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COMPLETE Elementary Physics 1 Notes (90% on final)

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Physics
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CAS PY 105
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Motion in one dimension We live in a 3-dimensional world, so why bother analyzing 1-dimensional situations? Basically, because any translational (straight-line, as opposed to rotational) motion problem can be separated into one or more 1-dimensional problems. Problems are often analyzed this way in physics; a complex problem can often be reduced to a series of simpler problems. The first step in solving a problem is to set up a coordinate system. This defines an origin (a starting point) as well as positive and negative directions. We'll also need to distinguish between scalars and vectors. Ascalar is something that has only a magnitude, like area or temperature, while a vector has both a magnitude and a direction, like displacement or velocity. In analyzing the motion of objects, there are four basic parameters to keep track of. These are time, displacement, velocity, and acceleration. Time is a scalar, while the other three are vectors. In 1 dimension, however, it's difficult to see the difference between a scalar and a vector! The difference will be more obvious in 2 dimensions. Displacement The displacement represents the distance traveled, but it is a vector, so it also gives the direction. If you start in a particular spot and then move north 5 meters from where you started, your displacement is 5 m north. If you then turn around and go back, with a displacement of 5 m south, you would have traveled a total distance of 10 m, but your net displacement is zero, because you're back where you started. Displacement is the difference between your final position (x) and your starting point (x )o: Speed and velocity Imagine that on your way to class one morning, you leave home on time, and you walk at 3 m/s east towards campus.After exactly one minute you realize that you've left your physics assignment at home, so you turn around and run, at 6 m/s, back to get it. You're running twice as fast as you walked, so it takes half as long (30 seconds) to get home again. There are several ways to analyze those 90 seconds between the time you left home and the time you arrived back again. One number to calculate is your average speed, which is defined as the total distance covered divided by the time. If you walked for 60 seconds at 3 m/s, you covered 180 m. You covered the same distance on the way back, so you went 360 m in 90 seconds. Average speed = distance / elapsed time = 360 / 90 = 4 m/s. The average velocity, on the other hand, is given by: Average velocity = displacement / elapsed time. In this case, your average velocity for the round trip is zero, because you're back where you started so the displacement is zero We usually think about speed and velocity in terms of their instantaneous values, which tell us how fast, and, for velocity, in what direction an object is traveling at a particular instant. The instantaneous velocity is defined as the rate of change of position with time, for a very small time interval. In a particular time interval delta t, if the displacement is , the velocity during that time interval is: The instantaneous speed is simply the magnitude of the instantaneous velocity. Acceleration An object accelerates whenever its velocity changes. Going back to the example we used above, let's say instead of instantly breaking into a run the moment you turned around, you steadily increased your velocity from 3m/s west to 6 m/s west in a 10 second period. If your velocity increased at a constant rate, you experienced a constant acceleration of 0.3 m/s per second (or, 0.3 m/s ). We can figure out the average velocity during this time. If the acceleration is constant, which it is in this case, then the average velocity is simply the average of the initial and final velocities. The average of 3 m/s west and 6 m/s west is 4.5 m/s west. This average velocity can then be used to calculate the distance you traveled during your acceleration period, which was 10 seconds long. The distance is simply the average velocity multiplied by the time interval, so 45 m. Similar to the way the average velocity is related to the displacement, the average acceleration is related to the change in velocity: the average acceleration is the change in velocity over the time interval (in this case a change in velocity of 3 m/s in a time interval of 10 seconds). The instantaneous acceleration is given by: As with the instantaneous velocity, the time interval is very small (unless the acceleration is constant, and then the time interval can be as big as we feel like making it). On the way out, you traveled at a constant velocity, so your acceleration was zero. On the trip back your instantaneous acceleration was 0.3 m/s for the first 10 seconds, and then zero after that as you maintained your top speed. Just as you arrived back at your front door, your instantaneous acceleration would be negative, because