![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
An object starts to move in a straight line with uniform acceleration from rest. The acceleration is A1. The movement time is T1. Then it moves in a uniform deceleration with acceleration A2. After T2, the speed is exactly zero. Then the average speed of the object in the whole process can be expressed as? Why are 0.5a1t1 and 0.5a2t2 also right
Because a1t1 = a2t2
In T1 time period, V-0 = a1t1, the speed accelerates uniformly from 0 to V, and in T2 time period, the speed decelerates uniformly from V to 0,0-v = a2t2
The average speed of T1 = a1t1 / 2, the average speed of T2 = a2t2 / 2, the average speed of T1 and T2 is equal, which is also the average speed of the whole process
An object starts to move in a uniform acceleration straight line with an acceleration of A1 from a standstill, and immediately changes to a uniform deceleration straight line with an acceleration of A2 after time T1 An object starts to move in a straight line with uniform acceleration of A1 at rest, changes to a straight line with uniform deceleration of A2 after time T1, and stops after time T2, then the average speed of the object in all time is () A．a1 t1/2 B．a2 t2/2 C．（a1 t1+a2 t2）/2 D．
You only provide three options
Maximum speed in the whole process v = a1t1 = a2t2
Then the average speed of the whole process is v / 2 = a1t1 / 2 = a2t2 / 2
A. B correct
C is wrong, (A1 T1 + A2 T2) / 2 = v
An object starts to move in a straight line with uniform acceleration from rest. The acceleration is A1. The motion time is T1. Then it moves in a uniform deceleration with acceleration A2. Then An object starts to move in a straight line with uniform acceleration from rest. The acceleration is A1. The movement time is T1. Then it moves in a uniform deceleration with acceleration A2. After T2, the speed is exactly zero. Then the average speed of the object in the whole process can be expressed as?
The average velocity of the whole process is equal to the total displacement s divided by the total time t
The displacement during acceleration is S1 = 1 / 2 * A1 * T1 ^ 2
The displacement during deceleration is S2 = 1 / 2 * A2 * T2 ^ 2
In addition, A1 * T1 = A2 * T2,
The average speed is v = (S1 + S2) / (T1 + T2). Bring in the above three equations,
V = 1 / 2 * A1 * T1 = 1 / 2 * A2 * T2 can be obtained
A particle starts at rest with the acceleration A1 Make a uniform acceleration linear motion for a period of time, and then make a uniform deceleration linear motion with the acceleration of A2 until it is stationary. The total time of particle motion is t, and calculate the total displacement of particle
Set the maximum speed during movement as VM,
Then the total time t = VM
a1+vm
a2 ①
Then the total displacement x = VM
2t1+vm
2t2＝vm
2t ②
The solution of simultaneous equations ① and ② is x = a1a2t2
2(a1+a2)．
A: the total displacement of the particle is x = a1a2t2
2(a1+a2)．
A particle starts to move in a straight line from rest. First, it moves in a straight line with uniform acceleration at the acceleration A1, and then it moves in a straight line with uniform deceleration at the acceleration A2 until it stops. It is known that the time of the whole process is t, then the total distance the object passes is
Let the total distance be s and the maximum speed be V (the final speed in the uniform acceleration stage and the initial speed in the uniform deceleration stage). Let the time in the uniform acceleration stage be T1 and the motion distance be S1; If the time of the uniform deceleration stage is T2 and the moving distance is S2, it is in the uniform acceleration stage: S1 = A1 * T1 ^ 2 / 2 and V = A1 * T1 at uniform deceleration
A particle starts to move in a straight line with uniform acceleration from rest, and the acceleration is A1. After a period of time, it then moves in a straight line with uniform deceleration until it stops. The acceleration is A2, and the displacement in the whole process is X. find the time of the whole process
Set the maximum speed during movement as VM,
Then the total time t = VM
a1+vm
a2 ①
Then the total displacement x = VM
2t1+vm
2t2＝vm
2t ②
The solution of simultaneous equations ① and ② yields t =
2(a1+a2)x
a1a2．
A: the total time of particle motion is t =
2(a1+a2)x
a1a2．
RELATED INFORMATIONS
- 1. A particle starts to move in a straight line with uniform acceleration from rest, and the acceleration is A1. After time t, it moves in a straight line with uniform deceleration, and the acceleration is A2. If it can just return to the starting point after time t, A1: A2 should be () A. 1：1 B. 1：2 C. 1：3 D. 1：4
- 2. If the maximum acceleration that passengers can endure in the subway train is 1.4m/s2. If the distance between two adjacent subway stations is 560m What is the shortest time for the subway train to travel between these two stations? The answer is 40,
- 3. Find indefinite integral ∫ [(e ^ (3x) - 1) / (e ^ (x) - 1)] DX How to get ∫ [(e ^ (x) - 1) (e ^ (2x) + e ^ (x) + 1) / (e ^ (x) - 1)] DX
- 4. On the sine theorem and cosine definite of triangle In triangle ABC, if sin2a = sin2b, what triangle must triangle ABC be? The answer is isosceles or right triangle!
