Physics and movement problems in the first year of high school, please ask the master, thank you!!

The easiest way to solve is to use the average velocity. If both B and C accelerate and decelerate with equal acceleration, then there is.

v A = v beginning.

v B = (vmax + v beginning) 2

v C = (vmin + v primary) 2

It can be seen that v B v A v C. From the meaning of the title, it can be seen that the displacement of motion is the same, so t and B arrive first. The acceleration is similar and the solution is completed.

Method 2: It is easier to find the same displacement by using the velocity-time image.

B's slowest speed = A's speed in the whole process = C's fastest speed So so, it will be clearer to draw a picture!

Consider the average speed.

A is always v average v

B is always greater than v and average greater than v

C is always less than v and average less than v

s v B is the smallest.

This is not clear here.

Let A, B, and C pass the first road sign is the speed of V, hey, I can't tell you clearly, you draw a map to analyze, or think about it intuitively... To have one’s hands tied.

It is known by the title. vo=vt

B accelerates the process v>voDeceleration v or "vo.".Passing = vt A is always equal to vo = vt

C deceleration vs A = s B = s C.

It can be seen from t=s v (v is the average velocity). B is the first to pass!

Because B's average velocity is greater than A's, C's average velocity is smaller than A's.

The whole question is based on the ground. When leaving the end of the platform to the treadmill, the instantaneous velocity is 4, and the acceleration a2 is also based on the ground (it can be imagined that the treadmill does not move, and the person moves to the right with acceleration a2=1). At this time, the acceleration A2 has reflected the factor that decreases due to the reverse motion of the treadmill), so the initial velocity of the treadmill is 4, and the person moves to the right with the acceleration relative to the ground A2.

People and the treadmill have friction, this friction is derived from the relative speed, but the friction force can not make the speed abruptly change, only the speed gradually, so when the person rushes to the treadmill, the instant speed is unchanged.

If you choose the ground reference system, your calculations are mainly based on the ground, so the speed of a person is 4 meters per second. If the treadmill is taken as a reference frame, the speed of a person is 3 meters per second. Here people and treadmills are independent. There is no correlation between their speeds, and the treadmill only affects the acceleration.

The speed of the person at the end of the platform is 4 (relative to the ground), and when he gets on the treadmill it is still 4 (relative to the ground), and the treadmill gives him a backward force. If he doesn't run after getting on the treadmill (assuming a wooden box), the speed of the wooden box will decrease from 4 to 0 due to the action of the treadmill, and then move in the opposite direction. That is, what the treadmill exerts on an object is a force, not a velocity.

Therefore, the speed of a person on the treadmill is still 4. However, this is a theoretical analysis, if it is a wooden box, it will be more accurate, and people run on two legs, and at a certain time with the treadmill, so you can abstract people to understand.

It should be 3 (because people run or walk with static friction--- the point is that ---if you want not to slip, you must ensure that the surface of the foot and the surface of the running foot are relatively stationary - "The initial velocity on the treadmill must become 3

The speed of the person mentioned above has nothing to do with the treadmill? (Unless the person is wearing "roller skates" - then the speed does not change).

Just like when --- block is placed on an absolutely smooth treadmill (no matter how fast the treadmill is), the block will not move, and the speed will remain at 0. At this point, if you give the block a speed such as "4", then let the treadmill speed up, slow down, or stop. The speed of the block remains unchanged at "4" (because it is always sliding relative to each other).

--But people need static friction to achieve running, and it is necessary to ensure that the surface of the foot and the surface of the running foot are relatively static (if there is a slide, the person will fall).

At the beginning, the platform was used as the reference system, and the acceleration time was 2s, v=a*t. When he gets on the treadmill it's definitely 4But after he rushed to the treadmill and the first hurdles, he shouldn't have been like that.

To use the treadmill conveyor belt as a reference, the acceleration remains the same, but the velocity becomes 3 relative to the conveyor belt. The main reason is that the reference unit needs to be changed according to the location. You can't always use the ground as a frame of reference.

It's been a long time since I studied middle school physics, so you can ask me any questions.

By the time point b is reached, the mechanical energy loss is 60-45=15J, which means that the frictional force has done work by 15J

So the work of friction is the answer of the potential energy 1 3

When it reaches the highest point, the work done by friction is equal to 100 4=25j, and when it comes down, it does 25j

So in total, the cavity early friction does 50j of energy.

So after coming down, the kinetic energy is 50j

Select Clear Tangerine A

According to the resultant external force of the curve motion points to the inside of the curve, and the two charge bands A and B have the same electrical properties and repel each other, so the A charge should be on the outside of the intersection of the tangent of point M and point N, as shown in the figure.

So a is correct, because a is a point charge, only when the distance from m and n to a is equal, the acceleration at these two points can be equal, if so, these two points should be on the same equipotential surface, in that case, the velocity of these two points must be equal, but the title says that the velocities of these two points are not equal, so the acceleration of m and n points cannot be equal. So b is wrong.

And because v v0, it means that ekm ekn, only by the electric field force, then the sum of kinetic energy and electric potential energy is constant, so the electric potential energy at point n The electric potential energy at point n. Because the kinetic energy decreases, the electric field force is negative, and the total work is negative, and cd is correct.

Correct Answer: ACD

According to the trajectory, the general direction of the force can be determined, and the trajectory must be between the angle between the initial velocity and the direction of the force, so a is correct.

Because v is conserved according to energy, the velocity at point n is small and the kinetic energy is small, so the electric potential energy is large, so c is correct.

Because the kinetic energy at point n is small, the total work done by the electric field force is negative during the whole process, d is correct.

Steps to write yourself.

1) There are also 3 sliders, i.e. 4 time interval t

Therefore, the movement time of the first slider is 4t

Find t=1 4t

2) The first slider is displaced

1/2at²=l

a=2l t direction along the inclined downward direction, i.e., in the opposite direction of velocity.

3) by the displacement formula.

l=(v²-0）/2a

The solution gives v = 2al under the root number

4) Displacement. Time 2t = 1 2t initial velocity v

So x=v (1 2t)-1 2a(1 2t) (self-calculated) velocity v end = v-a(1 2t) (self-calculated).

There is also a known amount missing here, is it that the little slider that happens to be taken away has a velocity of exactly 0? Without this condition, this topic cannot be done.