The energy is constant, and the velocity is related to the distance displacement .

s!=e f, according to e-fs=1 2mv 2, according to what is said upstairs, the velocity v=0

In this question, I think that if the landlord is talking about the total energy, then you can consider the sum of the kinetic energy of the first floor and the internal energy generated by friction. From an energy point of view, the total energy (i.e., the electrical energy that the electric vehicle is fully charged) has several final destinations: loss, kinetic energy of the car, and overcoming resistance.

According to the conservation of energy, e-fs can be obtained'(s'is the total distance that can be walked, not the displacement) is equal to the sum of 1 2mv 2, v is the instantaneous speed, which can also be said to be the average speed on the distance that the electric car can travel.

1 2mv 2 though is an instantaneous formula.

But when moving at a uniform speed, the kinetic energy remains the same, then the energy consumed is fs (distance) When accelerating, the energy consumed is converted into kinetic energy and fs friction When decelerating, there are two types, one is braking, and the other is not providing power Most of the braking kinetic energy becomes internal energy that cannot be calculated If kinetic energy is not provided, kinetic energy is converted into fs

So. When accelerating.

E consumption = 1 2mv 2 (v is the speed at which the acceleration finally becomes a constant velocity) + fs1 at a constant velocity. e consumes 2=fs2

When slowing down. E consumption 3 + 1 2 MV 2 = FS3

3 equations + e=fs provided that deceleration is to reduce power and not brake.

brakes. e=fs +1 2mv 2 (v is the constant velocity) has been fully modified, and then see what is wrong.

w=fs, how much work is done is only related to force and displacement, not speed, and power p=w t=fv, you get faster, the power will become larger, such as riding a bicycle, you will become tired, but t is also reduced, and you consume the same energy. The landlord's question didn't want to involve too complicated knowledge.

s=e f, this is a question of energy, and it is okay to think about it in terms of energy.

Summary. Hello, generally when finding the heat of an object, use relative displacement. Then find the velocity or kinetic energy of the object, and use the counterpoint shift when it changes.

When working on the topic of the kinetic energy theorem, when to use relative displacement and when to use counterpoint displacement? I'm a little confused.

Hello, generally when finding the heat of an object, use relative displacement. Then find the velocity or kinetic energy of the object, and use the counterpoint shift when it changes.

But doesn't the change of kinetic energy depend on friction, and the work done by friction is not relative displacement.

Friction work generates heat.

Therefore, when there is heat, relative displacement is generally used.

Whereas, the displacement of the object can change the kinetic energy of the object.

Got it, thank you.

You're welcome, if you're happy with my answer, please give me a thumbs up, thanks.

When the velocity of the object is proportional to the distance, the expression of the displacement of the object over time can be obtained.

Suppose the velocity of the object is v, the time is t, and the displacement is x. The stool is scattered.

From the known conditions, it can be obtained: the velocity v is proportional to the displacement x, i.e., v = kx, where k is the proportional acacia constant.

According to the definition of velocity, velocity v is equal to the derivative of displacement x versus time t, i.e., v = dx dt

Equalize the above two equations to get dx dt = kx By separating the variables and integrating, we can get dx x = kdt Integrating both sides of the above equation at the same time, we get (1 x) dx = k dtln|x|=kt + c (c is the integral constant) takes exponents on both sides of the equation to get |x|=e (kt+c) Since the displacement x is a scalar, the expression after taking the absolute value is strict, and the absolute value sign can be ignored

x = ae^(kt)

where a is a constant and can be determined by the initial conditions.

Thus, the above expression describes the change in the displacement of an object with time when its velocity is proportional to the distance.

I don't know if it can help you.

a, the relationship between displacement and distance, for example, a zigzag snake, the displacement is directly measured with a ruler, the distance from the snake's head to the snake's tail, and the distance is the distance from the snake's head to the snake's tail after the snake is straightened. Assuming that the object moves in a straight line according to a right-angled triangle, it is obvious that the displacement is the length of the hypotenuse and the distance is the sum of the two right-angled sides.

b.In one case, it is a circular motion with a uniform velocity, and the magnitude of his velocity is constant.

c.It is also an example, if you move in a circular motion with a uniform velocity, you have an acceleration, it is a centripetal acceleration, and your velocity is constant.

d, is correct, the velocity is the magnitude of the velocity, the velocity has one more direction than the velocity, it is a vector, and the velocity is a scalar.

The displacement is the straight-line distance from the initial position to the last position, and the direction is from the initial position to the last position.

Velocity is divided into instantaneous and average velocity, as well as acceleration.

Instantaneous velocity is the velocity of an object at a certain moment, which can be obtained by dividing the sum of the displacements at the front and rear ends at that moment by the time it takes, and the direction is the same as the direction of motion of the object at that time.

The average velocity is the ratio of displacement to time (sometimes distance, depending on the situation, but in high school, displacement is generally studied) in the same direction as the displacement that occurred during this time.

Acceleration is a physical quantity that describes how fast or slow the velocity of an object changes, the velocity of an object changes quickly, the acceleration is large, and vice versa is small. If the acceleration is greater than 0, it is the same as the direction of motion of the object, and less than 0, it is opposite to the direction of motion.

The distance is scalar, there is no direction.

Average velocity refers to the average motion of an object over a period of time and is the ratio of displacement to time. In single-line uniform motion, the average velocity is numerically equal to the instantaneous velocity, and in the case of variable speed linear motion, the magnitude of the average velocity is related to the selected time or displacement, and the average velocity is generally different in different time periods or displacements.

A is wrong, the distance is the length of the motion path, and the displacement is the physical quantity that describes the change of the position of the quality return handicap of the front beam.

b right, but not exactly. The displacement of an object is a directed straight line segment from the beginning to the end point.

a. Vector is a physical quantity that has both magnitude and direction, and displacement, velocity, and acceleration are all vector quantities, so a is correct

b. In linear motion, the displacement of the object is not necessarily equal to the distance, if the object is reciprocating in linear motion, the displacement is less than the distance; Only when the object is moving in a unidirectional straight line, the magnitude of the displacement is equal to the distance so b is wrong

c. The instantaneous velocity can accurately describe the variable speed movement, and the average velocity can only roughly describe the variable speed movement, so c is wrong

d. The velocity remains unchanged in uniform linear motion, so d is correct, so it is chosen: AD

A and B need t to meet, and all three have been walking for so long t

t=12/（5+3）=3/2

The total time that the puppy runs is also t

The distance of the puppy is 6*3 2=9km

At the time of the encounter, the dog and A and B both met in one place.

The displacement of the dog is the distance traveled by A = 5*3 2=15 2=

It's an interesting topic. Hehe.

First of all, the first step: find out the time when A and B meet. It should be 12 (5+3)=hours. That is to say, after this hour, two people A and B just met. At this time, A walked, B walked, the second step: determine the distance the dog walked, during this hour the dog always ran at 6km h, so the distance the dog wants to walk is. When A and B meet, the dog stops running back and forth.

Step 3: Determine the displacement of the dog's walking, since the dog starts from A, the final displacement of the dog's walking is the displacement of the A's walking, that is.

That's it. Think, if the dog starts from B, then the distance he walks is still 9km, and the displacement is...

a. Displacement is a physical quantity that describes the change of the position of the particle, and distance is a physical quantity that describes the trajectory of the object

b. The distance of the particle can also be a straight line, for example, when the object moves in a unidirectional straight line, so B is wrong;

c. Only in one-way linear motion, the magnitude of the displacement is equal to the distance so c is wrong d, and d is correct from the analysis of c

Therefore, choose D