Is it right to ask a physical formula? What do you say is the difference between these two physical

Yes, this is the distance formula for which the initial velocity is zero, and there is a non-zero initial velocity that adds vt.

This formula is only true when the initial velocity is zero in a uniform acceleration motion, you can push it by yourself by v1 2-v0 2=2as, when v0=0, s=v1 2 2a, and v1=at, so s=a*t 2 2

Correct, generally applied in free-fall movements.

It is a special deformation that is a special deformation with a formula for calculating the displacement of a linear motion with a constant acceleration.

The original formula is s=vt + 1 2*a*t 2

Because the initial velocity is zero, we get s=1 2*a*t 2, because the acceleration is g gives s=1 2*a*t 2, note that the s a g in the above formula all contain symbols (plus and minus signs, indicating direction).

That's right. But first, there is an object with zero muzzle velocity to accelerate uniformly, and the linear motion has s=1 2*a*t 2

Only then can there be s=1 2*g*t 2.

Yes, it is suitable for acceleration movements with an initial velocity of 0.

Generally, the initial velocity is not 0, that is, s=1, 2*a*t, 2+v0t

Exactly!! The application of the two formulas depends on the situation given by the question!!

That's right, the initial velocity is zero, and when it is not zero, use s=v0*t+(a*t 2) 2

These two formulas, w=fs is the formula for calculating work, which can be used regardless of any force, while w=gh is only applicable to gravitational work and is a case in doing work.

The physical meaning of these two formulas is the same, both calculate the work done, but there are differences. w=fs, where f represents the force exerted on the object, and s represents the distance that the object moves in the direction of the force under the action of this force; w = gh, g is the gravitational force at a certain location on the earth, is an invariant constant, h is the height at which the object falls or is lifted.

The high school physics teacher will give you the answer: w=fs is the definition of work. Its general expression is w=fscosa and w=gh is the idiom when calculating the work done by gravity. They are essentially the same, and they are both expressions of work in physics.

The difference between the two formulas is the scope of application.

w=fs is a universally applicable formula for calculating the work done by an object, where s refers to the distance that the object moves in the direction of the force.

w=gh is a special case of calculating the work done and is used to calculate the work done by gravity on the falling of an object, where h must be the distance the object falls.

The former is the general form, the latter is the special form of the former, w=fs, f is the force, s is the displacement, w = gh, g is the gravity is a type of force, h is the height, so w = gh is the work done by gravity.

The former is the work done by pulling or pushing, which is generally used to represent the total work, but the latter can only represent the work done by gravity, which is useful work.

There is no essential difference between these two formulas, both calculate the work done, where w=gh is used for free-fall motion.

The former is a formula for the general force to do work, and the latter is a special case and is a formula for calculating the work done by gravity.

In general, w=fs represents the work done by a horizontal or oblique force, and w=gh represents the work done by gravity in the vertical direction.

Those two five-five companies must not be able to go, and they are very big to go to others, each with its own position, and each with its own way to cut down.

The first floor is incomplete. Note that the question is an odd segment, not an even segment, and the odd segment data needs to go to the length of a paragraph, but not simply to the first or last paragraph. As for why, it is necessary to know the principle that the difference-by-difference method can reduce the error.

First of all, this method uses the difference-by-difference method.

However, secondary schools generally do not explain why the difference-by-difference method is used, and the specific reason involves error analysis, which is more than the average middle school student's mathematics level.

An illustrative approach is usually used to show that the difference-by-difference method is feasible.

For example, when there are multiple sets of data, the data should be processed with an averaging mindset. One is an image (straight line fitting), the other is addition and division n (arithmetic mean), and the difference-by-difference method is the latter.

And the usual addition and averaging, for example, is a1=(s2-s1) t 2;a2=(s3-s2)/t^2；a3=(s4-s3)/t^2；a=(a1+a2+a3)/3；We will find that a can be reduced to a=(s4-s1) (3t 2); That is to say, only the data at the two ends of S1 and S4 are used in the specific calculation. Experimental data were not used sufficiently. Therefore, the method of difference is adopted.

