The derivation process and reason of the formula of acceleration measured by physical clock in grade one of senior high school Why must a = [(S6 + S5 + S4) - (S3 + S2 + S1)] / 9t ^ 2 not be a group of three? What is the grouping standard? Is it n / 2? What if n is odd?

The derivation process and reason of the formula of acceleration measured by physical clock in grade one of senior high school Why must a = [(S6 + S5 + S4) - (S3 + S2 + S1)] / 9t ^ 2 not be a group of three? What is the grouping standard? Is it n / 2? What if n is odd?

First of all, the soul formula of this experiment is put on: Delta (x) = a * T ^ 2 [Delta (x) is the relative displacement, which is represented by D (x)]. First of all, answer your last two questions. The essence of this experiment is to calculate the acceleration with the measured relative displacement and the known period. If there is no error, we only need a set of data in theory. But it is impossible in practice, and the error must exist, and it is very large, How to reduce it? Yes, it is to measure several groups of data, the more the better. The textbook gives six groups of data, which is almost enough, the more the better. You may want to ask why, and you will understand when I finish the formula. In the same period, the displacement difference between each adjacent two ends is certain, that is, D (x), so if there is one interval in the middle? Naturally, it is the superposition of displacement differences, that is, s2-s1 = D (x), then s3-s1 = 2D (x), And so on: s6-s1 = 5D (x), s5-s2 = 3D (x), s4-s3 = D (x). You may want to ask, why are there so many intervals? It seems so annoying. It's not good to subtract directly adjacent to each other. In fact, experiments are not as inevitable as mathematical axioms. All the methods that can reduce errors are good methods. The farther the distance between the points, the smaller the influence of motion, so the more accurate the calculation is, Continue to deduce the formula ~ add the equal signs of the above three formulas on the same side, that is, s6-s1 + s5-s2 + s4-s3 = 9 * D (x) = 9 * at ^ 2, so a = [(S6 + S5 + S4) - (S3 + S2 + S1)] / 9t ^ 2

A formula for calculating the acceleration of a dot timer
It's the formula. There's no formula in some calculation books. It's the kind of acceleration calculation for 3-5 points. It's better to bring examples,

The difference of displacements in adjacent equal time is equal to the square of the acceleration of gravity times time

A uniformly accelerating object with non-zero initial velocity reaches point a through displacement X1 in time t, and then passes through displacement X1 in time t
For a uniformly accelerating rectilinear moving object with non-zero initial velocity, it reaches point a through displacement X1 in time t, and then reaches point B through displacement x2 in time T. what is the velocity and acceleration of the object in B?

The reverse process is as follows
The acceleration is a
From the title:
VT-1/2aT^2=x2
2VT-1/2a(2T)^2=x1+x2
V = (3 * x2-x1) / (2t)
a=(x2-x1)/T^2

When a particle moves in a straight line with uniform acceleration, its acceleration is a, it passes through point a at a certain time, and passes through point B through time t, and its displacement is S1;
After the third time t, it passes through point C; after the third time t, it passes through point D. the displacement in the third time t is S3
(please write clearly)

Take a line first, take four points of ABCD, write VA at point a, V is the speed, because the time interval is the same, then point B is V + at, and point C is V + 2at
OK, from S = VT + 1 / 2at ^ 2, S1 = VT + 1 / 2at ^ 2, S3 = (V + 2at) t + 1 / 2at ^ 2, ha ha, OK, according to mathematics, we can subtract the left side from the left side and the right side from the right side of the formula to get it

When an object starts to faint in a straight line with uniform acceleration at rest, it passes through the displacement S1 at the start time t to reach point a, and then passes through the displacement S1 at the same time t
When an object starts to faint in a straight line with uniform acceleration at rest, it reaches point a through displacement S1 at the start time t, and then reaches point B through displacement S2 at the same time T. why can the velocity of the object at point B be (3s2-s1) / 2T or 2 (s2-s1) / t?

Suppose acceleration a, then S1 = att / 2, S2 = a2t2t / 2 - att / 2 = 3att / 2, that is, S2 = 3s1
The velocity of a = 2S1 / TT at point B is v = a * 2T = 4s1 / T = (3s2-s1) / 2T = 2 (s2-s1) / T

In the arithmetic sequence {an}, let S1 = a1 + A2 + a3 + A4 +an,s2=a(n+1)+a(n+2)+.+a(2n),),s3=a(2n+1)+.+a(3n）
In the arithmetic sequence {an}, let S1 = a1 + A2 + a3 + A4 +An, S2 = a (n + 1) + a (n + 2) +. + a (2n), S3 = a (2n + 1) +. + a (3n), prove that S1, S2, S3 are also arithmetic sequence

We can prove the summation formula of the arithmetic sequence of S1, S2 and S3

s1=1/2² ,s2=(1-1/2²）*（1-1/3²）,s3=(1-1/2²）*（1-1/3²）*（1-1/4²）……
① Conjecture: S9 =?, Sn =?
② When n increases gradually, will the value of Sn gradually approach a certain value? Can Sn be equal to this value? Why?
S1 = 1 minus 1 / 2

S9=(1-1/2²）*（1-1/3²）…… （1-1/10²）
=(1-1/2）(1+1/2）*（1-1/3）（1+1/3）…… （1-1/10）*（1+1/10）
=(1/2）*（2/3）…… （9/10）*(3/2）(4/3）…… （11/10）
=1/10*11/2
=11/20
Conjecture, Sn = (n + 2) / (2n + 2)
When n increases,
Sn=（n+1+1）/（2n+2）
=（n+1）/（2n+2）+1/（2n+2）
=1/2+1/（2n+2）
We can see that Sn tends to be 1 / 2, but because 1 / (2n + 2) > 0, it will never be equal to 1 / 2

As shown in the figure, R1 = 10 Ω, R2 = 20 Ω, R3 = 30 Ω, the power supply voltage is constant, if the switch S1 is closed and S2 is open, the reading of the ammeter is 0.3A, find: (1) what is the power supply voltage? (2) When S1 and S2 are disconnected, what is the reading of the ammeter? (3) When S1 and S2 are closed, what is the reading of the ammeter?

① When S1 is closed and S2 is open, R2 access circuit: u = i2r2 = 0.3A × 20 Ω = 6V; ② when S1 and S2 are both open, r1r2 series access circuit: I1 = ur1 + R2 = 6v10 Ω + 20 Ω = 0.2A; ③ when S1 and S2 are both closed, R2R3 parallel access circuit: I3 = UR2 = 6v20 Ω = 0.3A A: (1) power supply voltage is 6V; (2) when S1 and S2 are both open, the reading of ammeter is 0.2A; (3) when S1 and S2 are both closed, the power supply voltage is 0.2A The flow meter read 0.3A

Sum: 1 ^ 2 + 2 ^ 2-3 ^ 2-4 ^ 2 + 5 ^ 2 + 6 ^ 2-7 ^ 2-8 ^ 2 + 9 ^ 2 How to sum up until 40 ^ 2?

Solution investigation n ^ 2 + (n + 1) ^ 2 - (n + 2) ^ 2 - (n + 3) ^ 2 = - 8n-12
So the original formula = - 120-8 (1 + 5 + 9 +...) +37）
=-120-8（1+37）*10/2=-1640

Sum: 1-2 + 3-4 + 5-6 + +（2n-1）-2n．

1-2+3-4+5-6+… +（2n-1）-2n=（1-2）+（3-4）+… +（2n-1-2n）=-n．