Physics Chase and Encounter Problem, Physics Chase and Encounter Problem Solving Skills

Car B first makes a uniform acceleration movement, and then moves at a uniform speed, and sets the time of his acceleration motion to ts, then his speed becomes 4+2t after ts, and this speed is also the speed at which he accelerates and moves at a uniform speed. Since they met, A caught up with B, and A had to go 84 meters more to catch up with B, that is, A12S traveled 84 meters more than B.

Then the distance of b is 4t+1 2 t 2+(4+2t)(12-t)=12 20-84, and this equation can be solved.

Let b accelerate to t, then the time to constant velocity is 12-t

When two objects meet after 12s, they move for the same time, and the relationship between displacements is.

12VA=VBT+1 2AT2+(VB+AT)(12-T)+84 to find the time.

t=18s (rounded) or t=6s

Solution: Let the acceleration time of car B be t, then the time of uniform motion is 12 t

After 12s, the two cars met, and B was 84 meters in front of A at the beginning, indicating that A had to travel 84 meters more than B.

The column formula is: 12x20 84 (4 2t)(12 t) 4t 1 2 x 2 x t 2

Solution: t 2-24-108=0

t-18)(t-6)=0

t=18 or t6 for a total time of 12s, so t18 is rounded.

1. In calculating the distance traveled by B, it is necessary to calculate the distance traveled by acceleration and the distance traveled at a constant speed by B.

Therefore, in the acceleration part, it is necessary to use the formula of uniform speed movement.

Second, because the title says that the acceleration suddenly becomes zero, at this time a 0, so the velocity is the original velocity is 4, before the uniform acceleration motion, indicating that there is a certain acceleration, when the acceleration is zero, the velocity is unchanged for the original velocity.

Hope it helps.

Physics Senior One Chase and Encounter Problem Solving Skills: Finding out that two objects are moving in the same straight line often involves problems such as chasing, meeting or avoiding collisions, and the key condition for answering such questions is: whether the two objects can reach a certain position in space at the same time.

The basic idea of problem solving

1. Study the two objects separately;

2. Draw a schematic diagram of the movement process;

3. List the displacement equations.

4. Find out the relationship between time and speed.

5. Solve the results and discuss them if necessary.

How to solve the problem

1. Critical condition method: when the velocities of the two are equal, the two are the farthest (closest) apart.

2. Image method: Liquid letter draws an X T image or a V T image, and then uses the image to analyze and solve.

3. Mathematical discriminant method: let the encounter time t, according to the conditional equation, obtain a quadratic equation about t, and discuss it with the discriminant formula, if δ> 0, that is, there are two solutions, indicating that they can meet twice; If δ 0, it means that you just caught up or met; If δ< 0, it means that you can't catch up or can't meet.

Pay attention to the problem of chasing and encountering

1. The analysis problem is one condition, two relationships.

One condition is the critical condition that is met when the velocity of two objects is equal, such as whether the distance between the two objects is the maximum or the minimum and whether it happens to catch up.

The two relationships are: the time relationship and the displacement relationship.

The time relationship refers to whether the motion time of two objects is equal, whether the two objects are moving at the same time or one after the other, etc.; The displacement relationship refers to the movement of two objects in the same place or one before and one behind, etc., among which finding the displacement relationship between two objects by drawing a motion schematic diagram is a breakthrough in solving the problem, so we must develop a good habit of drawing sketches to tell the problem of rotation analysis in learning, which is very beneficial to help us understand the meaning of the topic and enlighten our thinking.

xt image.

1) Physical significance: It reflects the law of the displacement of the object moving in a straight line with time

2) The significance of slope: the magnitude of the tangent slope of a point on the graph line represents the magnitude of the object's velocity, and the positive or negative slope represents the direction of the object's velocity

V t image.

1) Physical significance: It reflects the law of the change of the velocity of an object in a straight line with time

2) The significance of slope: the magnitude of the tangent slope of a point on the graph line represents the magnitude of the acceleration of the object at that point, and the positive or negative slope represents the direction of the acceleration of the object.

3) The meaning of "area".

The area enclosed by the plot line and the timeline represents the magnitude of the displacement in the corresponding friend time. If the area is above the time axis, the displacement direction is positive; If this area is below the timeline, the displacement direction is negative.

2. If the object being chased moves in a straight line with uniform deceleration, you must pay attention to whether the object has stopped moving before catching up. Review the question carefully, pay attention to the key words in the question, and fully dig out the implicit conditions in the question, such as "just right", "coincidence", "at most", "at least", etc. It often corresponds to a critical state and satisfies the corresponding critical conditions.

The problem of chasing and meeting is two types of problems often involved in the study of the motion of two objects on the same straight line in kinematics, and they are also the specific application of the law of linear motion with uniform speed in practical problems, the basic characteristics of both are the same, both are in the same position in the process of motion, and the processing methods are similar.