That's what the 1st floor says, a1=(s3-s1) 2t 2,a2=(s4-s2) 2t 2;a=(a1+a2)/2=(s3+s4-s2-s1)/(4t²)；But why should it be?

For example, you can define a1'=(s2-s1)/t^2；a2'=(s4-s3)/t^2；a'=(a1'+a2')/2=(s2+s4-s1-s3)/(2t^2)；Why is this bad, isn't it also used for each piece of data?

In fact, according to the error theory, it can be proved that the relative error of the latter is greater than that of the former (the difference-by-difference method).

a=(si-sii)/(4t^2)，si=s3+s4；sii=s1+s2

a'=s'i-s'ii/(2t^2)，s'i=s2+s4；s'ii=s1+s3

In the error analysis, it can be proved that in the above two formulas, the larger the difference between the molecules, the smaller the relative error, and in the difference-by-difference method defined in the textbook, the molecular difference is the largest (the displacement of the back minus the displacement of the front), so the relative error is the smallest. This is the difference method.

Therefore, in this question, because it is an odd number of terms, according to the analysis of the smallest error, it is reasonable to discard the middle paragraph, (but for the middle school stage with low requirements, you can also discard the first paragraph or the last paragraph of data, and generally discard the shortest paragraph), and then deal with it according to the above method, but pay attention to discard the middle.

The size of the actual interval between two sets of corresponding data, ie.

a1=s4-s1/(3t^2)

a2=s5-s2/(3t^2)

a=(a1+a2)/2=(s4+s5-s1-s2)/(6t^2)；(Ensure that the numerator is as large as possible, that is, the previous minus should be larger, and the subtracted number should be smaller).

That's right, in fact, this formula tells you a good way to understand it. Put s1

s2 as a group.

s3 and s4 are seen together, then these two distances are also adjacent, and the time is also equal, just know that the time at this time is 2t. Then the acceleration is equal to (s4+s3)-(s2+s1) (2t) 2, which is the formula you had better come up with. Again, if there are six distances.

Even the eight paragraphs can be compared by analogy.

a(2) According to the displacement formula s=v0*t0+(at0*t0) (1) Because the object moves uniformly and decelebly under the action of frictional force.

Finally, the final velocity is zero.

According to the acceleration formula a=(vt v0)t

t=(vt－v0)/a

where the terminal velocity vt 0

So t0 v0 2

Substituting t0 v0 a into the above displacement formula yields s=v0*t0+v0*v0

1/。The following is a general analysis of this major topic, and then roll on a semicircular smooth track, the radius of the circular track is r, and the height of the inclined plane is h:

The question generally gives the condition, which should be an expression for the conservation of mechanical energy; 2*mv 2 is the kinetic energy of the final state. The question generally asks about the velocity of the highest point, or the h-height of the highest point, etc. 】

The column formula is mgh=1 2*mv 2+mg2r, which is calculated according to the problem. You don't give a specific title and image, mg2r is the gravitational potential energy of the terminal state, and the ball slides down from a smooth slope.

Take the ground as the gravitational potential energy and the gravitational potential energy in the initial state.

According to P (power) = F (traction) * V (speed), it can be seen that when the power is constant, the traction force is inversely proportional to the speed, so the speed of the car is slow, the power of the car remains unchanged when decelerating, the traction force increases, and it is easy to go uphill.

As you said, your reasoning is based on the premise that the car has a certain power.

The power of the car engine is less than the external resistance (including ground friction and air resistance, etc.), which will slow the car down.

According to your formula, one physical quantity must be fixed before we can study the change relationship of the other two physical quantities. If all three physical quantities are changing, then it is difficult for you to draw a relationship between the two physical quantities. So we're going to use the control variable method to study things.

The premise is that the power must be certain! This means that when the car slows down, the pulling force must be increased to ensure a certain power; Similarly, when the speed increases, the speed has to be slowed down.

