Відео

20 is the value of the hypotenuse. So it’s sin30 = O/20, which is algebraically rearranged to 20sin30=O

Of course! I'm glad you found it helpful!

Yes. These MCQs have the same structure as questions you will see on the AP exam, just different numbers.

Glad you like them!

Thanks!

Yeah, sorry about that. I just circled the wrong one, the answer (0.25) is visible in the screen.

@@studentlearningcenter2014 no problem, i watched all your videos to help me prepare for my AP Physics 1 tests; they were all very helpful, thank you!

Thanks for the support! Good luck on all your tests!

No problem! Thanks for watching!

The equations are set up using the geometric and physical constraints set in the problem. One of the most important steps here is to check your equations to make sure you are optimizing the right thing. Next, you just solve it using algebraic techniques.

We can find the tangent line using point-slope form, with the derivative as the slope and the point of tangency on the curve as the point.

To have continuity, you need two different things. First, the left hand limit and right hand limit need to be the same. Next, the actual value of the function must equal these limits. We use limits here to express how the function reacts as we approach a point, and whether it it continuous as it actually crosses the point.

These FRQs have less parts since those parts are the only ones that you will be able to do at this stage in the course. In the later videos, you can see that we have full FRQs since by that point, you have the tools necessary to solve the whole FRQ.

The force of static friction is calculated as an INEQUALITY, not an equation. So, the force of static friction is less than or equal to the coefficient of static friction times the normal force. As the applied force increases, the static frictional force will balance it out, until the applied force gets too high.

Kinetic friction is an equation; it is the normal force times the coefficient of kinetic friction (a different coefficient), so it is always constant as long as the normal force is constant.

Yes, much of the course is applying similar content in many different ways. This is actually the reason I made the videos; since many people thought it was a lot of content and weren't focusing on how to apply it.

Yes, just about all problems in kinematics can be solved with just applying the kinematics formulas with gravitational acceleration and whatever variables given in the problem.

I am planning to make a video like this for the reading section and I will compile a list of strategies for both the reading and writing sections.

Yes, I am planning to make a video on all the strategies to ace the math section.

For any rope, the tension will be constant throughout. The mass of the counterweight determines the force of gravity on it, which can be used to determine the force of tension by Newton's Second Law (Net Force = Mass * Acceleration), and we knew both the mass and acceleration.

The normal force and gravitational force are not always equal, in any scenario. The normal force is simply a force that balances out the force along a certain axis. On inclines, the perpendicular direction to the incline will have no net force, since the normal force balances it out. Here, the upward force is the vertical component of the applied force plus the normal force, which equals the gravitational force. Therefore, the normal force is less, so the force of friction is less.

Yes, some questions will have parts like that. However, some AP Physics questions are also structured to have only one part. These are called the long answer questions and are based on explaining your answer in a long format with extensive writing instead of mostly math and steps.

For the long answer questions, how much do we have to write? Also, what is the best way to have a step by step process for the mathematical questions?

The long answer questions should be about a paragraph long. In the question, you will use some math, but the emphasis should be on explanation of concepts. For the mathematical questions, just be sure to show each step and what you are doing, and talk about when you are using a specific physics concept.

The kinematics equations will be located on the AP exam, but I think it is best to know them by heart to apply them quickly. Fortunately, the kinematics equations will be easier to apply and remember as you do more practice.

You can figure out which kinematics equation to use based on the variables you have. The x = vt + 0.5at^2 involves the variables displacement, initial velocity, time and acceleration. If you have three of those variables and need the fourth one, use this equation. You also know this based on which variable you DON'T have or don't need, which would be final velocity in this situation. You can do this for each kinematics equation.

It depends. This late in the admissions process, I'd recommend just not taking it again. A score of 1350 with good scores in some APs (most notably AP Calculus or APUSH/AP Lang) should be enough. It also depends on the score breakdown (Math vs. RWL), but I don't think it would make much difference compared to the rest of your application.

To find the equation, we relate the value they are giving us (like radius) to the value they want the rate of (the volume). We know this through geometric formulas. You don't have to memorize these, since the AP exam will give you a sheet of these formulas that you can use.

Are there broad topics of problems with related rates so I can understand how to approach each of them? Thanks.

Most related rates problems are based heavily on geometry. For problems like the shadow problem (rate of change of the length of shadow) and area/volume problems, just try setting up the equations and then solving. For more practice, you can visit Khan Academy's section on Related Rates.

If the force was not constant, we could either take the area under the curve of a force vs. distance graph, or perform an integral of the force with respect to distance. However, you will not have to deal with these topics in AP Physics I since it is Algebra-Based on not Calculus based.

In the second scenario, we have the hanging block as well as the block on the surface. The hanging block's net force is the force of gravity minus the tension. The surface block's net force is the tension minus the friction. To find the total, we add these together, causing the tension to cancel out.

Why do we need to find the total force? Could we not individually determine the acceleration for each one?

In an Atwood Machine, we know that the acceleration for the blocks is the same, so we add them together to cancel out the tension and solve more easily.

If the object was being represented as a point, then yes, we would draw all forces from the same point, since there would only be one point to draw from. However, very few things are actually just points, so we have to draw forces from where they are act. The gravitational force acts on the center of mass, so in the center. The force of friction acts between the ground and the object, so we draw it on an edge. The only other force that we draw not from the center is torque, since it acts rotationally and not transitionally.

How do we know if something is represented as a point? Will the question just tell us to assume it is a point?

Usually, you can assume something is a point for the sake of calculations. When drawing a free body diagram, the question will provide you with a shape to draw off of, and it will either be a point or a shape.

Also, why is the area under the velocity vs. time graph equal to the displacement? How do these relate?

We can think of the object has having two completely SEPARATE components: the horizontal and the vertical. Since they are separate, anything in one component does not affect the other. So, a vertical acceleration will not change the horizontal velocity. Since there is no horizontal force or acceleration, the horizontal velocity will be constant.

The area under the velocity vs. time graph is equal to displacement since you are continually multiplying the velocity times time to get small sections of displacement, which are then being added up. If you know calculus, this is essentially the same as a definite integral. Also, the area under the acceleration vs. time graph will be the velocity.

Thanks for the clarification! And thanks for the amazing content!!

Of course!

Thanks! I'm glad you found the video helpful!!

fr dude

Thank you!

1590. I scored 800 on Math, 400 on Reading, and 390 on Writing/Language.