Urgent high school physics to find the state of motion of objects

Answer: The state of motion of the objectThe motion of the object from point A to point B is an accelerated motion with increased acceleration.

x1= x2= x3= x4=

x2-x1= x3-x2= x4-x3=

The difference in displacement in adjacent equal time intervals is positive, and the object accelerates, and the displacement difference increases, and the object accelerates.

Is it adjacent to each other for equal time.

Assuming that the object moves at a uniform velocity in each time period, the initial velocity at point A is v0=0, the corresponding displacement at t= is s1=, and the time taken is t1. t=, the corresponding displacement is s2=, and the time taken is t2; t=, the corresponding displacement is s3=, and the time taken is t3; According to the formula s=v, the original t+, the default is taken in SI units, then there is:

s1=v0t1+，v0=0，a1=2*， v1=；

s2=v1t2+，a2=(s2-v1*t2)*2/t2^2=2×(，v2=v1+a2t2=;

s3=v2t3+，a3=(s3-v2*t3)*2/t3^2=2×(，v3=v2+a3t3=;

Because the last period does not know the time, assuming t=, then the corresponding displacement is s4=, and the time taken is t4, there is:

s4=v3t4+，a4=(s4-v3*t4)*2/t4^2=2×(，v4=v3+a4t4=

In summary: a1=; a2=10；a3=5；a4=15, so it doesn't seem to be a simple increase or a decrease in acceleration.

I'm sorry, I misunderstood, please ignore this! (If you delete it, you will also deduct points o o.)"…）

1 I think that the speed should change with time, and the embodiment of the displacement should be due to the influence of speed and time.

2 For example, the speed is 10 in 1s, the speed is 15 in 2s, and the speed is 20 in 3s (uniform units). Although it satisfies the uniform speed movement, it is still not the ratio of speed to time, to be precise, the speed is proportional to the time difference.

3 I think it's more complicated, so in order not to delay your time, let's pass.

The 510s touch is: we imagine that there are many hours in a second, and the last point in this period is the instantaneous speed, which is simply understood as the last speed.

But we calculated that the speed was already 0 at the time of the spring, and at the end of 10s, it was naturally 0, I thought. Thank you.

is over time.

The initial velocity is proportional to 0.

3.Let the length of the inclined plane be a, then from the second half, according to the displacement formula, S 2 V 2 2A

by the whole process of displacement formula.

s＝v。Band 2 2a

From these two formulas, it is known that it is dry:

v。=√2v

The average velocity of the uniform acceleration motion is half of the sum of the initial and final velocities, so the average velocity is (v. +v）/2

i.e. (2+1)v 2

The car has been parked) x = 1 2AT2 = 25mm).

Because there is a limit to the maximum speed in the question, the maximum speed of the car is 360 km/hIf you accelerate first and then decelerate, this limit will be exceeded, which can be solved with a V-T image.

If there is no part with a constant speed, it can only run 50x68=3400m according to the assumption, and the title says 4800m, which means that there must be a constant speed.

The part of the car at a constant speed reaches 360 km h, which is the maximum speed of the car, and of course it can no longer be accelerated.

I don't know, ask your brother and me.

A spring with a uniform mass when hung on the ceiling has an elongation of mg (2k) instead of mg (k).

That is, the answer to the first question is (f-mg 2).

1. The tensile force at the horizontal level is f, and the length after deformation is l; The elongation is effected by the action of f.

When upright, the elongation is determined by the weight of the object, the weight of the spring and the tension of the ceiling + the weight of the spring = the tension of the ceiling.

Together, they elongate the spring and have the same effect, i.e. 2(m+m)g=f

So m=f 2g -m

2. Some liquids are overweight; The liquid of the size of the iron block is overweight, the iron block accelerates the decline, and the same volume of liquid accelerates the rise, this part of the liquid is overweight.

Before the free fall, the iron is suspended and submerged in the liquid, and the number is W2+W1P P0

When the iron falls for its own use, the apparent weight is equal to the gravitational-buoyant force to get w1(1-p p0)g

From the weightlessness of the apparent weight g'=mg-ma, a=p p0g

The amount of liquid accelerating downward is as large as the amount of liquid accelerating upwards The volume of liquid with the same volume of overweight as iron is w1p p0

The state weight g'=w1p p0(g+a)=w1p p0(g+pg p0).

