How is the formula of the relationship between the displacement and the velocity of the uniform velocity linear motion derived? Hope to have the image and formula to push down together

How is the formula of the relationship between the displacement and the velocity of the uniform velocity linear motion derived? Hope to have the image and formula to push down together

The formula is v = 2As, which can be deduced from the formula v = at and S = 1 / 2 * at
V = at, v = a, t, and 2As = 2 * a * 1 / 2 * at = a, t
So V equals 2As. Understand?

The derivation process of the formula of the relationship between the speed and time of the uniform variable speed linear motion

The so-called quantitative change leads to qualitative change. If you can understand it, but you still can't, then the foundation is not good enough. In kinematics, uniform variable speed linear motion is the most basic. It's nothing more than the analysis of motion process and the application of several formulas. On the one hand, it depends on talent. On the other hand, it's also more important, it depends on hard training, In kinematics, there are quite a lot of problems with camouflage, which depends on the accumulation of your own experience and a solid grasp of basic skills
People can have instant epiphany, but it can be met but not sought. Besides, BeiChan also advocates that epiphany should be based on gradual epiphany. No matter what, more questions and more summaries are the basic methods. You can discuss the learning methods with your classmates
It's the most important to lay a good foundation when doing questions. The physical kinematics of college entrance examination is not particularly tricky. Everything is a little higher on the basis, and then open the score section. It's always the most important to lay a good foundation
Here's a technique for moving in a straight line with constant speed change
In terms of calculation:
1、 Acceleration is the core, almost any problem can encounter acceleration;
2、 Instantaneous velocity, such as initial velocity and final velocity, has a formula with displacement, which has nothing to do with time
V1^2 - V0^2 = 2as
3、 Average speed problem: the average speed of uniform variable speed linear motion is equal to the instantaneous speed in the middle of the whole process; it is also equal to (initial speed + final speed) / 2
4、 The concept of time is graphically represented by line segments on the number axis;
Analysis questions:
1、 Keep in mind that velocity, acceleration and displacement are vectors. What are vectors? Do you know what you need to pay attention to?
2、 Speed is the title, the average speed is not necessarily the size of the average speed, to remember;
3、 It's better to use the image flexibly. If you can't use the image well, don't use it
Well, having said so much, I wish you success as soon as possible!

The average velocity formula of uniform variable speed linear motion and the average velocity formula of variable speed linear motion
How is it deduced?

This conclusion can be deduced from the formula
Vt=V0+at
Vt^2-V0^2=2aS
Average velocity v = s / T = (VT ^ 2-v0 ^ 2) / 2at
Because VT = V0 + at
V=(V0^2+2atV0+a^t^2-V0^2)/2at=V0+at/2
So the average velocity is half of the sum of the initial velocity and the final velocity
square

How to deduce the displacement formula of uniform velocity linear motion
s=v0t+(1/2)at^2

Displacement = average velocity * time
The initial speed is v0
The velocity at T seconds v = V0 + at
So the average velocity = (V0 + V) / 2
So displacement S = average velocity * time = (V0 + V) / 2 * t = (V0 + V0 + at) / 2 * t = v0t + (1 / 2) at ^ 2

Derivation of the relationship between speed and time of uniform variable speed linear motion
The formula is v-v0 = 2aX
According to v = V0 + at and x = v0t + 1 / 2at
V is the final velocity. V0 the initial velocity. T time

Because V = V0 + at
The square of both sides is V2 = V0 ^ 2 = a ^ 2T ^ 2 = 2v0at. 1
And because x = v0t + 1 / 2at square. 2
So 2x = 2v0t + at square
Both sides multiply a by 1 to get V ^ 2-v0 ^ 2 = 2aX

The relationship between displacement and velocity
The formula of the change law of the speed of the uniform variable speed linear motion with time;
The formula of displacement changing with time in uniform speed changing linear motion;
The relationship between displacement and velocity of linear motion with uniform speed change;
The formula of average velocity of uniform speed change linear motion

The formula of the change law of the speed of uniform variable speed linear motion with time: v = at
The formula of displacement variation with time of uniform velocity linear motion: x = VT + 1 / 2at2 (is the square of time)
The relationship between displacement and velocity of uniform velocity linear motion: v2-v02 = 2aX
The average velocity formula of uniform variable speed linear motion: v = 1 / 2 (V0 + VT)
Do you understand? I really can't beat those things

Physics in grade one of senior high school
A, B, C three objects do uniform variable speed motion, when passing through point a, the speed of object a is 6m / s, the acceleration is 1m / S & # 178;; the speed of object B is 2m / s, the acceleration is 6m / S & # 178;; the speed of Object C is - 4m / s, the acceleration is 2m / S & # 178
A. When passing through point a, object a is the fastest and object B is the slowest
B. When passing through point a, Object C is the fastest and object B is the slowest
C. One second after passing point a, object B is the fastest and Object C is the slowest
D. None of the above is true
ABC analysis: according to the known conditions, when passing point a, object a is the fastest and object B is the slowest; 1 s before passing point a, object a is 5m / s, object B is 4m / s and Object C is - 6m / S; 1 s after passing point a, object a is 7m / s, object B is 8m / s and Object C is - 2m / s
Note: I know that the formula v = V0 + at is used, but it doesn't say whether it is acceleration or deceleration. How to calculate? I just calculated both of them, which is different from its answer. How to calculate?

