High School Physics Particle Motion Trajectory Problem

As you can see from your question, you also have a problem with understanding the relationship between motion and force, which force is the cause of acceleration, and in which direction the force is applied (resultant force) acceleration, and the direction perpendicular to this direction will be stationary or uniform motion, which can answer your first two questions. The third question shows that you don't understand the meaning of the image. The displacement time graph represents the relationship between displacement and time.

It is recommended that you work on the basics... After learning the basic knowledge, knowing the application of the basic knowledge, all kinds of problems will be solved naturally... Otherwise, it taught you this question or not another one...

Since velocity is measured in uniform motion, distance is proportional to time, that is, time can be determined by distance. For example, option A, the particle moves at a uniform speed on the x-axis, which means that the time can be determined by the coordinates of the x-axis, which is actually the time axis can be regarded as the time axis (this is the key to this problem). Then the graph becomes a displacement-time image, and then the speed of his movement on the y-axis and its change are determined by observing the change in slope.

In the above way, if the y-axis is moving at a uniform speed, the y-axis is considered the time axis.

You don't need to calculate this problem, just think about it. Here's how it works:

The key is to determine the position of the midpoint and the middle moment. 1.If the average velocity of the first half of the time must be smaller than the average velocity of the second half, then the displacement of the first half of the time must be less than that of the second half, that is, the position of the middle moment must be before the midpoint, and because it is a uniform acceleration, the motion speed of the middle moment is less than the speed of the midpoint.

2.If you do a uniform deceleration motion, the analysis idea is the same, that is, the movement speed at the middle moment is less than the movement speed at the midpoint.

Think about it, this topic should be AD

The air resistance is not counted, so the time of the upward throwing motion rising and falling is the same, but only one of them can be discussed, so the rise took 2 seconds, the velocity of point c can be calculated, the highest point is set to be distance from point c h, the column formula can be obtained h=, and t = 2 is substituted to get h = 20, because the air resistance is not counted, so it can be reduced to v=20 according to the conservation of energy mgh= substitution

The height difference between A and C is h-5=25, and the lost mechanical energy can be w, w=mg(h-5) and finally w=3000j (according to the law of conservation of energy).

Answer: The velocity at point C is 20m s and the mechanical energy lost in the orbit is 3000j.

After object A leaves the aircraft, it is subjected to air resistance, and the horizontal direction is a uniform deceleration motion, and object B is still moving at a uniform speed.

When B leaves the plane, the horizontal velocity of B is greater than that of A, and the horizontal acceleration of AB is the same.

Taking A as a reference, B moves in a straight line at a constant speed in the horizontal direction.

So: the man on the ground looks at the state of motion of two objects.

That is, the horizontal distance between the two objects is getting farther and farther away.

However, you must not forget that object A leaves the plane first, and it is subject to air resistance earlier than object B, so its velocity is smaller than that of object B. Positional relationship: object B is in front, object A is behind, of course, the distance is getting farther and farther away.

You can also see it from your diagram. At each moment, the velocity of object B is always greater than the velocity of object A, and the distance between them is increasing (the area of an oblique parallelogram).

There's one more second in between! At this point, A slows down, while B keeps moving.

In my opinion, if you are less than the critical velocity and there is no rope constraint, you are not doing circular motion. It will make a parabolic motion with an initial velocity of v, the trajectory is parabolic, and the magnitude of v depends on the trajectory of motion, but it is all parabolic.

Because the charged particles move in the magnetic field, they move in a uniform circular motion under the action of the magnetic field force, and the radius of motion is:

r=mv/(qb)

The figure shows that all particles have the same four quantities of MVQB, so their radius of circular motion in the magnetic field is the same, and their natural motion trajectory is the same.

With the right-angled vertices as the origin, the two right-angled edges are X, and the Y axis establishes a plane Cartesian coordinate system, then the three vertices of the triangle ABC are A(A,0)B(0,B)C(0,0) Let A move along the AC direction C moves in the CB direction, let the velocity be V, then the coordinates of point A are (A-VT,0) The coordinates of point C are (0,VT), then the trajectory of the point D coordinates in AC is ((A-VT) 2,VT 2)D, and the trajectory of D is Y=A 2-X

Straight lines, just consider a special case of isosceles right triangles.

Since the rope is not extensible, the velocity along the rope direction is 0

Therefore, it will do a uniform circular motion with a linear velocity of 4m s.

The trajectory is a circle (regardless of the presence of other external forces).

2 needs to be emphasized at any time, and 4 needs to be the same direction.

The correct answer is: B

Conservation of momentum in the horizontal direction, 0=mv+m(v+u) "1", let l be the displacement of m to position, m to position shift = m to position shift + m to position shift = l + h*cot , substitute into the inference of "1": 0=ml+m(l+h*cot)-l=-m*hcot (m+m), the symbol indicates that the l direction is reversed from the displacement of m to m (to the right), that is, to the left. To find the distance, you can remove the negative sign of the above result.