Analyze two classical problems in physics in the first year of high school

Question 1. Because 2a1s1=v 2

2a2s2=v^2

So a1s1=a2s2

So a1:a2=s2:s1

Let s1 correspond to t1 and s2 correspond to t2 t1+t2=ta1t1=a2t2

So a1t1=a2(t-t1).

Replace a1:a2=s1:s2.

i.e. s2t1=s1(t-t1).

Solve t1=s1t s1+s2

and then by 1 2a1t1 2=s1

Solve the second problem of a1=2(s1+s2) 2 s1t 2. 1.The landing speed is 5m s

Then the umbrella fell 224-h from the ground

The initial velocity of the fall v1 2=2g(224-h), so 2*

Find h and then v1

When moving, first accelerate and then decelerate, using the image method to solve, that is, immediately open the parachute, and then find the time.

1.Let the acceleration of the AB segment be A1, the time taken for the AB segment is T1, the acceleration of the BC segment is A2, and the time taken for the BC segment is T2, yes.

t1+t2=t

a1*t1=-a2*t2

1/2*(t1)^2*a1=s1

1/2*(t2)^2*a2=s2

Links are immediately available.

2.You start from the speed of the athlete when opening the parachute, this is the contact point between the front and back, the specific idea is: the initial speed limit is obtained from the landing speed limit and acceleration behind, and this initial speed is the speed of the athlete when opening the parachute.

1. Solve with V-T image. Let the maximum speed be V, and the time from A to B is t. Then s1 = 1 2vt, s2 = 1 2(t-t)v, a1=v t, a2=v (t-t).

Solve these equations together, we can get t=s1t s1+s2, v=2(s1+s2) t, and then a1 and a2 can be solved

2. (1) Let the height from the ground be h, the speed of opening the parachute is V, the speed of landing is Vm, the acceleration of gravity G, and the acceleration of deceleration A is 2G(H-H)=V2 -2Ah=VM2-V2, and H can be solved to be equivalent to falling from H2, then 2GH=VM2

2) Time t=t1+t2 v=gt1, vm=v-at2

Because after the airplane throws the object horizontally, according to Newton's first law, when the object is not subjected to force in the horizontal direction, the object will always maintain rapid linear motion, so in the horizontal direction, the object will maintain the same speed as the airplane. From the perspective of the aircraft, that is, taking the aircraft as the reference frame, assuming that the aircraft is stationary, the objects seen are relatively stationary with the aircraft in the horizontal direction, but they are used in free fall.

This involves the second characteristic of flat throwing motion, that is, after the object is thrown flat, there is only one gravitational force in the vertical direction, so the object does a uniform acceleration motion with g as the acceleration in the vertical direction, that is, free fall. Although the airplane and the object maintain the same speed in the horizontal direction, in the vertical direction, it is indeed in free fall, so from the perspective of the aircraft, the object falls vertically. The vertical distance from the plane is getting farther and farther away.

AB is not right.

CD is based on the ground as a reference frame, from the ground, that is, assuming that the ground is stationary, so what people on the ground see is that the object has been moving forward at the same speed as the aircraft in the horizontal direction, and is in free fall in the vertical direction. Therefore it is correct.

Question 2: It examines the concepts of displacement and distance, as well as the characteristics of vectors and scalars.

Displacement is a directional, straight-line distance from the origin of an object to the end point of motion of an object. It is a vector, and the displacement must be the same in size and direction. Therefore, to examine the displacement of a point on a coin, the displacement can be expressed by drawing the head of the line from the starting point to the investigation point, because the trajectory of the coin is not repeated, so the displacement is not the same.

The distance is the trajectory of the route actually taken by the particle, it is a curve, a scalar, and it is not put down. Because the coin rolling path is not repeated, the distance is not the same.

1。When viewed from an airplane, since the object has the same velocity as the airplane in the horizontal direction, and the object has acceleration in the vertical direction, the object is seen below and farther and farther away.

From the ground, it is needless to say that the object is moving, as mentioned earlier, the object and the plane have the same velocity in the horizontal direction, and they are getting farther and farther away in the vertical direction, so they are directly below.

