Is the ratio of displacements through the same time at2?

No The ratio of time taken through successive equal displacements is 1: Root 2 - 1: Root 3 - Root 2

x=(1/2)at1^2，t1=(2x/a)^1/2，t1=(2x/a)^1/2

2x=(1/2)at2^2，t2=(4x/a)^1/2，t2=t2-t1=[2(2x/a)]^1/2-t1=[(2^1/2)-1]t1

3x=(1/2)at3^2,t3=(6x/a)^1/2,t3=t3-t2=[3(2x/a)]^1/2-[2(2x/a)]^1/2=[3^1/2^1/2-2^1/2)])t1

t1:t2:t3...=1:[(root2)-1]:[root3)-(root2)].

1) The difference between the displacements within any two continuously equal time intervals t is a constant quantity, i.e.

s2-s1=s3-s2…=δs=at2

or sn+k-sn=kat2

A uniform acceleration motion with zero muzzle velocity has the following characteristics.

The ratio of the displacements that have passed in successive equal periods of time from the start of the motion is .

s1:s2:s3:…:sn=1:3:5:…:2n-1)(n
…)

Hello, the first thing you have to ask is whether it is the ratio of displacement in the same direction, if it is the ratio of displacement in the same direction then the answer is 1:3:5:7:9....If it's braking and then moving in the opposite direction, it's not the same value...

First, 1:3:5:

7:9...Why, because you can calculate in the simplest way, if the acceleration is a, the ratio of the displacement in 1 second...

That's a:(4a-a):(9a-4a):

16a-9a):(25a-16a):.

It's different for braking, hehe, I hope it will be beneficial to your judgment questions, physics is thoughtful.

Second, you are asking about the ratio of time spent on the same displacement, hehe, the same reason, also using examples.

Is the difference in displacement as at2 a times t squared.

According to the characteristics of the free fall motion, the displacement ratio in equal time is 1:3:5::

2n-1)。Let the equal time be t and the acceleration be a, then in the nth time period, the average velocity is (n-1 2)a and the displacement is (n-1 2)at, that is, the displacement ratio in equal time intervals is (1-1 2)at:(2-1 2)at:

3-1/2)at:.:n-1/2)at=1:3:

5:.:2n-1)。

The displacement in the first second is: s = meters. Then the displacement in the second should be 15 meters, but due to the known conditions, its displacement in the last second is 10, so it has landed in the second second, so the falling height is 5+10, so the falling height is 15 meters.

Features of free fall movement:

1) The object is stationary when it starts falling, i.e., the initial velocity v=0. If the initial velocity of the object is not 0, it cannot be considered free fall, even if it falls vertically.

2) In the process of falling, the object is no longer subjected to any other external force (including air resistance) or the net force of the external force is 0 except for the action of gravity.

3) When any object is in free fall at the same height, the falling time is the same.

The above content reference: Encyclopedia - Free Fall.

The formula can be derived from the bridger as follows: set the initial velocity v0 and t. for each time periodThen the first stage t displacement is v0*t+1 2*at 2

The velocity at the end of the first segment v1=v0+atThen the displacement in the second segment t is v1*t+1 2*at 2Subtracting the two gives a distance difference of at 2

Physics has to make use of mathematical expressions.

In addition to algebraic derivation, it is best to understand it from an intuitive physical process. Here's what I think is the simplest and most intuitive derivation:

First of all, draw a V-T diagram of the motion of an object moving at a uniform velocity, which is a ray of rays. It can be known that the displacement of an object over a period of time is the trapezoidal area enclosed by the image and the t-axis during this time. In this way, it is possible to make a displacement trapezoid with two ends adjacent to each other for the same time (set to t).

With knowledge of geometry, it is possible to discover (to accelerate motion evenly.

For example), the area of the second trapezoid is larger than the area of the first trapezoid, and the area of a long square is called. The length of the rectangle (along the t-axis) is t, and the width (along the v-axis) is at (which can be obtained by the slope of the straight line), so that the area difference, i.e., the difference in displacement of adjacent displacements in the same time period, is the square of at.

I don't know if I didn't say that Min Shu is clear, welcome to ask

Odd ratio Displacement in the first second: Displacement in the second second: Displacement in the third second: Displacement in the fourth second = 1:3:5:7......

The object moves in a uniform linear motion with zero initial velocity, let the magnitude of each equal displacement be x, and let the time for the object to pass through each segment of displacement t1, t2, t3 respectivelytn。Then x=1 2 at1 ,2x=1 2 a(t1+t2) ,3x=1 coarsely roll 2 a(t1+t2+t3)....Solution, t1:t2:

t3…tn=1：(√2-1)：(3-√2)：

n-√(n-1))。

If the magnitude is equal and the direction is the same, the displacement is equal, and if the magnitude is equal and the direction is different, the distance is equal. Displacement uses displacement to represent the change in the position of an object (particle). Defined as:

A directed segment from the beginning to the end. Its size is independent of the path, and the direction is from the start point to the end point. It is a physical quantity with magnitude and direction, i.e., a vector.

Velocity formula: v=v0+at (since v0 and a are fixed values, v is a primary function of t). Displacement Formula:

s=v0t+(at^2)/2。

Let the time interval be t, the acceleration be a, and the initial velocity at t=0 is vo.

then the displacement within time 0 t x1 vot at.

Time t displacement in 2t x2 2vot + a(2t) x1 vot + 3 2at.

Therefore, the displacement difference δx x2 x1 at.

Let the equality time be δt=δt1=δt2, then the displacement in the time of δt1 is s1=voδt+1 2aδt 2

After the time of δT1, the velocity becomes V1=VO+AδT and the displacement is in the time of δT2.

s2=v1δt+1/2aδt^2=(vo+aδt)δt+1/2aδt^2=voδt+3/2at^2

So δx=s2-s1=at2

Derived from three formulas of kinematics.