your velocity drops from 6 m/s west to zero in a small time interval. If you took 2 seconds to come to a stop, your acceleration is -6 / 2 = -3 m/s . Kinematics equations when the acceleration is constant When the acceleration of an object is constant, calculations of the distance traveled by an object, the velocity it's traveling at a particular time, and/or the time it takes to reach a particular velocity or go a particular distance, are simplified. There are four equations that can be used to relate the different variables, so that knowing some of the variables allows the others to be determined. Note that the equations apply under these conditions: 1. the acceleration is constant 2. the motion is measured from t = 0 the equations are vector equations, but the variables are not normally written in bold letters. The fact that they are vectors comes in, however, with positive and negative signs. The equations are: The equations above are all derived in section 2.5. ConstantAcceleration Applying the equations Doing a sample problem is probably the best way to see how you would use the kinematics equations. Let's say you're driving in your car, approaching a red light on CommonwealthAvenue. Ablack Porsche is stopped at the light in the right lane, but there's no-one in the left lane, so you pull into the left lane. You're traveling at 40 km/hr, and when you're 15 meters from the stop line the light turns green. You sail through the green light at a constant speed of 40 km/hr and pass the Porsche, which accelerated from rest at a constant rate of 3 m/s starting at the moment the light turned green. (a) How far from the stop line do you pass the Porsche? (b) When does the Porsche pass you? (c) If a Boston police officer happens to get you and the Porsche on the radar gun at the instant the Porsche passes you, will either of you be pulled over for speeding?Assume the speed limit is 50 km/hr. Step 1 - Write down everything you know. Define an origin - the stop line is a good choice in this problem. Then choose a positive direction. In this case, let's take the positive direction to be the direction you're traveling. Decide on a system of units...meters and seconds is a good choice here, so convert your speed to m/s from km/hr. Drawing a diagram is also a good idea. Origin = stop line Positive direction = the direction you're traveling Step 2 - Figure out what you need to solve for.At the instant you pass the Porsche, the x values (yours and the Porsche's) have to be equal. You're both the same distance from the stop line, in other words. Write out the expression for your x-value and the Porsche's. We'll use the equation: For you : x = -15 + 11.11 t + 0 2 2 For the Porsche : x = 0 + 0 + 1/2 (3) t = 1.5 t At some time t, when you pass the Porsche, these x values will be the same. So, we can set the equations equal to one another and solve for time, and then plug the time back in to either x equation to get the distance from the stop line. Doing this gives: -15 + 11.11 t = 1.5 t 2 Bringing everything to one side gives: 1.5 t - 11.11 t + 15 = 0 This is a quadratic equation, which we can solve using the quadratic formula: where a = 1.5, b = -11.11, and c = 15 This gives two values for t, t = 1.776 s and t = 5.631 s. What do these two values mean? In many cases only one answer will be relevant, and you'll have to figure out which. In this case both are relevant. The smaller value is when you pass the Porsche, while the larger one is when the Porsche passes you back. To get the answer to question (a), plug t = 1.776 into either of your x expressions. They should both give you the same value for x, so you can use one as a check. For you, at t = 1.776, x = 4.73 m. For the Porsche, at t = 1.776 s, x = 4.73 m. We've actually already calculated the answer to (b), when the Porsche passes you, which is at t = 5.6 s. To get the answer to part (c), we already know that you're traveling at a constant speed of 40 km/hr, which is under the speed limit. To figure out how fast the Porsche is going at t = 5.631 seconds, use: v = v o a t = 0 + (3) (5.631) = 16.893 m/s. Converting this to km/hr gives a speed of 60.8 km/hr, so the driver of the Porsche is in danger of getting a speeding ticket. Free fall Objects falling straight down under the influence of gravity are excellent examples of objects traveling at constant acceleration in one dimension. This also applies to anything you throw straight up in the air which, because of the constant acceleration downwards, will rise until the velocity drops to zero and then will fall back down again. The acceleration experienced by a dropped or thrown object while it is in flight comes from the gravitational force exerted on the object by the Earth. If we're dealing with objects at the Earth's surface, which we usually are, we call this acceleration g, which has 2 a value of 9.8 m/s . This value is determined by three things: the mass of the Earth, the radius of the Earth, and a number called the universal gravitational constant. We'll be dealing with all that later in the semester, though, so don't worry about it yet. For now, all you need to