How to get the x at square, how to push the physical formula x at2

Uniform variable speed linear motion.

The difference between the displacements of adjacent dates at equal time is x=atsquared=at*t

Let the adjacent equality time be t, the displacement in the first t is x1=vt+1 2at*t, and the displacement in the first two t is x2=v*2t+1 2a*2t*2t

The displacement that occurs in the second t is x=x2-x1=3 2at*t+vt

x=x1-x=at*t

Set the initial velocity v0

t, the speed is v0+at

The speed after 2t is V0+2AT

The first displacement in t v0t+at2

The second intra-t displacement (v0+at)t+at2

Subtract to get at2

The uncertainty of v0 generalizes to the arbitrariness of two t's.

So it was proven.

x=at2

Solution: An object moves in a straight line with uniform variable speed, the displacement in the first second is x1=v0+1 2a, and the displacement in the second second is: x2=x (displacement in the first two seconds)—x (displacement in the first second) = 2v0+1 2a4-v0+1 2a=vo+3 2a

So the difference between the displacements in adjacent times = x2 - x1 = at2

First of all, "x" is the displacement "s", which can be derived with the help of images. Look at the picture.

For a uniformly accelerated linear motion with acceleration a, the velocity changes to at after time t. In the V-T graph, the displacement is represented by area, so its displacement in time t is s at*t*1 2=1 2at2

The formula should be x=1 2at2, which is the displacement calculation formula of a uniform acceleration linear motion with zero initial velocity, its initial velocity is zero, and the final velocity is at, because it is a uniform acceleration motion, so its average velocity is half of the sum of the initial velocity plus the final velocity, which is 1 2at, multiplied by t is the displacement x=1 2at2

v=at

The two ends are scored for t.

s=∫vdt=1/2at^2+c

If the initial event is 0, it can be obtained.

s＝1/2at^2

The formula is that s is equal to one-half of the square of at.

x=at square, which is the difference in displacement between adjacent equal time intervals, if not adjacent time intervals, it is at 2 multiplied by the number of segments of their interval time.

x=at square 2, you should be talking about a uniform acceleration linear silver auspicious motion with an initial velocity of 0.

The only formula for calculating the cherry blossoms, if the initial velocity of the cherry blossom is not zero, is x=vt+1 2at 2

Displacement x, acceleration a, time t, initial velocity v0, x=v0t+at 2, when the initial velocity is 0, x=at 2,

This is called the difference-by-difference method.

Here's how it works.

If there are two displacements of equal duration, S1 and S2, and the equal time interval between these two displacements is n, the length of time is equal, note that the time and defeat of each displacement are equal, and the interval n is the number of displacement segments between S1 and S2).

Then there is the difference-by-difference method to find the acceleration.

Equation s1-s2=(n+1)*a*t 2

The book should say that when s1 is adjacent to s2 (i.e. n=0), the formula is s1-s2=a*t 2 (which is the most common formula in the book).

The formula is the difference in displacement x=at 2 in a uniform acceleration linear motion over adjacent times of equal time

This is called the difference-by-difference method.

Here's how it works.

If there are two displacements of equal duration, S1 and S2, and the number of equal time intervals between these two displacements is n (equal time means that the length of time is equal, note that the time of each displacement is equal here, and the interval n is the number of displacement segments between S1 and S2).

Then there is the difference method to find the acceleration formula: s1-s2=(n+1)*a*t 2, the book should say that when s1 is adjacent to s2 (i.e., n=0), the formula is s1-s2=a*t 2 (which is the most common formula in the book).