Here we are talking about diagrams. diagrams are great tools based on the quote one picture because 1000 words. They provide a wealth of information but can be used to describe physical meanings. In terms of math. Let me explain the two main processes a student must follow when he has to do with diagrams. Part one first, how to decode a diagram.
Number one, focus on the quantities shown on the axis. Maybe it seems simple and obvious, but it isn't. I will show you an example of kinematics later. Number two, try to understand the mathematical function or relationship between the not always easy video number three, record low formulas, etc. but may connect the above quantities. Number four.
Here's your knowledge. Which is relative to the connected quantities in order to produce the necessary results. I said something earlier about an example in kinematics, which is relative to the first goal I have set. If I give you the following diagram, which gives the variation of the quantity against time determines the kind of motion Think about it. The answer is depending on which quantity is represented on axis y. For example, if we have position x, then we have an object address, but it was have velocity V, then we have uniform motion.
And finally, if we have acceleration a and we have uniformly accelerated motion. The three other points of interest could be described through another diagram in kinematics again. In this example, we have a diagram of lacing against the linear form of the graph shows that the motions of constant acceleration. In other words, they are uniformly accelerate. The formulas connecting the velocity, time and displacement for this kind of motion are the following. Don't forget about the slope of such a graph represents acceleration and area gives us the displacement correct.
Now, we can apply all this info we have records and go on like this. In this diagram, we have x is y for velocity and axis x for time, we can easily realize that we have a linear function, which means that motions are all uniformly accelerated. The formulas we have to record for this type of motion are the following. Now taking all this into account, we can go on and write down the dates and we can call it For each motion, first of all, we must determine the types of motion. For example, from zero to two seconds, we have uniformly accelerated motion with initial velocity of four meters per second, and acceleration at one, delta V over delta T two meters per square second, displacement equal to the area, delta x one, which is 12 meters. Okay.
Now, from two seconds to six seconds, we have uniformly decelerate motion with initial velocity of eight meters per second. And this is a ratio of a to delta V over delta t, negative one meters per square circle. displacement delta x two equals 16 meters. Okay. Now, Part B joining us right now needs the following steps first Find the function which connects quantities on x and y axes. Second, recall your math knowledge in order to determine the type of curve or line, you must draw according to the fraction you have found of the first step.
Third, complete the diagram by putting the numerical values on it, especially the points on the axis. Would you like to give you an example. It's one object of mass 100 grams, is in contact with the one end of a horizontal spring of constant k equals 10 Newtons per meter, as shown below. we compress the spring by five centimeters and then release the object. sketch it that now showing the variation of the magnitude of velocity of the object as a function of time. The horizontal surface is frictionless to date direction to the right as the positive okay.
First of all, we have to realize, but this code object will undergo a symbol harmonic motion for the time interval, but it keeps being in touch with the spring. In other words, for one quarter of a period, after that, it will move at constant speed. Why? Because the surface is frictionless. That means that we will have a double factional velocity against time, which will be the following one, first V equals V max sine omega t, and then V equals V max equals constants. After this step, we have to calculate the magnitudes of all the quantities involved in the previous formula.
First, the period is given by the formula t equals two times pi times square root of mass over k, etc and it gives us by over five seconds maximum velocity V V V max equals omega time say etc. It gives us one meter per second. And putting all the above values on the previous for all t, v equals one times sine 10 times T and V equals one meter per second is it after all the above calculations, it is rather obvious that the correct diagram will be the follow up. Take a look at closing this session, I want to give you the following remark trying to kiss the vt diagram again if the object is attached on the screen, instead of being in contact with it, okay, contact me if you cannot answer it, and join me in the next lecture. Bye