It is known that the kinematics equation of a particle is x = 10cos (π T), y = 10sin (π T), (SI unit)

It is known that the kinematics equation of a particle is x = 10cos (π T), y = 10sin (π T), (SI unit)

I don't know what you want, but I can tell you
The particle equation is x ^ 2 + y ^ 2 = 100, which is obviously a circle. The radius of the circle is 10, the angular velocity is 2, the linear velocity is 20, and the acceleration is 40

Given that the motion equation of a particle is r = 6T ^ 2I + (3T + 4) J, then the orbit equation of the particle is?

R = √ {36t ^ 4 + 9t ^ 2 + 24t + 16} I don't know if it's right. I hope you can join me and see if my answer is right

Given the kinematics equation of the particle, how to find the acceleration and trajectory of the particle?

Can you find derivative? Because it's not clear when derivative was learned. The first derivative is velocity and the second derivative is acceleration

The two passenger and freight cars are running at the same time from A. B. the speed of the freight car is 3 / 4 of that of the passenger car. It takes six hours for the passenger car to complete the whole journey. After the two cars meet, they continue to move forward and return to the destination immediately. After a few hours, they meet for the second time

The two passenger and freight cars are running at the same time from A. B. the speed of the freight car is 3 / 4 of that of the passenger car. It takes six hours for the passenger car to complete the whole journey. After the two cars meet, they continue to move forward and return to the destination immediately. After a few hours, they meet for the second time
The speed of a freight car is 3 / 4 of that of a passenger car, which means that the speed ratio is 3:4 and the time to finish is 4:3
The second time the two cars met was three times. The whole journey was from a to B for passenger cars and from B to a for freight cars. When they met for the second time, they walked one section altogether, so it was three sections
It takes 6 hours for the bus to complete the whole journey, from which the number of the whole journey can be calculated
The bus speed is 4 * 6 = 24 shares, 24 * 3 / (3 + 4) = 10 and 2 / 7
I am a teacher

I've come across a practical problem in primary school. It's very distressing,
Simplified questions:
Once upon a time, there were six children who wanted to buy toys. Each of them could buy five toys at most. There were only a and B toys in the store, and the additional condition was that the six children did not necessarily buy toys at the same time, and they could only buy one of AB toys at a time
How many possibilities are there to buy these toys

This is a problem related to probability and statistics, involving step-by-step multiplication formula and classified addition formula. The calculation is a bit complicated... It needs to be classified

The perimeter of the cross section of a piece of wood is 25.12 cm. What is the area of its cross section in square cm?

Radius: 25.12 ÷ (2 × 3.14) = 25.12 ÷ 6.28 = 4 (CM); area: 3.14 × 42 = 3.14 × 16 = 50.24 (CM); answer: its cross-sectional area is 50.24 cm

Teacher called Mark, I am afraid of wrong, so to a pair
Xiao Ming deposits 400 yuan in the bank and withdraws it as a lump sum for two years. If the annual interest rate is 2.79%, how many yuan is the interest after maturity?
If the annual interest rate is 3.33%, how much is the principal and interest after maturity?
Mr. Li deposits a sum of money in the bank. Calculated according to the annual interest rate of 2.79% of the two-year deposit, the interest will be 223.2 yuan after maturity. How much is the principal of Mr. Li deposited in the bank?
Xiaoqiang's income in October is 5000 yuan, leaving 2400 yuan as living expenses. The rest of the money is deposited in the bank for one year. If the annual interest rate is 2.25%, how much interest will there be after maturity?
The school bought "Ping An insurance" for 1200 students. The insurance period is one year. According to the insurance rate of 0.4%, the whole school paid a total premium of 6000 yuan. How much is the insurance amount per person?
Wang Hong deposited 300 yuan of her lucky money in the bank for one year. When it was due, she even donated the money and interest to the children in the disaster area. If the annual interest rate is 2.25%, how many yuan did she donate?
Write the quantity in unit one

These questions do not need to write unit "1" quantity? According to the calculation formula of interest, it's OK

Among them, there are 71 young pioneers. It is known that the number of young pioneers in the first class accounts for three-quarters of the whole class, and the number of young pioneers in the second class accounts for five sixth of the whole class. How many people are there in each class?

If there are x people in the first class, there will be 90-x people in the second class
3X/4+5(90-X)/6=71
9X+10(90-X)=852
9X+900-10X=852
-X=-48
X=48
Class 1 = 48
Class 2 = 90-48 = 42

How to solve a system of linear equations with two variables? To be more specific, write out all the solutions to the system of linear equations with two variables,

Definition of bivariate linear equation
If an equation contains two unknowns, and the degree of the unknowns is 1, then the integral equation is called bivariate linear equation, which has infinite solutions. If the condition is imposed, it has finite solutions. For bivariate linear equations, there is generally one solution, sometimes no solution. The general form of bivariate linear equation is ax + by + C = 0 (a, B are not 0)
Here's a simple example:
（1)x-y=3
（2)3x-8y=4
（3)x=y+3
From (1) we get x = y + 3 and substitute (2) to get x = y + 3
3×（y+3)-8y=4
You can calculate y = - 1
So x = 2
The solution of this system of linear equations of two variables
x=2
y=1
There are also some solutions as follows
Case 1, 13X + 14y = 41 (1)
14x+13y=40 (2)
(2) - (1) get
x-y=-1
x=y-1 (3)
Substituting (3) into (1) leads to
13(y-1)+14y=41
13y-13+14y=41
27y=54
y=2
Substituting y = 2 into (3) yields
x=1
So: x = 1, y = 2
Features: add and subtract two equations, single X or single Y, so that the next substitution elimination is applicable
（2） Substitution method
Example 2, (x + 5) + (y-4) = 8
(x+5)-(y-4)=4
Let x + 5 = m, y-4 = n
The original equation can be written as
m+n=8
m-n=4
The solution is m = 6, n = 2
So x + 5 = 6, y-4 = 2
So x = 1, y = 6
Features: the two equations contain the same algebraic formula, such as x + 5, y-4 and so on
(3) Alternative exchange
Example 3, X: y = 1:4
5x+6y=29
Let x = t, y = 4T
Equation 2 can be written as: 5T + 6 * 4T = 29
29t=29
t=1
So x = 1, y = 4

How to solve the system of quadratic trigonometric equations?
Known:
ax=400 ay=300
bx=200 by=100
Equations
bx=axcosθ-aysinθ
by=axsinθ+aycosθ
After substituting the specific value, ask friends to help solve the result, and also give the calculation process

Ax = 400, ay = 300
bx=200 by=100
Put it in, get it
200=400cosθ-300sinθ ==>2=4cosθ-3sinθ…… (1)
100=400sinθ+300cosθ==>1=4sinθ+3cosθ…… (2)
From (1) × 4 + (2) × 3, 11 = 25 cos θ = = > cos θ = 11 / 25
Then we substitute cos θ = 11 / 25 into (1) to get sin θ = - 2 / 25
Supplement: whenever there is sin & sup2; θ + cos & sup2; θ = 1 (this is axiom, always correct),
But from the result of this problem, Sin & sup2; θ + cos & sup2; θ = 1 / 5 ≠ 1
There is no problem in the method of solving the problem. The reason is that the author made a fatal mistake, he ignored the condition of Sin & sup2; θ + cos & sup2; θ = 1