remember is that g is 9.8 m/s at the surface of the Earth, directed down. Atypical one-dimensional free fall question (free fall meaning that the only acceleration we have to worry about is g) might go like this. You throw a ball straight up. It leaves your hand at 12.0 m/s. (a) How high does it go? (b) If, when the ball is on the way down, you catch it at the same height at which you let it go, how long was it in flight? (c) How fast is it traveling when you catch it? Origin = height at which it leaves your hand Positive direction = up (a) At the very top of its flight, the ball has an instantaneous velocity of zero. We can plug v = 0 into the equation: This gives:0 = 144 + 2 (-9.8) x Solving for x gives x = 7.35 m, so the ball goes 7.35 m high. (b) To analyze the rest of the problem, it's helpful to remember that the down half of the trip is a mirror image of the up half. In other words, if, while going up, the ball passes through a particular height at a particular velocity (2 m/s up, for example), on its way down it will pass through that height at the same speed, with its velocity directed down rather than up. This means that the up half of the trip takes the same time as the down half of the trip, so we could just figure out how long it takes to reach its maximum height, and then double that to get the total time. Another way to do it is simply to plug x = 0 into the equation: This gives 0 = 0 + 12 t - 4.9 t 2 Afactor of t can be canceled out of both terms, leaving: 0 = 12 - 4.9 t, which gives a time of t = 12 / 4.9 = 2.45 s. (c) The answer for part (c) has to be 12 m/s down, because of the mirror-image relationship between the up half of the flight and the down half. We could also figure it out using the equation: v = vo + a t which gives: v = 12 - 9.8 (2.45) = -12 m/s. Graphical analysis; and Vectors Drawing good pictures can be the secret to solving physics problems. It's amazing how much information you can get from a diagram. We also usually need equations to find numerical solutions. Graphs are basically pictures of equations. Learning how to interpret these pictures can really help you understand physics. Let's return to our last example, a ball thrown vertically upward with an initial speed of 12 m/s. The only acceleration we have to worry about is the acceleration due to gravity, 9.8 m/s down. This acceleration is constant, so it's easy to plot on a graph. If the time T represents the time when the ball returns to your hand, the area under the curve must equal -24.0 m/s, because we know the velocity changes from 12.0 m/s to -12.0 m/s. This allows us to solve for T, using: T = -24.0 / -9.8 = 2.45 s, agreeing with what we calculated previously. What about the velocity graph? The equation for velocity is: v = vo + at Plugging in the initial velocity and acceleration here gives v = 12.0 -9.8t The velocity graph can give all sorts of information: 3. The slope of the velocity graph is the acceleration, while the area under the curve is the displacement. 4. For this example of a ball going up and then back down, the graph confirms that the time taken on the way up equals the time taken on the way down. 5. Calculating the area under the curve for the ball on the way up (the positive area on the graph) gives the maximum displacement. The area is just a triangle, with a a base T/2 = 1.2245 s and height of 12 m/s. This gives an area of 0.5(1.2245)(12) = 7.35 m.Again, this agrees with the maximum height calculated previously. 6. The positive and negative areas cancel each other out, meaning the net displacement is zero. This is correct, as the ball returns to its starting point. The graph of position as a function of time is a plot of the equation: 2 2 x = xo + vot + 1/2 at In this case, that's x = 0 + 12t -4.9t The slope of the position graph gives the instantaneous velocity. This is positive but steadily decreasing on the way up, zero at the very top, and then becomes more and more negative on the way down. Vectors and scalars We'll move on from looking at motion in one dimension to motion in two or three dimensions. It's critical now to distinguish between two kinds of quantities, scalars and vectors.Ascalar is something that's just a number with a unit, like mass ( 200 g ) or temperature ( -30°C). Avector has both a number and a direction, like velocity. If you came to campus on the T today, at some point you may have been traveling 20 km/hr east. Velocity is a combination of a scalar (speed, 20 km/hr) and a direction (east). Examples of scalars : mass, temperature, speed, distance Examples of vectors : displacement, velocity, acceleration, force One crucial difference between scalars and vectors involves the use of plus and minus signs.Ascalar with a negative sign means something very different from a scalar with a plus sign; +30°C feels an awful lot different than -30°C, for example. With a vector the sign simply tells you the direction of the vector. If you're traveling with a velocity of 20 km/hr east, it means you're traveling east, and your speed is 20 km/hr.Avelocity of -20 km/hr east also means that you're traveling at a speed of 20 km/hr, but in