How to find the instantaneous speed of a cut paper tape to accelerate evenly?

Find a point on the left and right of the point you are asking for, ask for the same time interval, and then find the average velocity, which is the instantaneous velocity of this point.

Point C is left, point B is right, and point D is 12

The total travel time is:

So the speed is.

Find the two points closest to the point where you require speed, divide the distance between those two points by the time it takes to hit those two points (if there is no relevant explanation on the question, it is generally a four-digit counting point:, a hit point:, you should use the first data to divide by two numbers and one division is the instantaneous speed of the point required.

I don't know how to ask again.

Take the adjacent three consecutive points a, b, and c, and let the time of ab s1bc s2 be t;

then the acceleration a (s2 s1) tt.

Take the adjacent three consecutive points a, b, and c, and let the time of ab s1bc s2 be t;

Then the instantaneous velocity of the acceleration a (s2 s1) ttb point v (s1 s2) 2t.

Remember the formula: δx at is fine, where "δx" is the displacement difference of the distance of adjacent equal points, and "t" is the number of adjacent points.

Time interval (e.g., take two adjacent points x1 and x2, δx x2 x1, t 0 02s).

Take the adjacent three consecutive points a, b, and c, and set ab=s1 bc=s2 for every two burning points;

Then the acceleration a=(s2-s1) the instantaneous velocity of the point b of the cave tt v=(s1+s2) 2t

Method for finding instantaneous velocity and acceleration of paper tape:

1. According to the linear motion of uniform variable speed, the velocity at the midpoint of time is equal to the average velocity during this time, vn=(xn+xn+1) 2t

2. According to the equal displacement difference in the adjacent and equal time interval of the linear motion of uniform variable speed, the acceleration a= x t 2 is obtained by the difference-by-difference method

In fact, when finding the acceleration of the paper tape, the difference method is generally used. For example, there are 3 segments in this question, the acceleration is a1, a2, and a3, and the displacement of each segment is s1, s2, and s3 respectively

a=1/3(a1+a2+a3)=1/3(s1/t^2+s2/t^2+s3/t^2)

We need to consider the problem of the size of the error, because the acceleration on the paper tape is the same, but because the displacement of the AB segment is smaller, the calculated error is larger. Later, as the velocity increases, the resulting displacement also increases. So s1 should be discarded

So put (vd-vb) 2t

instead of (vd-va) 3t(va=0).

It's okay, as long as the VA is known.

The acceleration of the strip – (s2-s1) = at 2

Instantaneous velocity - v=(sn-s1) nt, how to find the average velocity - v=(s2+s1) 2t

Let the initial velocity be u

The acceleration is constant a

The elapsed time is t

The velocity at any point is v

v=u+at