Analysis of Acceleration Formula in senior one physics

Analysis of Acceleration Formula in senior one physics

V=Vo+at
There are several important corollaries
(1) Vt2 - V02 = 2As (uniform acceleration linear motion: A is positive, uniform deceleration linear motion: A is positive)
(2) Real time speed in the middle of segment a and B:
Vt/ 2 == A S a t B
(3) Real time velocity of displacement midpoint of AB segment:
Vs/2 =
Uniform speed: VT / 2 = vs / 2; uniform acceleration or deceleration linear motion: uniform acceleration linear motion with VT / 2 initial speed zero, in 1s, 2S, 3S The displacement ratio in NS is 12:22:32
…… N2; in 1s, 2S, 3S The displacement ratio in NS is 1:3:5
(2n-1); within the first meter, within the second meter, within the third meter The ratio of time in the nth meter is 1:1
…… (
No matter whether the initial velocity is zero or not, the position of a particle moving in a straight line with uniform velocity in a continuous and adjacent equal time interval
The displacement difference is a constant: S = at2 (a - acceleration of uniform variable speed linear motion, T - time of each time interval)
Vertical upward motion: upward motion is uniform deceleration linear motion, and downward motion is uniform acceleration linear motion
It is a uniform deceleration linear motion with initial velocity VO and acceleration G

The formula of gravity and velocity displacement in physical formula
S = 1 / 2GT ^ 2, then the deformation is v = 1 / 2GT?
How can it become V ^ 2 = 2GH?
In addition, how can I see some people write v = 1 / 2GT ^ 2 = 2GH?

V ^ 2-V ^ 2 = 2As (the former V is the final velocity and the latter is the initial velocity). When the initial velocity is 0, we can get V ^ 2 = 2GS if h is the displacement ~ V ^ 2 = 2GH
V = 1 / 2GT ^ 2 = 2GH has not seen, should not be a formula ~ may be a relationship written in specific circumstances

The solution of the inequality about X: the x power of 9 - the X + 1 power of 3 - 4 > 0

Let 3 ^ x = t, t > 0, then the inequality can be T ^ 2-3t-4 > 0, ∫ T > 4 or T < - 1, ∫ T > 0, ∫ T > 4, ∫ 3 ^ x > 4
The base of 3 is 4 in ﹥ x > log, and the base of 3 is 2 in ﹥ x > 2log

1. The turnover of a food store last month was 240000 yuan, which is 20% higher than that of last month. How many million yuan is the turnover of this month? If the business tax is paid by 5%, how many million yuan should be paid?
2. Dad saved 20000 yuan of Education Fund for the army, which will mature in three years. If the annual interest rate is 3.24%, how many million yuan can dad get after maturity? (because the army is still in the stage of compulsory education, he has to pay 20% interest tax.)
We must calculate the formula!

Turnover of this month: 24 × (1 + 20%) = 288000 yuan
Business tax: 28.8 × 5% = 14400 yuan
2 × 3 × 3.24% × (1-20%) + 2 = 215552 (ten thousand yuan)

Distance between points
The line L of point P (1,2) is cut by two parallel lines l1:4x + 3Y + 1 = 0 and l2:4x + 3Y + 6 = 0 to obtain the line length | ab | = 2. The equation of line L is obtained

Answer: let the line passing through point P (1,2) be Y-2 = K (x-1), y = K (x-1) + 2
The distance between 4x + 3Y + 1 = 0 and 4x + 3Y + 6 = 0 is d = | 6-1 | / √ (4 & # 178; + 3 & # 178;) = 1
The length of the line y = kx-k + 2 cut by two parallel lines AB = 2
So: the angle c between the straight line and the parallel straight line satisfies SINB = D / AB = 1 / 2
So: C = 30 degree
The slope of parallel line k = Tana = - 4 / 3
Then the inclination angle B of the straight line satisfies B = a + C = a + 30 ° or B = a-c = A-30 °
tanb=tan(a+30°)
=(tana+tan30°)/(1-tanatan30°)
=(-4/3+√3/3)/(1+4√3/9)
=(25√3-48)/11
Or:
tanb=tan(a-30°)
=(tana-tan30°)/(1+tanatan30°)
=(-4/3-√3/3)/(1-4√3/9)
=-(25√3+48)/11
So: the straight line is y = K (x-1) + 2 = (25 √ 3-48) (x-1) / 11 + 2 or y = - (25 √ 3 + 48) (x-1) / 11 + 2

Given that x ∈ R, a = x2 + 12, B = 2-x, C = x2-x + 1, we try to prove that at least one of a, B, C is not less than 1

It is proved that if a, B and C are all less than 1, that is, a < 1, B < 1 and C < 1, then there is a + B + C < 3 and a + B + C = 2x2-2x + 12 + 3 = 2 (x − 12) 2 + 3 ≥ 3, which is contradictory; therefore, at least one of a, B and C is not less than 1

Find the second derivative d ^ 2Y / DX ^ 2 of the following parametric equation
x=t^2/2
y=1-t
The answer is 1 / T ^ 3
I don't know how to cut it out
I got 1 / T ^ 2

The second derivative of parametric equation is often wrong in the second step
(1) Dy / DX = [dy / dt] / [DX / dt] = - 1 / T this step has less errors
(2) D ^ 2Y / DX ^ 2 = D (dy / DX) / DX. Most people who make mistakes regard it as D (dy / DX) / dt. That's your answer
= d(dy/dx) / dt * dt /dx
= 1/t^2 * 1/ t
=1/t^3

The formula of projective theorem

Formula: for the right angle △ ABC, ∠ BAC = 90 degrees, ad is the height on the hypotenuse BC,
Projective theorem,
（AD）^2=BD·DC
（AB）^2=BD·BC
（AC）^2=CD·BC
So ad / BD = CD / ad
So (AD) ^ 2 = BD · DC

Let f (x) = {X & # 178;, | x | ≥ 1 x, | x | < 1, G (x) be a quadratic function. If the range of function f (g (x)) is [0, + ∞), find g (x)
I've read your explanation before. I don't understand why g (x) can't take the interval from negative infinity to negative one

f(x)=｛x²,|x|≥1
{ x,|x|＜1
F (x) is a piecewise function
|When x | ≥ 1, f (x) = x & # 178; ≥ 1,
When | x|

Find the derivatives of the following functions: 1) y = ln (x2 + 3) 2) y = excosx 3) y = √ X-1 / x 4) y = x2 / X-1

1) Y = ln (x2 + 3) y '= 1 / (X & # 178; + 2) * (X & # 178; + 3)' = 2x / (X & # 178; + 3) 2) y = e ^ xcosxy '= e ^ xcosx + e ^ x (- SiNx) = e ^ x (cosx SiNx) 3) y = √ X-1 / XY' = 1 / (2 √ x) + 1 / X & # 178; well, y = √ [(x-1) / x] y '= 1 / 2 * √ [x / (x-1)] * [(x-1) / x]' = 1 / 2 * √ [x / (x-1) / x] '= 1 / 2 * √ [x / (x-1) / x]