If there are positive and negative, it can be judged that the direction of velocity and acceleration of a and B is the same acceleration, and the direction of velocity and acceleration of C is different from deceleration
A is bigger than size 6. A 5-b-4-c-6 one second before B, so B 7-B 8-c-2 one second after C

Analysis problem: the motorcycle is required to walk through a section of straight road in the shortest time from standstill, and then drive into a semicircular curve, but the speed should not be too fast when driving on the curve, so as not to deviate from the lane due to centrifugal effect. Find the shortest time for the motorcycle to drive on the straight road. See the table for relevant data. Starting acceleration a14m / S2, braking acceleration a28m / S2, maximum speed of straight road The maximum speed of V 140m / S curve V 22m / s straight length s 218 m a student's solution is as follows: in order to make the motorcycle take the shortest time, the motorcycle should first accelerate from standstill to the maximum speed & nbsp; v 1 = 40 m / s, and then decelerate to V 2 = 20 m / s, T 1 = V 1A1 = ；  t2=v1−v2a2=… ; & nbsp; & nbsp; t = T1 + T2 do you think this student's solution is reasonable? If it is reasonable, please complete the calculation; if it is unreasonable, please explain the reason and use your own method to calculate the correct result

According to this classmate's solution, we can get that: T1 = v1a1 = 10s, T2 = V1 − v2a2 = 2.5s, so the acceleration distance S1 = v12t1 = 200m, S2 = V1 + v22t2 = 75m, total displacement S1 + S2 = 275m ＞ s

Exercises related to the relationship between the speed and time of uniform variable speed linear motion
The truck used to drive at a constant speed on the straight road at the speed of 10 meters per second. Because there was a red light at the intersection, the driver started to brake from a far place to make the truck slow down. When the car slowed down to 2 meters per second, the traffic light turned green and so on. The driver immediately braked and accelerated to the original speed in only one third of the deceleration time. It took only 12 seconds from the beginning of braking to the recovery of the original speed, Calculate the instantaneous speed at the end of 2 seconds and the end of 10 seconds after the acceleration starts to brake during deceleration and acceleration

If it is one third, please use the following formula structure. Because it shares 12s, the deceleration time T1 and acceleration time T2 are 12st1 + T2 = 12s, T2 = 1 / 2t1, so T1 = 8s, T2 = 4svt = V0 + at2 = 10 + A * 8A = - 1m / S ^ 2, VT = VO + at10 = 2 + 4a2a2 = 2m / ^ 2vt = V0

Several calculation problems of the relationship between the speed and time of uniform speed change linear motion in physics of grade one of senior high school
1. A car braking test is carried out, and the speed is reduced from 8m / s to 0 in 1 second. The sliding distance of a car braking at the specified speed of 8m / s shall not exceed 5.9m. Suppose that the car decelerates evenly when braking, ask whether the car's braking performance meets the requirements?
2. The car starts to move in a straight line at a constant speed from a standstill. At the end of the fourth second, the engine is turned off, and then it stops for six seconds. The car has driven for a total of 30 meters?
3. A small ball moves along a smooth inclined plane with a speed of 20 m / s, and the acceleration is 5 m / s. if the inclined plane is long enough, what is the size and direction of the ball's speed after 6 s?
4. When an aircraft takes off and taxis, it starts to make uniform acceleration linear motion from static state, and the acceleration is 4m / s and 178; when the taxiing speed of the aircraft reaches 85m / s, it leaves the ground and takes off. If the aircraft suddenly receives the command from the command tower to stop taking off when it reaches the take-off speed, the pilot immediately brakes the aircraft, and the aircraft makes uniform deceleration linear motion, and the acceleration is 5m / s and 178; How long does it take for the plane to take off and stop?
5. The normal driving speed of the car is 30 m / s. After the engine is turned off, the car starts to decelerate evenly, and the speed at the end of 12 s is 24 m / s. The calculation is: ① the acceleration of the car; ② the speed at the end of 16 S; ③ the speed at the end of 65 s

1. The acceleration of the car is - 8m / S ^ 2, the braking distance of the car is 4m, less than 5m, so it meets the requirements
2. Suppose that the acceleration of the car in the acceleration phase is A1, and the acceleration of the car in the deceleration phase is A2, so the formula can be expressed as VT = 5A1, S1 = 25 / 2 * A1, - 5A1 = 6A2, S2 = 30a1 + 18A2, S1 + S2 = 30, and the above formulas, and finally the answer can be solved, in which the maximum speed is vt
3. From the meaning of the question, when t = 4S, the ball reaches the highest point, so after 6 seconds, the ball's movement direction is downward along the slope, and then from the formula VT = V0 + at, we can know that the speed of the ball is 10m / s
4. The acceleration is 4m / S ^ 2. If the speed reaches 85m / s, the time required is 85 / 4. At this time, the speed is 85m / s. when the acceleration is - 5m / S ^ 2, the time required is 17s, so the total time is 85 / 4 + 17 = 38.25 seconds
5. According to the formula, VT = VO + at, so the acceleration is - 6 / 13m / S ^ 2, the speed at the end of 16S is v = 22.154m/s, the car has stopped at 65s, so the speed of the car is 0 at 65th second