The conclusion of the experiment of measuring the acceleration of the car with the clock

The conclusion of the experiment of measuring the acceleration of the car with the clock

The dot timer is a machine that can continuously dot on the paper according to the same time interval. When the car drags the paper tape for acceleration motion, the dot timer will dot on the paper tape according to the same time interval. Because the paper tape is moving, and the dot machine is static, the dots on the paper are spaced, and because the car is doing acceleration motion, So the distance between these points is not equal, and the distance between these points is the displacement of the car at different times and at the same time interval. Then you can calculate the acceleration of the car. For example, when the car moves for a period of time, the puncher hits four points on the paper tape, and the total distance of these four points is 100cm, The interval of the dot machine is 0.5s to make a dot. Then you can make a dot according to the formula a = 2S / T # 178; = 200 / (0.5 × 4) # 178; = 50cm / s # 178;, of course, the acceleration should be uniform

The acceleration of the car is 4.41m/s *. In the test
If other conditions remain unchanged and only the frequency of alternating current is increased from 50 Hz to 60 Hz, the measured acceleration is 0

If the frequency of alternating current is 50 Hz, the acceleration A1 is calculated by using the interval T1 = 1 / 50 = 0.02 seconds; if the frequency of alternating current is 60 Hz, the acceleration A2 is calculated by using the interval T2 = 1 / 60 seconds

May I ask what factors will cause errors when using the dot timer to measure the acceleration of the car?

The main reason is the friction between the table and the car, followed by some other energy conversion and loss of kinetic energy

The time of acceleration difference method of dot timer
Like this formula
a=〔(S6+S5+S4)-(S3+S2+S1)〕/9T^2
How did 9t ^ 2 come from

ruxia

Generally, six intervals are taken as an example to calculate acceleration by successive difference method, and the average value of acceleration measured is a = (S4 + S5 + s6-s1-s2 - S3) / 9t * t. then, what about four intervals? When calculating acceleration of each segment, is it (s2-s1) / 9t * t or (s3-s1) / 9t * t? And the teacher said in class that if the interval obtained is odd, the interval in the middle should be omitted, Why? I don't understand these places. Thank you very much!

First of all, let's calculate the principle of a single A. let's set two intervals SX. If sy sets the initial velocity through SX as VX, then SX = vxt + at ^ 2 / 2sy = [VX + a (Y-X) t] t + at ^ 2 / 2sy SX = a (Y-X) T ^ 2, sy, SX, y, x, t are all known, so we can find a = (SY SX) / (Y-X) T ^ 2. Take six intervals for example: S1, S2, S3, S4, s

I read what others said on the Internet: "let the displacement of an object in several adjacent equal periods of time t be S1, S2, S3, S4, S5, S6. Then there are S4-S1 = 3At ^ 2, s5-s2 = 3At ^ 2, s6-s3 = 3At ^ 2. The above three relations can be obtained: a = (S4 + S5 + S6) - (S1 + S2 + S3) / 9at ^ 2." why must it be S4-S1, s5-s2, s6-s3, S6-s2. What's the principle of it? I don't understand it very well

The purpose is to use every data equally, so as to reduce the error,
According to your data, S2 is used twice. If S2 is changed to S4 according to your previous regularity, S4 is used twice. This is unreasonable. For example, if you want to measure the diameter of a wire, you should measure it in different places to get the average value. You can't measure it in the same place twice, and measure it in other places to get the average value again

Senior one physics with the difference by difference formula for acceleration, thank you!

How to calculate acceleration by difference method

Such a problem is generally to give a strip of paper and select it from the relatively clear points. Every five points is a counting point. In this way, the time interval between each counting point is Δ t = 0.1s. Generally, about seven points are selected. The first point is marked as o point. The distance between adjacent two points is measured as S1, S2 S6 according to uniform speed change

Under what circumstances is acceleration calculated by successive difference method
For example, do the questions about paper tape to find the acceleration. Under what circumstances, use the successive difference method to find the acceleration

In the uniform velocity linear motion, the adjacent is equal, and there is such a relationship in time △ x = at2
If the time is equal, the difference by difference method can be used

How to calculate acceleration by successive difference method

Let's suppose that an object is accelerating uniformly, the displacement passing through in the first t is S1, the displacement passing through in the second t is S2, and the displacement in the third t is S3. Then the acceleration of the object can be calculated as a = (s2-s1) / T square or (s3-s2) / T square. T is any time, t must be the same!