the direction opposite to east : 20 km/hr west, in other words. With a vector, the negative sign can always be incorporated into the direction. Note that a vector will normally be written in bold, like this : A.Ascalar, like the magnitude of the vector, will not be in bold face (e.g.,A). Components of a vector Avector pointing in a random direction in the x-y plane has x and y components: it can be split into two vectors, one entirely in the x-direction (the x-component) and one entirely in the y-direction (the y-component).Added together, the two components give the original vector. The easiest way to add or subtract vectors, which is often required in physics, is to add or subtract components. Splitting a vector into its components involves nothing more complicated than the trigonometry associated with a right-angled triangle. Consider the following example. Avector, which we will call A, has a length of 5.00 cm and points at an angle of 25.0° above the negative x-axis, as shown in the diagram. The x and y components of A, A andxA are fyund by drawing right-angled triangles, as shown. Only one right-angled triangle is actually necessary; the two shown in the diagram are identical. Knowing the length of A, and the angle of 25.0°,A andAy cax be found by re-arranging the expressions for sin and cos. Note that this analysis, using trigonometry, produces just the magnitudes of the vectors A xnd A . yhe directions can be found by looking on the diagram. Usually, positive x is to the right of the origin; A pxints left, so it is negative: A x -4.53 cm in the x-direction (or, 4.53 cm in the negative x-direction) Positive y is generally up; A isydirected up, so it is positive: A y 2.11 cm in the y-direction Adding vectors It's fairly easy to add vectors that are parallel to each other, or at right angles. In general, however, the angle between vectors being added (or subtracted) will be something other than 0, 90, or 180°. Consider the following example, where the vector C equals A + B. A has a length of 5.00 cm and points at an angle of 25.0° above the negative x-axis. B has a length of 7.00 cm and points at an angle of 40.0° above the positive x-axis. If C = A + B, what is the magnitude and direction of C? There are basically two ways to answer this. One way is to draw a vector diagram, moving the tail of B to the head of A, or vice versa. The vector C will then extend from the origin to wherever the tip of the second vector is. The second way to find the magnitude and direction of C we'll use a lot in this course, because we'll often have vector equations of the formC =A + B. The simplest way to solve any vector equation is to split it up into one-dimensional equations, one equation for each dimension of the vector. In this case we're working in two dimensions, so the one vector equation can be replaced by two equations: In the x-direction : C = A + B x x x In the y-direction : C = y + B y y In other words, to find the magnitude and direction of C, the vectors A and B are split into components. The components are: A x -4.532 cm in the x-direction A y 2.113 cm in the y-direction B x 7.00 cos40 = 5.362 cm By = 7.00 sin40 = 4.500 cm B x 5.362 cm in the x-direction B y 4.500 cm in the y-direction The components of C are found by adding the components of A and B: C x A + x = (-x.532 + 5.362) cm in the x-direction = 0.83 cm in the x-direction C y A + y = (y.113 + 4.500) cm in the y-direction = 6.61 cm in the y-direction The magnitude of C can be found from its components using the Pythagorean theorem: The direction of C can be found by taking the inverse tangent of C /C : y x inverse tan of 6.613 / 0.830 = 82.8°. Combined, this gives C = 6.66 cm at an angle of 82.8° above the positive x-axis. Note how the calculations and the diagrams go hand-in-hand. This will often be the case; it is always a good idea to draw diagrams as you go along. Note also that we could have used the cosine law to get the length of C, and then applied the sine law, with a bit of geometry, to get the angle. It's worth trying that for yourself, just to convince yourself that the number is correct. Motion in two dimensions Extending things from 1 dimension In 1 dimension, we wrote down some general equations relating velocity to displacement, and relating acceleration to the change in velocity. We also wrote down the four equations that apply in the special case where the acceleration is constant. We're going to do the same thing in 2 dimensions, and the equations will look similar; this shouldn't be surprising because, as we will see, a two (or three) dimensional problem can always be broken down into two (or three) 1-dimensional problems. When we're dealing with more than 1 dimension (and we'll focus on 2D, but we could use these same equations for 3D), the position is represented by the vector r. The velocity will still be represented by v and the acceleration by a. In general, the average velocity will be given by: The instantaneous velocity is given by a similar formula, with the condition that a very small time interval is used to measure the displacement. Asimilar formula gives the average