An explanation of pursuit and encounter in high school physics

You also. There are several scenarios for high school physics chase encounters:

1. Catch up with the problem:

The equal velocity (co-direction) of the two objects being chased is a critical condition for whether they can catch up and the distance between the two is extreme.

The first category: the speed is large, decelerates (such as uniform deceleration and linear motion), and the chasing speed is small (such as uniform deceleration and linear motion).

When the velocity of the two is equal, the displacement of the pursuer The displacement of the pursuer is still less than the displacement of the pursued, and it will never catch up, and there is a minimum distance between the two at this time.

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

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

In the specific solution, it can be solved by using the condition of equal velocity, or by using the knowledge of quadratic functions, and by using images.

The second category: the one with the small velocity accelerates (such as the uniform acceleration linear motion with zero initial velocity) and the one with the large chasing speed (uniform linear motion).

There is a maximum distance when the velocity of both is equal.

When the displacement of both is equal, it catches up.

The specific solution method is similar to the first type, that is, the analysis using equal velocity can also use quadratic function images and image images.

2. Encounter problems.

Two objects moving in the same direction chase and meet.

Objects moving in opposite directions meet when the sum of the magnitudes of their respective displacements is equal to the distance between the two objects at the beginning.

When the speed of the two vehicles is equal, the distance is the closest. (Before that, the speed of the front A was small, and the distance between the two cars was shrinking; After that, because the acceleration of the brakes of the rear car is definitely greater than that of the front car, the speed of the rear car B is small, and the distance between the two cars begins to increase).

Therefore, as long as there is no collision when the velocity is equal, it is impossible to collide again in the future.

Velocity relation: 8-2t = 16-at

Distance relationship: 8t-1 2*2*t 2+8=16t-1 2*a*t 2t=2s

a=6m/s^2

The algorithm upstairs should be right.

A stops when they meet, in this case B has the smallest acceleration) But you first imagine A and B as a particle, according to this idea, B will stop 24 meters in front of you, and then A will stop at the place where B stops. B overtakes A, and A catches up with B. Already collided.

For example, A is chasing B, according to the conditions you said, A's initial velocity is greater than B's, but A is slowing down, that is, the speed is getting slower and slower, and B's velocity remains the same.

Direction of movement – >

Stage 1: A ——— B (A speed 1>B speed).

Stage 2: - A ——— B (A speed 1> A speed 2> B speed because A is slowing down) 3 stages: - A B (A catches up with B, A speed 1> A speed 2> A speed 3 >B speed) 4 stages:

B - A (A exceeds B, and because A has been slowing down, so at this moment A's velocity 4 = B's velocity, after which B's velocity will exceed A's velocity 4, becoming B chasing A).

Stage 5: - B A (B catches up with A, at this time B speed "A speed 5).

Draw a velocity-time graph, and you can explain it well. In terms of physical methods, when a large deceleration chases an object moving at a uniform speed, the object that is decelerating when catching up is greater than the constant velocity, then this is the first time that two objects meet; Then the object that is decelerating runs ahead, but as the speed decreases (and eventually comes to rest), the object moving at a uniform speed will catch up with the decelerating object again, i.e., the second encounter.

When slowing down and chasing at a constant speed, if the speed of the chase is greater than that of the chase, then the chasing person will run to the front of the chased after the encounter. Because of the deceleration relationship, when the speed decreases to a slower than the car at a constant speed, it becomes a constant speed chasing and deceleration, so the gap between the two will be narrowed, and finally the decelerated car will be caught up by the car at a constant speed.