- 5. What is the second definition of ellipse, hyperbola and parabola?
- 6. Power function exponential function logarithmic function trigonometric function and inverse trigonometric function are collectively referred to as basic elementary functions. Can't you remember clearly
- 7. The railway between the two stations is 1464 kilometers long. Two trains leave from the two stations at the same time. It is known that the speed of one train is 63 kilometers and the speed of the other train is 59 kilometers per hour. How many kilometers did each line travel when they met?
- 8. A 200 meter long train passes through a tunnel at a speed of 4 meters per second for 240 seconds. How long is the tunnel? Calculation problem: a 200m long train passes through a tunnel at the speed of 4m / s for 240 seconds. How long is the tunnel? [write down the process as required
- 9. The Qing government appointed Zhan Tianyou as the chief engineer to build the railway from Beijing to Zhangjiakou!
- 10. When did Zhan Tianyou build the Beijing Zhangjiakou Railway
- 11. A 200m long train passed the bridge. It was measured that it took 1 minute for the train to get off the bridge. The duration of the whole train on the bridge was 40s. Calculate the length of the bridge and the speed of the train
- 12. A train is 200 meters long and passes through a 1.8 km tunnel. It takes 2 minutes to calculate the train speed To be more detailed, there must be known, unit conversion,
- 13. Someone estimated the acceleration of the train with a watch. He observed it for 3min and found that the train had advanced 540M. After 2min, he observed it for 1min and found that the train had advanced 1620m If the train moves in a straight line with uniform acceleration within 6min, the speed of the train is? A.0.01m/s ² B.0.1m/s ² C.0.6m/s ² D.1.8m/s ²
- 14. Friction force F and sin α cos α Reference relation of equal trigonometric function That is, when to use sin when the inclination angle on the inclined plane has F α Or cos α For example, what does f = mgcosa mean, and what is the difference between uphill and downhill angles?
- 15. Find the indefinite integral x ∫ (1-x ^ 2) / X DX under the root sign of X ∫ (1-x ^ 2) / X * x DX under the root sign
- 16. Solving indefinite integral ∫ 1 / (2x-3) ^ 2DX by substitution method
- 17. Find the transformation relationship between the initial three trigonometric functions sin Tan cos. For example, sin40 is equal to Cos 50
- 18. An object moves in a straight line with uniform acceleration, the initial velocity is 0.5m/s, and the displacement in the 7th s is 4m more than that in the 5th s Find: (1) the magnitude of the acceleration a of the uniformly accelerated motion of the object (2) The displacement of the object in the first 5S
- 19. A vehicle traveling along a flat and straight road has a speed of 36km / h. The acceleration obtained after braking is 4m / S2. Find: (1) Speed at the end of 3S after braking; (2) The speed at which the vehicle coasts half the distance from the start of braking to the stop
- 20. A train moves in a straight line at an initial speed of two meters per second and an acceleration of 0.5 meters per second. Find (1) what is the speed of the train at the end of three seconds? (2)... A train moves in a straight line at an initial speed of two meters per second and an acceleration of 0.5 meters per second. Find (1) what is the speed of the train at the end of three seconds? (2) What is the average speed in the first four seconds?