acceleration: Again, the instantaneous acceleration is found by measuring the change in velocity over a small time interval. The constant acceleration equations When the acceleration is constant, we can write out four equations relating the displacement, initial velocity, velocity, acceleration, and time for each dimension. Like the 1D equations, these apply under the following conditions: 7. the acceleration is constant 8. the motion is measured from t = 0 9. the equations are vector equations, but the variables are not normally written in bold letters. The fact that they are vectors comes in, however, with positive and negative signs. If we focus on two dimensions, we get four equations for the x direction and four more for the y direction. The four x equations involve only the x-components, while the four y equations involve only the y-components. One thing to notice is that the time, t, is the only thing that doesn't involve an x or a y. This is because everything else is a vector (or a component of a vector, if you'd rather look at it that way), but time is a scalar. Time is the one thing that can be used directly in both the x and y equations; everything else (displacement, velocity, and acceleration) has to be split into components. This is important! Something that probably can't be emphasized enough is that even though an object may travel in a two-dimensional path (often following a parabola, in the standard case of an object moving under the influence of gravity alone), the motion can always be reduced to two independent one-dimensional motions. The x motion takes place as if the y motion isn't happening, and the y motion takes place independent of whatever is happening in the x direction. One good example of this is the case of two objects (e.g., baseballs) which are released at the same time. One is dropped so it falls straight down; the other is thrown horizontally. As long as they start at the same height, both objects will hit the ground at the same time, no matter how fast the second one is thrown. What we're ignoring We will generally neglect the effect of air resistance in most of the problems we do. In some cases that's just fine. In other cases it's not so fine.Afeather and a brick dropped at the same time from the same height will not reach the ground at the same time, for example. This has nothing to do with the weight of the feather compared to the brick. It's simply air resistance; if we took away all the air and dropped the feather and brick, they would hit the ground at exactly the same time. So, remember that we're often analyzing ideal cases, especially this early in the semester. In reality, things might be a little different because of factors we're neglecting at the moment. An example Probably the simplest way to see how to apply these constant acceleration equations is to work through a sample problem. Let's say you're on top of a cliff, which drops vertically 150 m to a flat valley below.You throw a ball off the cliff, launching it at 8.40 m/s at an angle of 20° above the horizontal. (a) How long does it take to reach the ground?(b) How far is it from the base of the cliff to the point of impact? It's a good idea to be as systematic as possible when it comes to analyzing the situation. Here's an example of how to organize the information you're given. First, draw a diagram. Then set up a table to keep track of everything you know. It's important to pick an origin, and to choose a coordinate system showing positive directions. In this case, the origin was chosen to be the base of the cliff, with +x being right and +y being up. You don't have to choose the origin or the positive directions this way. Pick others if you'd like, and stick with them (an origin at the top of the cliff, and/or positive y-direction down would be two possible changes). Now that everything's neatly organized, think about what can be used to calculate what. You know plenty of y-information, so we can use that to find the time it takes to reach the ground. One way to do this (definitely not the only way) is to do it in two steps, first calculating the final velocity using the equation: This gives v =y2.873 + 2 (-9.8) (-150) = 2948.3 m / s . Taking the square root gives: v y = +/- 54.30 m/s. Remember that the square root can be positive or negative. In this case it's negative, because the y-component of the velocity will be directed down when the ball hits the ground. Now we can use another equation to solve for time: So, -54.30 = 2.873 - 9.8 t, which gives t = 5.834 seconds. Rounding off, the ball was in the air for 5.83 s. We can use this time for part (b), to get the distance traveled in the x-direction during the course of its flight. The best equation to use is: So, from the base of the cliff to the point of impact is 46.0 m. Apoint about symmetry At some point in its flight, the ball in the example above returned to the level of the top of the cliff (you threw it from the top of the cliff, it went up, and on its way down it passed through a point the same height off the ground as the top of the cliff). What was the ball's velocity when it passed this height? Its