1. Features: 1) The main condition for catching up is that two objects are in the same position when catching up: first, the object with zero initial velocity is in uniform acceleration to catch up with the object moving at uniform speed in the same direction, and it must be able to catch up, and there is a maximum distance before catching up (the condition is that the velocity is equal v1=v2), and the two are in the same position when catching up; The second is that the object of uniform velocity motion in the same direction may or may not catch up, and there is a critical condition that just catches up or just can't catch up (the speed of the two is equal, when the two objects are in the same position, if v chase is greater than v to be chased, it can catch up, v chase is less than v to be chased, then it cannot catch up, if it is always unable to catch up, then the distance is the smallest when the speed of the two is equal); The third is that an object moving at a uniform deceleration speeds up with an object moving at a uniform speed in the same direction, and may or may not catch up, and the situation is similar to the second situation.

2) When they meet, they are in the same position, and if they do not meet, the speed of the two objects is exactly the same.

2. Treatment:

1) Grasp "one condition, two relations": one condition is the critical condition that the speed of the two meets, such as the maximum and minimum distance between two objects, which happens to be the first to catch up and just catch up, and the critical condition is that the speed is equal. The two relationships are the time relationship and the displacement relationship, which is the breakthrough point for solving the problem, and it is best to draw a good sketch analysis to find out the time and displacement relationship.

2) Carefully review the question and grasp the key words in the question, such as just, coincidence, most, at least, etc., because they often correspond to a critical state and meet the critical condition.

3) The solutions are roughly physical methods and mathematical methods, the physical methods are commonly used by the critical condition method and the image method, and the mathematical methods are commonly used by the discriminant method.

In fact, there is a lot of information about this on the Internet, which can be solved by looking it up online.

The car starts to swim quickly from a standstill with an acceleration of 1m s square, and at the same time, a person 60m behind the car chases the car at a certain speed and wants to get on the car. People have been less than 20m away from the car, and the duration is 2s to shout stop, so that the parking message can be transmitted to the driver. Q: How big is v0 (human speed) at least?

If V0 is 19m s to catch up with a car, what is the minimum distance between people and vehicles?

1.This problem does not take the ground as a reference frame, but a car as a reference frame, the relative initial velocity of a person is v0, and the relative acceleration is -1m s 2

When the speed of the person is zero, the closest to the car, the movement in the previous second and the next second is decelerated from 1m s to 0, and then accelerated from 0 to 1m s, respectively, so the distance is equal.

If the person is closest to the car at the time of t, then as long as he is 20 meters away from the car at t'=t-1s.

The distance traveled by a person during the time from t' to t is (1 2)a(t-t') 2=, so movement is required in the period from 0 to t.

And because (v0) 2=2as, v0=9m s

When the relative displacement of people and vehicles is (vt")-1/2)at"2=361m>60m, so he can catch up with the car in 19 seconds, and the minimum distance is 0

List a quadratic function about the distance between two objects.

Chase problem: The same velocity (co-direction) of the two objects being chased and pursued is the critical condition for whether they can catch up and the distance between the two has an extreme value, and the chase problem is usually divided into two categories:

1.The one with the big speed slows down and chases the one with the small speed:

1) When the velocity of the two is equal, if the displacement of the pursuer is still less than the displacement of the pursued, it will never catch up, and there is a minimum distance between the two.

2) If the displacement of the two is equal and the velocity of the two is equal, it can catch up, which is also the critical condition for the two to avoid collision.

3) If the displacement of the two is equal, the speed of the pursuer is still greater than the speed of the pursued, then the respondent has another chance to catch up with the pursuer, and the distance between the two has a large value when the velocity is equal.

The drawing method uses a quadratic equation.

To summarize it for you, the chase problem in high school physics is mainly to accelerate and chase at a constant speed, or to chase and decelerate at a constant speed, and there is a same condition that when the two velocities are equal, the distance reaches the maximum. It is also possible to ask about the problem of not being able to catch up, which can be solved with displacement and time. The idea is mainly whether the distance between the two is equal in a straight line (or in the same position), and the speed comparison (who is larger, chasing more than the chased, or less than), and the problem can be solved with a simple analysis.

In high school physics, it is often seen that the speed is equal.

Buy a workbook and you're done.

After setting the elapsed time t, the two cars go to the same position. The car in front of it moves at a constant speed s1=v2 t, and the thing behind moves at a uniform deceleration, which can be seen as the opposite uniform acceleration motion, v1=a t, s2=1 2 a t squared. In addition, s2 = s1 + x.

If all equations are set up, we get v2 (v1 a)+x=v1 squared 2a, and a=v1 x (v1 2-v2). Then a should satisfy a v1 x (v1 2-v2).

I have written about the process and ideas, this kind of physics problem has Chinese, English and mathematical symbols, it is very troublesome to write, you add 5 points too little.

It's easier to calculate with relative velocity. Assuming that the carriage is stationary, that is, using it as a reference, then the speed of B is 20m s, and the distance is s, then the two meet when the speed of B decreases to zero. Time t, at=20, t=, s=100m