What's the difference between the two formulas of compulsory physics: x = v0t + 1 / 2at ^ and x = (VO + VT) / 2 × t?

x=v0t+1/2at^2
It is suitable for uniform motion (uniform addition or subtraction) and knows VO, a, t
And x = (VO + VT) / 2 × t
It is suitable for any linear motion. Whether it is uniform or variable speed, as long as it is linear motion. And we know that VO, VT, t

How to deduce the displacement formula s = VO + 1 / 2at ^ 2 in physics

In a section of displacement s, the initial velocity is v. since the acceleration is constant a, the final velocity is v. + at, so the average velocity is v '= [(at + 2 VO) / 2], so the displacement S = V't = VO + 1 / 2at ^ 2
Using the image method, the time axis is regarded as a very small segment. In this small segment, because s = V * t, the area of the image can represent the displacement. Because the V-T image is a trapezoid, its area s = (upper bottom + lower bottom) * high / 2, that is, the displacement S = (V0 + V0 + at) * t / 2

How to deduce the physical formula VT ^ 2 + V0 ^ 2 = 2As in senior one

Your formula is wrong, not VT ^ 2 + V0 ^ 2 = 2As! But VT ^ 2 - V0 ^ 2 = 2As
By using two basic formulas: VT = V0 + a T, s = V0 * t + (a * T ^ 2 / 2) and eliminating the time t, we get
Vt^2－V0^2=2as　.

The formula VT ^ 2-v0 ^ 2 = 2As, why can we deduce V0 ^ 2 = 2As when the final velocity is 0?
Why not - V0 ^ 2 = 2As

For the formula:
Vt^2-V0^2=2as
When VT = 0, it's natural to get - V0 ^ 2 = 2As, so you're right
Here, a and s are both vector representations, that is, both size and direction
Moreover, it is easy to see from the formula that the product of displacement and acceleration is negative
But we often use scalar to calculate, so it is convenient to write V0 ^ 2 = 2As
As for the direction, we can only have a clear idea of it, or if necessary, we can supplement it with words
If you have any questions, please ask!

How to get the formula VT ^ 2-v0 ^ 2 = 2As?

S = 1 / 2at ^ 2 VT = VO + at

How to deduce VT ^ 2-v0 ^ 2 = 2As?

vt=v0+at
vt^2=v0^2+2v0at+a^2t^2
vt^2-v0^2=2v0at+a^2t^2=2a*(v0+at^2/2)=2as

Vt ^ 2 = V0 ^ 2 + - 2As what is a formula?

The formula of uniform variable speed linear motion (VT is the final velocity, V0 is the initial velocity, a is the acceleration, s is the displacement)

What is the physical meaning of the displacement formula v = v0t + 1 / 2at square and 1 / 2vt square

If the object moves in a straight line, then the displacement formula is VT, right
We divide its motion process into two parts: 1. Uniform velocity part (assuming no acceleration) 2. Uniform acceleration with initial velocity of 0
For 1, the displacement is vt
The displacement of 2 is at / 2 * t, which is 1 / 2at. You see, it is the acceleration displacement

In the square of the formula s = v0t + 1 / 2at, when t = 1, s = 13, when t = 2, s = 42, then t = 5, the value of S is obtained
In the square of the formula s = v0t + 1 / 2at, when t = 1, s = 13, when t = 2, s = 42, then t = 5, we can find the value of S

∵ when t = 1, s = 13, when t = 2, s = 42,
∴ 13=V0+(1/2)a ,
42=V0*2+(1/2)a*2² ,
By solving the equations, we get the following results
a=16 ,v0=5 ,
∴ S=5t+8t² .
When t = 5,
S=5*5+8*5²=225 .

The formula of displacement v = v0t + 1 / 2at square what is the physical meaning of 1 / 2vt square? Why is it not 1 / 3

In fact, the displacement formula of uniform acceleration motion is: average speed multiplied by total time. [VO + (VO + at)] * 1 / 2 is the average speed x = (2VO + at) * 1 / 2 * t = VOT + at * t * 1 / 2. The mobile phone is too difficult to type

What's the difference between the two formulas of compulsory physics: x = v0t + 1 / 2at ^ and x = (VO + VT) / 2 × t?