2。Displacement is a vector quantity, which is the line of the initial and final positions, while the distance is the actual trajectory of the object. Therefore, the displacement of each point of the coin is a straight line parallel to the table, and the distance is a curve that changes up and down.

The first question, after reading the above answers, I believe you have no doubt, that is, the horizontal velocity of the flat throwing object does not change but the vertical speed has a speed.

Regarding the second question, I personally feel that the answer is wrong. Let's talk about displacement first, from the point of view of the beginning and end state, each point has completely turned 10 times, so the connection line of the beginning and end position of each point is parallel to the desktop, and the displacement size is the distance of the center of the circle at the beginning and end position. So the displacement is the same.

Speaking of the distance, since each point on the circle has been completely turned 10 times, the position of each point on the circle has no effect on the distance. Because the circle is highly symmetrical, going from a point of symmetry below to a point of symmetry above is practically the same as going from a point of symmetry above to which point below. Moreover, each point goes through a full circle, so in fact the displacement and distance are the same.

Question 1: If you use the ground as a reference system. Of course, the object is in motion, because the plane drops an object, the plane already has velocity, so according to the law of inertia, the object being placed will also have the same initial velocity as the airplane.

Therefore, after the object is put down, it does a flat throwing motion (on the one hand, it moves in a straight line at a constant speed in the horizontal direction, and the speed is equal to the speed of the aircraft, and on the other hand, it moves in free fall). So the ground looks undoubtedly right to move, think of an airplane dropping a bomb. That's it.

In addition, although the object falls, it seems that the object is always under the plane until it falls to the ground, regardless of air resistance. To add to the mistakes of A and B: because the object has the same velocity horizontally as the airplane, the object will only see the object falling vertically when looking at the object from the airplane.

Question 2: I think the person above has explained this almost perfectly. Because the problem is not easy to explain.

But as long as the coin goes around the edge of the east and west, there will always be an extra circle. For example, if the circumference of a coin is 1, and the circumference of something to be wound is 3, most people will think that the coin goes around 3 times, but the actual coin goes around 4 times. This results in different distances and displacements.

The length of the carriage is L, and there are a total of N carriages. When driving out of the station, do a uniform acceleration linear motion with zero initial velocity s=1 2*at2 to get l=2a, nl=18a, so a total of 9 boxes.

1) Acceleration, a 0 6m s2 (2) The car slows down and brakes with an acceleration greater than 0 643m s2.

Analysis: (1) Accelerate and pass L+X less than or equal to VT+ AT2 2 before the train passes through the intersection

The time it takes for the train to pass through the intersection.

t=x1/vv1②

by a 0 6m s2

2) The car slows down and stops before the train arrives, and the time required for the train to pass through the intersection t = (l1 + x1) v1 = 25s

V2 2·t = 175m

t = 350 15 x = 23 35 25s then the car should be satisfied with the deceleration movement.

v2)2=2a′·x2

a = (v2)2 2x2 = 0 643m s2 accidents can also be avoided when the car brakes with an acceleration greater than 0 643m s2.

1. Let the length of the carriage be l, the acceleration is a, and when there are n carriages, the first carriage passes: l= t (t=2s), and obtains: l=2a;

When the whole vehicle passes: nl =, gets: nl = 18a, n = 9.

2.(1) According to traffic regulations and common sense: cars should slow down and wait for the train to pass before crossing the intersection. The train passage time is (200+300) 20=25s, 175 25=7m s, and the car can be slowed down to 7m s (without considering the deceleration process).

2) Those who are not afraid of death can speed up the passage. Train arrival time is 200 20 = 10s, 175 10 =;

The car passes through the intersection at high speed at a speed greater than (excluding the acceleration process).

1.From the uniform acceleration linear motion formula s=, it can be seen that the ratio of the first carriage to the whole carriage is 2*2 : 6*6, so a total of 9 carriages (EMU?) Remember to buy your own insurance ...... before you ride）

2.This is a very brave ...... questionEither hit the brakes or the accelerator ......If you apply the brakes, you can stop within 175m. Calculate the acceleration accordingly and do the math yourself.

