Can the formula be used for the average velocity of variable acceleration motion? (V0+VT)/2

Can the formula be used for the average velocity of variable acceleration motion? (V0+VT)/2

The question is not accurate
In general, we can't, because in general, the so-called average speed refers to the average speed in time
It is not ruled out that the average velocity of a certain physical quantity is (V0 + VT) / 2 under some special conditions
In a word, it can't be done in most cases

On the formula of average velocity of linear motion with uniform speed change
Average velocity = half VO + vt
This formula is proved to be correct by the teacher through tedious formula transformation, but can it be proved by listening to the logical understanding? The following is my understanding, please help correct: within △ T, and the speed increases evenly, then the average speed should be equal to the middle speed
Is that ok?

You're right. Intermediate speed is intermediate time
This law is only applicable to the motion of straight line with uniform speed change

The problem of uniform acceleration motion and acceleration in Physics
From o point, the car moves in a straight line with uniform acceleration on a straight road. After 9s, it passes through P point and arrives at Q point. It is known that the distance between P and Q is 60m, and the speed of the car passing through Q point is 15m / s
(1) What is the acceleration of the car?
(2) What's the speed of the car passing p?
Please write the steps in white. Thank you for your easy understanding
This question won't help. If it's good, I'll just add a number and write out the reason and why

When 9s arrives at Q point, the velocity of Q point is 15m / s, the initial velocity is zero, and the acceleration a = V / t a = 5 / 3 can be obtained directly
Is it easy to solve VP with vq-vp = 2aX?

Is the acceleration formula applicable to the motion of non-uniform variable speed?
Is the acceleration calculated by Acceleration Formula instantaneous acceleration?

The acceleration formula is not applicable in the non-uniform variable speed motion. The instantaneous acceleration is obtained by the acceleration formula. Because the uniform variable speed motion is a motion with constant acceleration, the instantaneous acceleration obtained by the acceleration formula is its acceleration, which is constant. If the non-uniform variable speed motion, its acceleration always changes, Then we can't use the original acceleration formula to solve it
If you don't understand, you can ask me. If you are satisfied with the answer,

The application of the average velocity formula of uniform variable speed linear motion
The car starts from the ground a and moves in a straight line to the ground C, with B at the midpoint of the two places. The speed of the car from the ground a to the ground B is 60km / h, and then from the ground B to the ground C, with 120km / h. The average speed of the car from the ground a to the ground C is calculated
If u means = (U beginning + U ending) / 2, can it be found? How does the reference book answer 45 come from?

If a process is a uniformly accelerating linear motion, u means = (U beginning + U ending) / 2, this relation can be used, that is, a can be used without changing
Because your problem is two uniform acceleration process, not necessarily the acceleration of the former and the latter is equal, so it can not be used directly in the whole process
The average speed of the first half is 60 / 2 = 30km / h, and the average speed of the second half is (60 + 120) / 2 = 90km / h
If the half displacement is s, the first half time is s / 30 and the second half time is s / 90
Vplane = 2S / (s / 30 + S / 90) = 45km / h

Derivation of the formula of force composition and decomposition
Please deduce the formula F = √ (F1 ^ 2 + F2 ^ 2 + F1 * F2 * cos φ)
(for example, pictures can be described in words)

If you draw an auxiliary line, you can see that there is a parallelogram. At this time, the opposite side of F1 is regarded as F1, which is actually the same. This opposite side, F2 and f form a triangle. By using the cosine theorem, you can get the formula: f ^ 2 = F1 ^ 2 + F2 ^ 2 + F1 * F2 * cos φ
The above formula can be obtained by square root

The thought method and concrete rule of force composition

The concept of decomposition of force (1) component force: the effect produced by the joint action of several forces is the same as that produced by the original force. These forces are called the component force of the original force. (2) the decomposition of force is to find a component force of a known force

What is the formula for calculating the composition of forces?

The combination of forces is actually vector addition
Because force is a vector in space
Parallelogram rule or triangle rule of vector addition
In the coordinate system can be calculated with coordinates, so that is the decomposition of force calculation, and the calculation of vector in mathematics is exactly the same

Explain the formula for the composition of forces
F=(F1＾2+F2＾2+2F1＊F2cosα)1/2
α is the angle between F1 and F2
F is the resultant force of F 1 and F 2

F=(F1＾2+F2＾2+2F1＊F2cosα)1/2
The cosine formula is used
You first draw the diagram of the output

How to calculate the impact force produced by the collision of two objects? Is there a formula?

There seems to be something wrong with the above formula. It should be I = ft. I is the impulse, f is the force (in Newtons), and t is the time (in seconds)
When an object collides, impulse and momentum are equal. The formula of momentum is p = MV, where m is mass (in kilogram) and V is velocity (in meter / second)
Because I = P, there is: ft = MV, that is, f = MV / T (in fact, f = MA)