What are the 5 included in the physical kinematics of the third year of high school

Answer: It can be roughly divided into two categories: (1) uniform variable speed motion: force f is constant (1) variable acceleration motion: force f is changing.

1) Uniform variable speed motion can be divided into: (1) Uniform linear motion: FV is in a straight line. (2) Uniform speed curve motion: FV is not in a straight line, for example: flat throwing motion.

2) Variable acceleration motion can be divided into: (1) non-uniform variable speed linear motion (the direction of velocity is changing all the time, and the magnitude of velocity is also converted between maximum and minimum), such as simple harmonic vibration (single pendulum, fluctuation) (2) circular motion: the direction of velocity is changing all the time.

Motion is divided into linear motion, curvilinear motion, and vibration, linear motion is divided into uniform linear motion and non-uniform linear motion, non-uniform linear motion is divided into uniform variable speed linear motion and non-uniform variable speed linear motion, uniform variable speed linear motion is divided into uniform acceleration linear motion and uniform deceleration into linear motion.

Curvilinear motion is divided into circular motion and flat throwing motion, and circular motion is further divided into uniform circular motion and non-uniform circular motion.

Vibration is divided into simple harmonic vibration and general vibration.

Motion is divided into linear motion, curvilinear motion, and vibration.

This question is a bit difficult to understand, and it is analyzed as follows.

First of all, I set the distance from downstairs to upstairs is s;

The length of each staircase is m;

The speed of the escalator is v.

A: t1=s (; The distance traveled by A relative to the escalator is t1*(;

Similarly: B: t2=s (; The distance traveled by A relative to the escalator is t2*(;

Then sort out the equations;

A:(s (;

B:(s (;

The two equations are divided.

I get it after eliminating m,s finishing.

v 2+, and finally v=-9 v= so v=;

When asked the second time, add v=

Bring in (s (;

Got s m=63 people.

In terms of the critical state, that is, it is assumed that it happens to collide.

A: vt 2-v0 2=2as

S A 18m T A 6s

According to the assumption, then S C = S A + 10 = 28m

vt^2-v0^2=2as

It is also possible to solve a C to be equal to or add a negative sign in front of it.

Learning physics is very process-oriented, a process of cognition, understanding, and application.

1.Cognition: Use the things or phenomena around you or even some examples narrated by the teacher to help yourself fully understand it and become interested in it.

2.Comprehension: Use understanding to memorize formulas, theorems, experiments, etc. It can be understood and memorized in clever ways such as visual thinking. For example, what is a vacuum can be understood in this way: a vacuum is really empty, there is nothing.

3.Application: One is to cope with the exam, and the other is to explain some physical phenomena around you.

So, when studying, first of all, don't be afraid, because your previous experience of not learning well may suggest something to you, which may lead to a vicious cycle for you. Try to tell yourself, "I can do it!! In fact, psychological suggestion is very useful!

However, in order to increase your confidence, it is best to do a good job of pre-study and be aware of it.

Secondly, you should follow the teacher's train of thought in class, take some notes appropriately, and write down some knowledge points that are not clearly stated in the book or even omitted and are easy to make mistakes. Take time out of class to practice more, don't prevaricate for any reason, thus giving up the best time to practice, which can only lead to tragedy in the end.

Last but not least, you must make sure to make a timely summary. For example, the paper of the last exam has been sent down, and although it has been carefully corrected, you still have to think about why it was wrong. How is the correct answer calculated?

Will it be wrong if I get it again next time? Wait a minute.

I think that through these study methods, I will definitely be able to learn physics well.

Formulas can't be memorized, must not be memorized, must be understood and memorized, the reason why I can learn physics well is because I am very familiar with the use of formulas.

Kinematics is actually the relationship between four quantities: s v t a

Look at what quantity is known and what quantity is not known, and then choose the appropriate formula.

The maximum value of v should be found first, that is, when the speed of the flatbed and the cargo box are the same, the cargo box just moves to the rear end of the flatbed.

Set the warp time t, the flatbed car and the cargo box speed is the same.

Then: ug·t (cargo speed) = v - at (flatbed speed) ....1）

vt-1/2a·t·t）- 1/2ug·t·t = l ……2）

Flatbed Truck Movement Distance Cargo Box Movement Distance.

From the above two equations, the maximum value of v can be obtained by substituting the numerical value.

Or (2) is directly changed to (vt-1 2a·t·t)-1 2ug·t·t < l

Less than or equal to.

Set: the speed of the vehicle is V at the moment before braking; When the cargo box and the car reach the same velocity v'The displacement of the car relative to the braking is S2, and the displacement of the cargo box is S1.

s2 - s1 <= l

s1 = v*v/2ug

s2 = (v'*v'- v*v) 2a consists of the above three formulas, Synlips:

l>= v*v/8 - 3v'*v'/8

i.e. l+ 3v'*v'8 > = v*v 8 to ensure that the above is absolutely true, only if.

l >= v*v 8 is to meet the requirements.

The length of the bottom plate of the trolley is l=

Let the acceleration of the trolley be a, then when the rope is broken, the speed of the car and the slider v=2a, the relative sliding distance of the slider in the first three seconds is s1=2a*3=6a, and the relative sliding distance of the trolley in the first three seconds is s11=2a*3+, then the relative distance between the slider and the trolley in the first three seconds is s =s11-s1=solution acceleration a=1

Therefore, when the v=2 slider starts to move and leaves the rear of the car, the acceleration time of the slider is 2s, and the uniform motion time is 6s, so the distance of the slider relative to the ground is s=

Because the title says that the same amount of time has elapsed and the distance is the same, the average speed is the same. Let the velocity at f1 be removed as v, then (0 + v) 2 = v + 8) 2Thus, the velocity of f1 can be removed at 4m s.

You're right.

After removing f1, first do a uniform deceleration, and then a uniform acceleration.