Step on the accelerator and accelerate to the intersection (180 meters) before the head of the train reaches the intersection (20 seconds). Calculate the acceleration accordingly and do the math yourself.

In fact, you can also play the steering wheel ......

The length of the first carriage is 1 2a2 2 = 2a (a is the acceleration), and the length of all carriages is 1 2a6 2 = 18a

18a 2a = 9 carriages.

The time for the car to pass through the intersection takes (175+5) 15=12s, the time from the head of the train to the big intersection is 200 20=10s, and it takes (200+300) 20=25s for all to pass, so the car must slow down to avoid colliding with the train. If the time taken to decelerate is not taken into account, then he should immediately decelerate to a speed below 180 25 = 36 5m s.

The first question has an inference about the time of motion of an object that starts to accelerate uniformly at rest, and it seems that the first carriage passes through the space in front of the person from a standstill, and the second is ...... than the thirdEqual to 1: (root number 2-1) :(root number 3 - root number 2): ....

The second question should be counted as a time, and this time should be the maximum. It means how much time it takes for the driver to brake to avoid the accident. But don't you know how your train is going, in the opposite direction?

1.Let the length of the carriage be l, the acceleration is a, and there are n carriages in total: l=1 2*a*2 2, n*l=1 2*a*6 2 obtain:

n=92.You don't give the friction coefficient of the ground, and you can't calculate the maximum acceleration driver can be a uniform acceleration motion, a uniform acceleration motion, a uniform circular motion, or a uniform acceleration circular motion.

n1=mg+ma=m(g+a)=60*12=720n(2)When the elevator accelerates and descends with an acceleration and deceleration of 2m s, how many Nius is read on the scale?

It must be a slowdown, huh? n2＝mg+ma=m(g+a)=60*8=480n

3) When the elevator rises at a constant speed of 2m s, how many Nius is read on the scale?

Evenly accelerated ascent, right? n2＝mg+ma=m(g+a)=60*8=480n

Decompose f horizontally to the right and vertically upwards into f1, f2;

2) Then the gravitational force of the object on the ground is g'=mg-f2, ground resistance f=g'*Left;

3) the horizontal rightward tensile force f'=f1-f, acceleration a=f' m; The displacement of the first five seconds is s=1 2*a*5 2(

The acceleration a is downward, the acceleration is formed by the supporting force of the scale on the person (upward) and the gravity of the person (downward), the supporting force of the scale on the person n upward, and n "g, n=g-ma=m(g-a) direction upward;

2) When the uniform deceleration is decreasing, the acceleration a is upward, and n》g, n=g+ma=m(g+a) direction is upward;

3) At constant velocity, there is no acceleration, a=0, n=g=mg.

1. According to the decomposition of the force, the horizontal force of the object is 400cos37°, so the acceleration of the object is 244 100=

The displacement of the object in 5s is 1 2*

2、（1）w=ma=60*（10-2）=480n（2）w=ma=60*（10+2）=720n(3)w=ma=60*10=600n

The upward component of f is f1 = f*sin37 = 240 N, the right component of f is f2 = f*cos37 = 320 N, the net force of the object in the vertical direction is f3 = mg-f1 = 760 N, (g = 10), friction f Mo = 760*

So the net force to the right is 320-76=244 N.

f=maa=The displacement of the first five seconds is squared, and the answer is meters, and the direction is to the right.

Question 2. g = 10, done with equivalent gravitational acceleration.

Your question is weird. It seems to have been copied by mistake.

When accelerating downward, the gravitational acceleration is divided into 2m s +8m s, 2 is accelerated, and 8 is flattened by the scale, so the scale power is 60*8=480

If you accelerate upward, the gravitational acceleration is divided into -2m s + 12m s, as above, 12*60=720

If there is a constant velocity, there is no acceleration and there is no force, so there is only a gravity g of 10, which is directly 60*10=600