The derivation process of high school physics formulas, senior one physics formulas and derivation f

That's what I want to know.

The physical formula and derivation formula are as follows:

The movement of the particle - the linear motion grip elimination.

Average velocity v-flat = s t (defined) 2Useful inference vt 2 v0 2=2as

Intermediate moment velocity vt 2= v flat = (v t + v o) 2

Intermediate position velocity vs 2 = [(v o2 + v t2) 2] 1 2

Displacement s = v flat t = v o t + at2 2=v t 2 t

Acceleration a=(v t - v o) t is in the positive direction of v o, a is in the same direction as v o (accelerates) a>0;If it is reversed, it will be a<0

The experimental inference is that δS=at2 δs is the difference between the displacements of adjacent succession equal time (t).

Main physical quantities and units: muzzle velocity (v o): m s plus section huaizhi velocity (a): m s2 final velocity (vt): m s

The law of conservation of energy.

The gravitational and repulsive forces of the intermolecular explicit argument (1) rr=r0, f-induced = frepulsion, f-molecular=0, e-molecule potential energy = emin (minimum).

r>r0, f repulsion, f molecular force behaves as gravitational force.

r>10r0, f induction = f repulsion 0, f molecular force 0, e molecular potential energy 0

Let the time of each time period be t, and the initial velocity of each time period v0=at is obtained according to the formula s=v0t+1 2at 2v=at

Note: t represents the time consumed from the start of the exercise to the time period consumed) s1 = 1 2at 2

s2=(at)t+1 2at 2=3 2at 2s3=(2at)t+1 2at 2=5 2at 2sn=(nat)t+1 2at 2=(2n-1) 2at 2In summary, we get s1:s2:s3:

sn=1:3:

5:…:2n-1)(n
…)

We can first find that the motion displacement of the object reaches s2s3s

ns.

The initial velocity is 0s=1 2at 2

Time taken t1=t=(1-0)t

2s=1/2a(√2t)^2

Time taken t2 = 2t - t1 = ( 2 - 1)t

3s=1/2a(√3t)^2

Time taken t3 = 3t - (t1 + t2) = (3- 2) tns = 1 2a( nt) 2

Time taken tn= nt-(t1+......t2)=(n- n-1)t, we get t1:t2:t3:....:tn=(√1-0):(2-√1):(3

2):…n-√n-1)

It's actually quite simple.

It's just not good to fight.

and the square has a root number.

So you set the distance for 2s

V1v2V3 three speeds are the start.

The last three velocities in the middle.

v2^2-v1^2=2as

v3^2-v2^2=2as

A is equal to v2 2-v1 2=v3 2-v2 2 moved over and resolved.

I got it.

Have you ever learned.

v^2-v0^2=2ax

Ah, it's easy to have.

Set in order. v1v2

v3v2^2-v1^2=2ax

v3^2-v2^2=2ax

Because the acceleration a is constant, and the displacement between each velocity is equal, then v2 2-v1 2=v3 2-v2 2 2 shift term is obtained 2*(v2 2)=v1 2

v3^2ps:v^2-v0^2=2ax

It is derived from v=v0+at

T = (v-v0) a is brought in x=v0t+1 2*at 2, then x=v0(v-v0) a+(1 2)*a(v-v0) 2 a2 is multiplied by 2a on both sides of the equation, simplify:

2ax=2v0*v-2v0^2+v^2-2v0*v+v0^22ax=v^2-v0^2

Let's take an example.

According to the equations s s =v t and v=at, they have a t which is a common physical quantity, and then and and . Available:

v²=2as

It is by joining the same variables that many formulas can be deduced

、...... at the end of the first second and the end of the second secondThe ratio of velocity at the end of the nth second v1:v2:v3......：vn=1：2：3:……n。

、...... within the first 1 second and within the first 2 secondsThe ratio of displacements in the first n seconds s1:s2:s3:......sn=1：4：9……：n2。

、...... in time t and time 2tThe ratio of the displacement at the first time n is s :s defense :s ......sn=1：3：5：……2n－1）。

Through the pre-s, pre-2s, pre-3s ......, the ratio of the time required in the first ns t1:t2:......tn=1：√2：√3……：n。

Over 1s, 2s, blind loss 3s、......The ratio of the time required for the first ns t:t:t ......tn=1: grinding the god of the stove (2 1):( 3- 2) ......n－√n－1）

、...... at the end of the first second and the end of the second secondThe ratio of velocity at the end of the nth second v1:v2:v3......：vn=1：2：3:……Judgment Swift: n.

Derivation: v1=at, v2=a*2t, v3=a*3t....vn=a*nt

、...... within the first 1 second and within the first 2 secondsThe ratio of displacements in the first n seconds s1:s2:s3:......sn=1：4：9……：n2。

Derivation: s1=at*t 2 , s2=a(2t)*(2t) 2=2at*t s3=a(3t)*(3t) 2=9at*t 2

.sn=a(nt)*(nt)/2=n*nat*t/2

、...... in time t and time 2tThe ratio of the displacement in the first time nt is s:s :s ......sn=1：3：5：……2n－1）。Pin this flush search.

Derivation: From the knowable:

si=s1=at*t/2 , sii=s2-s1=3at*t/2, siii=s3-s2=5at*t/2...

sn=sn-sn_1=n*nat*t/2-(n-1)*(n-1)at*t/2 =(2n-1)at*t/2

The last three are all obtained by adding and subtracting the first three, which is very easy, so I won't say more.

1. Uniform acceleration of the linear motion of the nuclear disturbance The continuous mass is equally displaced.

s1=s2=s3=sn From vo=o, we can see that s=s is proportional to t*t under the condition of equal acceleration.

Obtain t1 square t2 square t3 square tn square = 1 2 3 n conversion to obtain three billy formulas.

Two. Uniform acceleration or linear motion for continuous equal time.

v = at ti t2 t3 tn=1 2 3 n v is proportional to t In the same way, three other formulas can be obtained by s=at*t.

The derivation of six proportions of uniform acceleration linear motion with a physical initial velocity of 0 is obtained.

1.This ratio is only suitable for uniform acceleration linear motion with zero initial velocity.

According to s = , then within 1s, within 2s, within 3s 、......The displacements of ns are ,,ah, so the ratio is 1:4:9:....:n^2

2.、... in the 1st, 2nd, and 3rd sThe displacement ratio of the ns is 1:(4-1):

9-4）：（16-9）……n^2 - n-1)^2] = 1:

2n-1)3.This one is the easiest. The instantaneous velocity at the end of n seconds is an, so the velocity ratio at the end of 1s, 2s, and 3s is 1:

2：3……There is also a scale, that is, from moment zero, the displacement ratio in equal time intervals is 1: (with 2 -1):

with 3 - with 2) :(with 4 - with 3): ....Heel n - Heel (n-1)).

This is the formula for uniform acceleration motion with an initial velocity of 0.

For example, the acceleration is 1m s and the end velocity is 1m s and the displacement in one second = (end velocity 1 + initial velocity 0) multiplied by time 1 = displacement in 2 seconds = (end velocity 2 + 0) multiplied by 2 = 2m, and then push it down by yourself to help yourself.

The displacement in 1 second is the displacement in 2 seconds.