters.

Â This matches what our displacement graph in earlier calculations found.

Â >> Let's look at an example now in which the object in motion does not start and

Â end at the same position.

Â A rock is thrown vertically upward from a window with a speed of

Â 6 meters per second.

Â It hits the ground 5 seconds later.

Â What is the height of the window above the ground?

Â Notice that in problems like these, the verb used in the description is important.

Â Words like thrown, launched, or kick imply initial velocity of some kind.

Â Words like dropped, fall, or released typically imply an initial velocity of 0.

Â As always, pay careful attention to your positive and

Â negative signs when setting up these problems.

Â So to help me solve this problem, I'm going to start off with

Â a quick sketch of the scenario to help me visualize what's going on.

Â We have a ball being thrown up and coming back down and hitting the ground.

Â But that window is some height above the ground and that's what we want to know.

Â Well, this is different from some of the other problems we've solved,

Â because, the starting location is not the same as the finishing location.

Â And that might seem confusing at first.

Â But what's helpful about this, is that, our equations, things like x equals

Â v naught t plus one half a t squared.

Â Well this equation is a displacement.

Â It's a vector.

Â And so, I can plug in any point in time this function, and

Â it will tell me where the object is located at any point in time,

Â even if that's above, below, or at the starting location.

Â And so that's my approach to this problem.

Â I'm going to start off.

Â And I know my initial velocity, it was moving upward with a speed here of 6 m/s,

Â so that's a positive 6.

Â I know that I am looking for

Â its location 5 seconds later, so I'm going to plug in time plus one half.

Â Anything that's in free fall, we're talking about the vertical direction.

Â Has a -10 meters per second squared acceleration.

Â And again, I'm plugging in my 5 seconds and not forgetting to square this

Â term here, which is, a common mistake when it comes to doing these calculations.

Â And so plugging those numbers into my calculator and, and

Â running through that calculation, I get a -95 meters.

Â Which is a good result.

Â That's what we were looking for.

Â This window was above the ground,

Â and the ball ends up lower, below, that window by 95 meters.

Â Our window is 95 meters above the ground, and

Â that would be my answer to this problem.

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