speed will be the same as the initial speed, 8.40 m/s, and its angle will be the same as the launch angle, only measured below the horizontal. This is not just true of the initial height.At every height the ball passes through on the way up, there is a mirror-image point (at the same height, with the same speed, and the same angle, just down rather than up) on the downward part of the path it come out the same. Relative velocity Most people find relative velocity to be a relatively difficult concept. In one dimension, however, it's reasonably straight-forward. Let's say you're walking along a road, heading west at 8 km/hr.Atrain track runs parallel to the road and a train is passing by, traveling at 40 km/hr west. There is also a car driving by on the road, going 30 km/hr east. How fast is the train traveling relative to you? How fast is the car traveling relative to you? And how fast is the train traveling relative to the car? One way to look at it is this: in an hour, the train will be 40 km west of where you are now, but you will be 8 km west, so the train will be 32 km further west than you in an hour. Relative to you, then, the train has a velocity of 32 km/hr west. Similarly, relative to the train, you have a velocity of 32 km/hr east. Using a subscript y for you, t for the train, and g for the ground, we can say this: the velocity of you relative to the ground = = 8 km/hr west the velocity of the train relative to the ground = = 40 km/hr west Note that if you flip the order of the subscripts, to get the velocity of the ground relative to you, for example, you get an equal and opposite vector.You can write this equal and opposite vector by flipping the sign, or by reversing the direction, like this: the velocity of the ground relative to you = = -8 km/hr west = 8 km/hr east The velocity of the train relative to you, , can be found by adding vectors appropriately. Note the order of the subscripts in this equation: Rearranging this gives: Asimilar argument can be used to show that the velocity of the car relative to you is 38 km/hr east. the velocity of you relative to the ground = = 8 km/hr west = -8 km/hr east the velocity of the car relative to the ground = = 30 km/hr east The velocity of the train relative to the car is 70 km/hr west, and the velocity of the car relative to the train is 70 km/hr east. Relative velocity in 2 dimensions In two dimensions, the relative velocity equations look identical to the way they look in one dimension. The main difference is that it's harder to add and subtract the vectors, because you have to use components. Let's change the 1D example to 2D. The train still moves at 40 km/hr west, but the car turns on to a road going 40° south of east, and travels at 30 km/hr. What is the velocity of the car relative to the train now? The relative velocity equation for this situation looks like this: The corresponding vector diagram looks like this: Because this is a 2-D situation, we have to write this as two separate equations, one for the x-components (east-west) and one for the y-components (north-south): Now we have to figure out what the x and y components are for these vectors. The train doesn't have a y-component, because it is traveling west. So: The car has both an x and y component: Plugging these in to the x and y equations gives: Combining these two components into the vector gives a magnitude of: at an angle given by the inverse tangent of 19.3 / 63.0, which is 17 degrees. So, the velocity of the car relative to the train is 66 km/hr, 17 degrees south of east. Newton's laws of motion Force We've introduced the concept of projectile motion, and talked about throwing a ball off a cliff, analyzing the motion as it traveled through the air. But, how did the ball get its initial velocity in the first place? When it hit the ground, what made it eventually come to a stop? To give the ball the initial velocity, we threw it, so we applied a force to the ball. When it hit the ground, more forces came into play to bring the ball to a stop. Aforce is an interaction between objects that tends to produce acceleration of the objects. Acceleration occurs when there is a net force on an object; no acceleration occurs when the net force (the sum of all the forces) is zero. In other words, acceleration occurs when there is a net force, but no acceleration occurs when the forces are balanced. Remember that an acceleration produces a change in velocity (magnitude and/or direction), so an unbalanced force will change the velocity of an object. Isaac Newton (1642-1727) studied forces and noticed three things in particular about them. These are important enough that we call them Newton's laws of motion. We'll look at the three laws one at a time. Newton's first law The ancient Greeks, guided byAristotle (384-322 BC) in particular, thought that the natural state of motion of an object is at rest, seeing as anything they set into motion eventually came to a stop. Galileo (1564-1642) had a better understanding of the situation, however, and realized that the Greeks weren't accounting for
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