In the end, the corresponding indication for this part is g' g=w1p p0+w1(p p0) 2

Combine the two parts to get the final result.

w2+w1p/p0+w1(p/p0)^2

1.The force required to make the spring length l is f, and now it is mg, then the force of f-mg is also required.

The mass is (f-mg) m=f m-g

2.The system of iron, beakers, liquids is only supported by gravity w=w1+w2 and the bench scale n. Equilibrium, n=w1+w2

The support force of the system and the pressure of the system on the scale are the relationship between the action force and the reaction force, so the pressure n'=n=w1+w2 is the indicator.

I will teach you a simple method to close together, and it is also the most subtle method used in the linear motion of uniform variable speed.

Theoretical basis: The average velocity of a linear moving object with uniform velocity for a certain period of time is equal to the instantaneous velocity at the intermediate moment of that time.

This problem assumes that the displacement of the first stage at the beginning is 0 moment, then the velocity at the middle moment, i.e., 4s, is v4=24+64 8=11m s

Then according to v=v0+at, you can find the speed at any time.

Places are limited. I'll give you a hint. You draw it out. To three equations.

Set the time from departure to 24m, muzzle velocity and displacement.

The ridge uses 24m and 64m displacement plus the displacement of the first section to train the two equations. There is also the relationship between initial velocity and time acceleration. It can be solved after the linkage.

No. The dot-up timer has cycles. t＝

s＝vt＋1/2at^2

Han Zai can chain his hands and count the old suspicion of the ...... out of VIt's time for college! Remember correctly

Because tell you what the time is"The shortest"

This means that both the start and the brake have reached their maximum acceleration.

If there is no intermediate uniform process.

Only the distance traveled by acceleration and braking, so the total displacement is insufficient, I don't know if you are satisfied with this?

Solution: Let the acceleration be a(m s 2, the maximum speed is v(m s), the time required to accelerate from rest to the maximum speed is t1(s), the time required to decelerate from the maximum speed to rest is also t1(s), and the time required to move at the maximum speed is t2(s), then there is.

s=a*t1 2 2+v*t2+v*t1-a*t1 2 2=v(t1+t2), i.e., 4800=360, t1+t2=48s, and t1+t2+t1=68s, t1=20s, t2=28s.

a=v/t1=100/20=5m/s^2。

The explanation upstairs is completely correct, but I don't understand why the maximum acceleration that a person can withstand is only 5 meters and seconds squared. You know, the acceleration of a free fall is also 10 meters and seconds squared, but aren't people who play extreme bungee jumping unscathed?

A change in the motion of an object means a change in velocity.

When an object moves at a uniform speed, the velocity does not change, so the force does not change.

But if it is subjected to a resultant external force, the direction of the resultant force remains the same, and the acceleration it provides is constant, so the velocity will always increase or decrease. At this time, the velocity changes, the state of motion changes, but the resultant external force is constant.

The most obvious is the uniform circular motion, the velocity is always changing, but the resultant external force is always directed towards the center of the circle, so it can also be explained.

1) The velocity of B when it moves to point n is exactly 0, and according to the kinematic formula, vb -0 = 2 gl, vb = 10m s

Let the right direction be positive, the momentum is conserved, 0=MVB+MVA, VA=-10M s, and the negative sign indicates the direction to the left, and the magnitude is 10M S

2) B returns from point n and the velocity changes from 0 to 2m s and the displacement is v 2 g = <10m

So b will move in a straight line at a uniform speed, and the velocity when leaving the conveyor belt is v=2m s

There is no energy loss when A collides with P, so the velocity of A does not change after the collision, and the direction changes.

Let the velocity after AB collision be v common, which is conserved by momentum, MVA-MV=2mv co-friction grip=4m s, and the direction is to the right.

3) After AB slides on the conveyor belt, it first decelerates to 0 and then accelerates in reverse, and the process between AB and the conveyor belt depends on friction, so the friction work is the additional electrical energy consumed by the conveyor belt.

The deceleration displacement s=v is 4 g=<10m, so AB does not slide out of the N-end.

w1=1 2*2mv total =16j

Reverse acceleration displacement s'=v Blind Brother Qing 4 g=w2=1 2*2mv =4j

Total electrical energy e=w1+w2=20j