Physics problem Three ballerinas equilateral triangles

I'll give you a diagram and you'll understand.

Let's start by drawing a diagram at a moment when the velocity of A is drawn vertically upward, then B is directly above A and the velocity of B is obliquely downward. At this time, the upward velocity of a is v, and the downward component of the velocity of b is vcos60, which is the velocity close to each other. Gathering at the center is an inevitable consequence of symmetry.

Assuming that the meeting point is closer to A, and the motion of BC and A is the same, then why should it be farther away from BC? So be sure to meet in the center.

The first one is explained by advanced mathematics, calculus, I don't remember much, let's solve it by a master.

Three people stand at the vertex of an equilateral triangle and the equation of motion is the same, indicating that the three people are completely symmetrical, so the concentration point must be at the center of the triangle, that is, the center of gravity. On the contrary, if you don't concentrate on the center of gravity, then who will you favor and why should you favor him?

First of all, from the three positions, their movements are exactly the same, and according to the symmetry of the motions, it can be seen that the instantaneous positions of the three form equilateral triangles, and the velocity direction is also connected along the line between them. The final point of encounter is definitely the center of gravity of the equilateral triangle.

Then according to the concept of relative motion, it can be seen that the velocity of a with respect to b is va-vb. But there are two motions here, radial motion, and the magnitude is the projection of (va-vb) on the ab line (i.e., va):

VN = VA-VBCOS120° (120° angle between VA and VB) Tangential rotation, the magnitude is the projection of (VA-VB) in the perpendicular direction to VA: VT = VBSIN60°.

Their radial starting distance is d, so the time is d vn.

This question is a bit like a physics contest question.

You can refer to it.

Both questions are the same.

Take a closer look at the original question, is that the meaning of the question?

Drawing a diagram will solve the problem.

Draw an equilateral triangle and mark the velocity (the direction along the edge of the segment), it can be seen that the velocity of the two people along the direction of the edge is v(1+cos(60)), because in the process of motion, the triangle is always equilateral, that is, the speed of the side shortening is always v(1+cos(60)), so the time is t=l (v(1+cos(60))).

This question mainly tests the decomposition of speed and the selection of reference system. Taking the triangle formed by three people as a reference, it can be seen as three people running towards the middle, and the speed is the minute velocity of the person facing the center of the triangle relative to the ground speed. i.e. v·cos30°

a 2 cos30° = v·cos30°·t so that the time comes out.

Then S = v · t and you're done.

I've seen this question in the high school physics competition tutorial book, and I don't know who adapted whom. However, cultivating physical thinking is the central purpose of physics problems, and this student deserves to be commended for being so active in seeking answers through the Internet. I hope to work hard.

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=

The last three people must be gathered at one point at the same time, so as long as you know when A and B meet, so it is enough to ask for the relative velocity, and the relative velocity of their opposite motion is equal to v(1+cos60°)=3 2*v, so the time is divided by this velocity for the side length a, and the distance is equal to this time multiplied by the velocity, so the answer is 2 3*a

I've done a similar question: the position of the three snails forms a regular triangle with a side length of l=60cm, the first one climbs towards the second, 2 climbs towards 3, and 3 climbs towards one. v=50cm/min。Ask how long to meet.

It's the same as you, except that when you turn around, it's the other way around. Fold the path of the three people in your question symmetrically once every even number (that is, after the first turn to before the third turn), that is, to fold the figure and come back, which is the same as the path of the snail in my question. In polar coordinates, the equation for the snail's path is r=(l root number 3)*e (negative root number 3*fai) power.

In short, this problem uses the idea of transformation (to transform the path into this) and the idea of calculus (to detai x).

v‖=vcos30°

After t=oa v =2l 3v=2a 3v.

The l in my question is the side length of the triangle, which is l=a

o is the center of a regular triangle.

My topic, High School Physics Contest Problem Solving Methods Mechanics Part, Part 1 Motion of Objects, Lecture 2 Important Models and Special Topics Point 5, Reflections on the Trajectory of Particles Meeting Dynamic Polygon Vertices. (pp. 13-15).

According to the right-hand theorem, the magnetic induction intensity B1 at C is perpendicular to the AC edge downward, and the magnetic induction intensity B2 is perpendicular to the BC edge at Sock digging C.

Because the nuclei B1 and B2 are equal in size, the induced intensity is vertically downward by the isosceles triangle midline theorem.

It's good to go to the elite to consult about learning, and the teacher will give you an analysis.

The high line coincides with the bisector of the angle, indicating that the three lines are one, the triangle is an isosceles triangle, and the vertical foot is k, then oa-ak=absolute(1 3) 2=1km and ak=

then oa=h =oa -ak gets the grip h=2km