The landlord's inference that when the car slows down, the power decreases?

Don't you want that premise of yours?

In fact, the limitation of this formula is relatively large, only applicable to some physics problems, if you use it in middle school, it is estimated that there is no problem, and you can analyze the situation yourself.

When going uphill, the car slows down, but the gravitational potential energy increases, and the power of the car is mainly used for the gravitational potential energy. So the power has not decreased.

The power is determined by the oil you burn, the car slows down but the force becomes larger, the power will not necessarily decrease, I hope it will help you.

1. Yes, it refers to angular acceleration, which is also written in some textbooks. is the angular velocity derivative of time.

2. Because the angle satisfies = t, and s=vt is symmetrical; The angular velocity satisfies = t and v=at symmetry. So all the relevant equations are in the same form, and they can be compared one by one, and the symbol names can be replaced.

If we use the knowledge of advanced mathematics, the real reason is that they are both solutions of the same differential equation d 2(s) dt 2=a (i.e., acceleration is constant, because acceleration is the second derivative of s versus time t), and this solution is only mathematically derived, regardless of what symbol is used (whether it is angle or distance s).

3. It is not possible to change to tangential acceleration, because the circular motion is not a uniform acceleration motion, but a variable acceleration motion with a magnitude of r 2 and a constant rotation and change direction, and the above formula is only applicable to uniform acceleration motion, so it cannot be used. However, in the above problem, although the linear acceleration is changing, the angular acceleration is constant, so this formula can be used.

4. Maz = 0 means that there is no acceleration in the z direction. This is because the circular or curvilinear motion discussed here is confined to only one plane, so the velocity and acceleration in the z-direction are zero. Therefore, the force in the z direction is also zero.

At the same time, for curvilinear motion, ar in the r direction must be w 2 r, which has nothing to do with uniform circular motion or not, but is only a characteristic of the moment. In other words, with a certain instantaneous velocity v, the magnitude of r at this moment depends on the radial resultant force f, and the greater the f, the smaller r (because in the denominator...).v 2 r), which means that the turn is sharper and the arc is greater, that is, the greater the force perpendicular to the direction of motion, the faster the turn...In the special case of uniform circular motion, where r is fixed, then the radial resultant force f must also be kept constant, i.e. w2 r.

Whereas, AT is the tangential acceleration which is equal to the force along the direction of motion divided by the mass m. at=0 in uniform circular motionIf the angular acceleration in a uniform circular motion is constant, then the tangential at= r.

En: I hope I can help if I have any questions.

Cute little A

Does a in the red box refer to angular acceleration?

Why does the formula for displacement also apply here?

If you change angular velocity to linear velocity and a to tangent acceleration, is that still true?

Finally, there is what that maz=0 means.

The velocity of motion of dimension z is 0

A in the red box refers to angular acceleration.

Because it is analogous to the general displacement formula, this is the angular displacement (change of angle) is not valid, because angular acceleration is used to describe the speed of the change of angle, so it cannot be changed maz=0 means that in the Cartesian coordinate system, the Z direction is not affected by the force and is expected to be adopted.

The second solution is wrong in the application of the kinetic energy theorem, the initial kinetic energy of the whole process is 1 2mv 2, (the initial kinetic energy is not zero, the kinetic energy theorem is equal to the work done by the resultant force by subtracting the initial kinetic energy from the final kinetic energy) The final kinetic energy is 1 2mv2 2

It should be noted that your two solutions are the velocity obtained by the charged particle in the vertical direction, and the final kinetic energy I said is 1 2mv2 v2 in 2 is the final velocity, which is the combined velocity of the horizontal velocity v and the vertical velocity, I don't know if you can understand it!

If you want to use energy, you can't just consider one direction, that is, if you use the kinetic energy theorem of "w", you should consider the initial and final kinetic energy (including v), then the whole equation will become very complicated (because you need to subtract because it includes the square), but it should be possible.