A question in the kinematics part of the physics competition

The area in the V-T diagram represents the distance, you should know that.

Your question can't just be seen from the diagram.

According to you, even if you count the area of the rectangle formed by ABMJ.

It should also be the car leading the truck.

The area of the ABE is the amount of distance that the car travels longer than the truck, as long as this part is less than S1, which is the key. Since you want to use the area to represent the distance, it is more intuitive to represent s0 in the graph.

As for the representation in your diagram, you can just understand it.

Think of it this way: from T0 to T1, there's no car, no truck.

I walked s0 by myself, and then the car came out. As for why the truck is ahead of the car, the title doesn't say, you can think about it.

In fact, the focus of this question is not here, but the paragraph analyzed in the book is the most critical.

No need to look at the picture. I'll tell you the easiest way to do it.

1 2at 2 = s1 ( 2 for squared).

at=3vo

Calculate the acceleration of the trolley through these 2.

When the two cars meet, the elapsed time t reaches the same speed... This way they don't collide.

so+v0t>3v0t-1/2at^2

Only t is unknown. Anything can be solved.

Only the method was said. Didn't count.

It's too complicated. It's better to have a diagram of what you're doing.

The distance traveled by the car at this point = the distance traveled by the truck + the distance.

vqi*1/2at2<=vkat+s

Just bring it in.

There are many ways to do it, and I'll tell you the easiest one.

At each moment, 3 people are on an equilateral triangle that rotates and shrinks, and finally shrinks to a point in the center.

So we know that the 3 points all have a velocity pointing towards the center and it doesn't change because the total velocity doesn't change.

v=，s=lcos30°*2/3。So, t=s v.

The distance is s=

1. The supersonic aircraft flies at a constant speed V along a straight line ob. An observer looks at the airplane from point A, boa= and is considered unchanged during the observation time. The aircraft successively emits two pulse segments of sound waves, one small and one large, with an interval of .

So under what conditions can the observer record the strong pulses first and the small ones early? OA=L is known and the speed of sound is V.

2. All particles slide down from point o at the same time along the smooth without muzzle velocity at different inclination angles, 1. If the points of movement of each particle on the inclined plane are connected into a line, the nature of the connection is .

2.If you connect the points with the same rate of movement of each particle on the inclined plane into a line, choose that answer again.

a arc b parabola c horizontal d slash i know the answer but don't know how to prove it please elaborate on how to prove it.

F-FSIN = m*ay

fcosθ=m*ax

So: v= (0, )aydt= (0, )f m)(1-sin )dt

0=VO- (0, )AXDT=VO- (0, )F M)DT (where VO=1M S).

Further analysis is expected to yield: v=

It's not the original question, it's a wrong question.

1. It is not said that the two children are of equal mass, "He leaned his back against a large tree and pushed each other horizontally at a speed of v = 1m s. "You can't use the momentum theorem.

2. "There is enough ice on the ground that the child will slide down the mountain at a constant speed as long as the slightest force is applied, as shown in the picture." "The sliding force will be greater than the frictional force, and the velocity will increase.

Third, the picture is gone.

Momentum is conserved, and the two children with equal mass score velocity.

It is slightly different from the calculation results upstairs, please identify:

Kinetic energy at the first time rolling from the left inclined plane to point O: MGH1(1-COT).

Height from point o to the highest point of the right slope: h1(1-cot) (1+'cotθ')

Kinetic energy when returning from the right inclined plane to point O: mgh1(1- *cot)(1-'cotθ')/(1+μ'cotθ'(Note that this formula is different from the upstairs).

Return again to the height of the left bevel: h1(1-cot)(1-'cotθ')/(1+μcotθ)(1+μ'cotθ'(Note that this formula is different from the upstairs).

The total distance of a round trip: H1 SIN +2H1(1- Cot ) Sin'(1+μ'cotθ')+h1(1-μcotθ)(1-μ'cotθ')/sinθ(1+μcotθ)(1+μ'cotθ'(Note that this formula is different from the upstairs).

At a glance, you can see that it is a problem of work and displacement, and half an hour is pressure, hehe, I didn't think about it.

h(1/sinθ+1/sinθ')/(μ/tgθ+μ'/tgθ')

I gave up, half an hour was too stressful, and I cried.

Someone actually said that this question is not difficult, I have been thinking about it for almost an hour, and I have participated in many physics competitions, and I cried.

It would be nice if today's young people really had that high IQ, I just hope they don't be too arrogant.

In the first round, the kinetic energy to point O is mgh1(1-U tg).

The height to the right vertex is h1(1-u tg) (1+u).'/tgθ')

The kinetic energy returned to point O is mgh1(1-U tg) (1-U'/tgθ')/(1+u/tgθ)

Return to the left at a height of h1(1-u tg) (1-u.)'/tgθ')/(1+u'/tgθ')(1+u/tgθ)

From this, a cycle is formed.

Calculate the distance of this cycle.

h1/sinθ+2h1(1-u/tgθ)/sinθ'(1+u'/tgθ')+h1(1-u/tgθ)/(1-u/tgθ')/sinθ(1+u/tgθ)(1+u'/tgθ')

2h1/sinθ(1+uu'/tgθtgθ')/(1+u/tgθ)(1+u'/tgθ')+2h1(1-u/tgθ)/sinθ'(1+u'/tgθ')

This distance is then divided by the lost kinetic energy and multiplied by the total kinetic energy.

The loss of kinetic energy is 2mgh1uu'/tgθtgθ'/(1+u/tgθ)(1+u'/tgθ')

Divide (2+uu'/tgθtgθ'-(u/tgθ)^2)/(mguu'/tgθtgθ')

Then multiply by the kinetic energy as h1(2+uu'/tgθtgθ'-(u/tgθ)^2)/(uu'/tgθtgθ')

Solution: The force analysis of the box is shown in the figure above, in which ma is the inertia force generated by the acceleration of the car, 1) the friction force f can not provide the inertia force required for the accelerated movement of the box ma when the box slides.

f=mgμma>mgμ

a>gμ

2) If the box is turned over, it can be sure that it will be turned around the back and bottom of the box. The reason for turning over should be that the resultant moment around the lower edge of the back is not zero.

mah-mgl>0

a>lg/h

3) According to the correlation with l h, you can judge whether the box slides or overturns first.

If when AWhen g When A>lg H, the box slides while tipping over if >l h, then:

It can be judged that the box was tipped over first.

When A is LG H when A > g, the box overturns and the box slides at the same time.

When B is jetted, A makes the gas in it eject to the left at high speed, and the gas produces left momentum, which is known by the conservation of momentum, and the object A needs to produce rightward momentum, that is, the right velocity.

When inhaling, B inhales the gas, and after the gas enters, it stops moving by acting with B, and from the outside, the gas is stationary at first, and then it is also stationary, and the momentum does not change, so B does not produce motion.

A will experience a force to the right, so it moves to the right.

B will experience a force to the left, but it is small, it may not move, or it may be left.

As shown in the figure, let the red part have a virtual displacement δ and by the principle of virtual work, t( )rδ -mg δ r sin =0, then t( )=mg sin

The maximum tension is mg