Senior 1 Physics Questions A B Two cars are driving in the same direction on a straight road

In 2, the first two terms are the distance traveled by b to accelerate the linear motion uniformly, the formula s=vt+1 2at 2, after elapsed time t, the velocity of b is vb+at, the time used is the result of substituting formula 3, and formula 5 is the result of simplification of formula 4.

1 2at 2+vt is the displacement of car b when it accelerates uniformly, which is derived from the formula s=vot+1 2at 2, v+at is the final velocity of car b accelerated to by time t, 12-t) is the time of uniform motion after car b accelerates t t seconds (because the total time is 12 seconds), so (12-t) (v+at) is the displacement of car b moving at a constant speed, so s=1 2at 2+vt+(12-t) (v+at) is the displacement of car b's accelerated motion plus the displacement of uniform motion, That is, the total displacement.

is the total distance traveled by b, vb*t+1 2*a*t is the distance traveled by b when the acceleration is a, (vb+at) is the speed at the moment when the acceleration is canceled, after the acceleration is canceled, b does a uniform linear motion, (vb+at)*(t0-t) is the distance traveled after the acceleration is canceled.

There is no special meaning, just substituting and substituting to , and then simplifying it.

It is obtained by substituting each known number into .

Because car A has been moving at a constant speed of 20 ms, we chose car A as the reference system. Then the initial velocity of car B is -16m s, and the acceleration is 2m s. After the acceleration is 0, car B travels at a constant speed. If the acceleration time is t, the constant velocity time is 12-t

The solution is t=or you can choose the ground as the reference frame, the solution is similar, but it is more troublesome to calculate. The difference in displacement between the two is 84 meters.

20*12-84=4t+

You can understand it this way, it is the instantaneous speed of car B, that is, the instantaneous speed that car B starts to start when car A reaches 84 meters from car B.

The diagram is shown below: A---84 m---B (start acceleration).

a (constant velocity) - b (gradually accelerating).

1）156m (2) 6s

Question Analysis: Let the speed of car A be <>

The acceleration time of car B is t, and the two cars are in <>

Moment to moment. Then there are:

1) According to the title, the displacement of car B in 12 s <> so <>

<> <> of them

<> is the distance traveled before the A and B cars meet. According to the title, there are:

<>

Lianli solution: <>

Solution: <>

<> are: <>

where t is in s).

Solution: <>

<>It's not on topic, drop it).

Therefore, the acceleration time of car B is 6s

Note: As long as the other solutions to the subjective question are in line with the meaning of the question, points can be given as appropriate.

Comments: When analyzing and chasing the problem, we must pay attention to grasping the two relationships, that is, the time relationship and the displacement relationship The time relationship refers to whether the motion time of the 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 the movement of one before and one after the other, etc., among which finding the displacement relationship between the two objects by drawing the motion schematic diagram is the breakthrough point for solving the problem, so we must develop a good habit of drawing sketches to analyze the problem in learning.

Time of 12s, displacement of a xa=va·t=8 displacement of 12=96b xb=vot+ at

Catch-up displacement relation: xb-xa=96m

vot+½at² -96=96 【vo=4 t=12 】a=2

Ditto, just to get the job done.

Let the acceleration of car B when accelerating is a, and the acceleration time is t, which is determined by the title.

vb+at=vb′①

vat0＝vbt+

at2+vb′(t0?t)+x0 ②

Lianli and substitute the data to obtain t=6s, a=2m s2 Answer: (1) The acceleration time of car B is 6s

2) The acceleration of car B is 2m s2

Let the acceleration time t1 and the constant speed time t2

84+( 4t1 + 4 + a*t1)*t2 =20*12...1

t1+t2=12...2

Simultaneous 1,2 can be solved: t1=6s, t2=6s