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Suppose that an object starts to move in a straight line when its initial velocity is 1, and the acceleration at any time t is known to be (2 root sign T) + 1. Try to express the displacement s as a function of time t
A series of differential equations
A = DV / dt = (2 radical T) + 1
The solution is: v = (4 / 3) t radical T + T + C
Because when t = 0, v = 1
So C = 1
So v = (4 / 3) t radical T + T + 1
Both sides multiply DT to get: VDT = [(4 / 3) t radical T + T + 1] DT
The integral on both sides is s = (8 / 15) (T ^ 2) radical T + (1 / 2) T ^ 2 + T + C
Because when t = 0, s = 0
So C = 0
So s = (8 / 15) (T ^ 2) radical T + (1 / 2) T ^ 2 + T
The graph is the V-T graph line of the object motion. Starting from t = 0, the moment of the maximum displacement to the origin is? The time period when the acceleration and velocity direction are the same is?
The time of maximum displacement of the origin is T2, because T1 accelerates before T1 and T1-T2 decelerates, but the direction of motion remains unchanged. When T2 is reached, it moves in the opposite direction. Therefore, T2 is the farthest time. The time period of acceleration and velocity in the same direction: 0-t1, T2-T3
An object moves in a straight line with constant acceleration at initial velocity V0 and acceleration A. if the acceleration of the object gradually decreases to 0 from time t
Explain it in detail
When the initial velocity is 0, the acceleration will not change with the increase of velocity. The acceleration will decrease from t, but the velocity will increase all the time, only the increment of velocity will decrease in unit time
If the length of the spring is the same at T1 and T2, then it is certain that the oscillator will vibrate at these two moments ()
A. Equal displacement and opposite direction B. equal velocity and same direction C. equal acceleration and same direction D. equal velocity and opposite direction
A. The spring vibrator vibrates harmonically on a smooth horizontal plane, and the spring length is the same at T1 and T2, which indicates that the displacement and direction of the vibrator passing through the same position are the same, so a is wrong. B, D and velocity are the same, but the direction is not necessarily the same, so BD is wrong. C and vibrator passing through the same position, the displacement and direction are the same, according to a = - kxm The magnitude and direction of acceleration are the same, so C is correct
How to calculate the acceleration of the dot timer? What is the formula?
a=ΔS/T^2
In this question, TAD = TDG = t = 0.02 * 3 = 0.06
So sdg-sad = Δ s = 0.0390-0.0210 = 0.0180
a=ΔS/T^2=0.0180/0.06^2=5.00m/s^2
How to calculate the acceleration according to the paper tape made by the dot timer
1. Find the velocity (approximate method)
The average velocity of a displacement near a point is used to replace the instantaneous velocity
2. Calculate acceleration (successive difference method)
In the motion of uniform velocity change, in the equal time between the front and the back, the displacement difference is at ^ 2
For example, s2-s1 = at ^ 2, s3-s1 = 2at ^ 2, we can use this to calculate the acceleration
Of course, the acceleration can also be calculated by (V 2 - v 1) / T, but the error is less than that of the successive difference method
Acceleration (paper tape of dot timer)
How to calculate the acceleration with the paper tape of the timer? Why do I often get a big answer based on x = at ^ 2?
The formula you used is correct
(Note: let's see if you are observing the time point or the counting point)
Using the paper tape produced by the dot timer to calculate the acceleration
Successive difference method Δ s = at ^ 2
In physics, how to calculate the acceleration of the point on the paper tape of the dot timer
According to the uniform linear motion, the displacement difference of adjacent equal time intervals is equal
x2-x1=x3-x2=x4-x3=.xn-xn-1=aT^2
In order to reduce the error, 4 or 6 segments are selected
a=[(x3+x4)-(x1+x2)]/4T^2
Unidentified questioning
What is the acceleration formula of the clock experiment?
According to △ s = at * t, there are s4-s3 = (s4-s3) + (s3-s2) + (s2-s1) = 3At * t. similarly, s5-s2 = s6-s3 = 3At * t, calculate A1 = (S4-S1) / 3T * TA2 = (s5-s2) / 3T * TA3 = (s6-s3) / 3T * t, and then calculate the average value
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