![](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>)
The car starts to make a uniform acceleration linear motion with the acceleration A1 from the standstill, and then accelerates with the acceleration A2 after a period of time The car starts to make uniform acceleration linear motion with the acceleration of A1 from standstill, and then makes uniform deceleration linear motion with the acceleration of A2 after a period of time. It comes to a standstill after a total distance of L, so as to find the total time of car motion There should be a process. Thank you
Let t 1, T 2 be the first two segments, s 1, s 2 be the distance traveled, and V t be the highest speed reached. Then the initial speed in T 1 is 0 and the final speed is V T, so a 1 times T 1 = V t.1
In T2, the initial velocity is VT and the final velocity is 0
Vt + A2 times T2 = 0 (Note: A2 is negative at this time, due to deceleration movement)
And S1 = [A1 times (T1) square] / 2.3
S2 = VT times T2 + [A2 times (T2) square] / 2.4
S1+S2=L.5
The equation of 1 2 3 4 5 can be used
RELATED INFORMATIONS
- 1. The car starts to make a uniform acceleration linear motion with the acceleration of A1 from static, and after a period of time, makes a uniform motion with the acceleration of A2. The time of the whole process is t. please write down the detailed process of solving the problem. It is also useful to discuss the size of A2 instead of the direction?
- 2. A particle starts at rest, first makes a period of uniform acceleration linear motion with the acceleration A1 & nbsp; for a period of time, then makes a uniform deceleration linear motion with the acceleration A2 until it stops. The total time of the particle motion is t, and the total displacement of the particle is calculated
- 3. The object is moving in a straight line with uniform acceleration, the acceleration is 4m / S2, and the displacement is 12m within 2S
- 4. When the acceleration is 2 m / S2 and the displacement is 20 m in the first four seconds, the initial velocity is 0___ M / s, the initial velocity in the second second second is___ m/s The acceleration is 2m / S2 and the displacement is 20m in the first four seconds Then the initial velocity is______ m/s, The velocity at the beginning of the second second is______ m/s, The final velocity in 4S is______ m/s, The ratio of the first 2 s displacement to the second 2 s displacement is 0______
- 5. When an object moves in a variable speed linear motion, the velocity is 4 m / s at a certain moment, and becomes 10 m / s after 2 seconds, then the average acceleration of the object in these 2 seconds is less than 10 m / s? Why? I know the answer 3m / S2 and 7m / s2
- 6. A. Car B stops at the same place. At a certain moment, car a starts at a constant acceleration of 2m / S ^ 2. After 2 seconds, car B starts at a constant acceleration of 4.5m/x ^ 2 in the same direction (1) How long does car B leave to catch up with car a? (2) Before car B overtakes car a, what is the maximum distance between the two cars? Draw a V-T image
- 7. When an object is moving in a straight line with uniform speed change, the velocity is 4m / s and 178; at a certain time. After 2 seconds, the velocity becomes 8m / s and 178; to calculate the acceleration of the object
- 8. When an object moves in a straight line with constant speed, the velocity at a certain moment is V1 = 4m / s. after t = 2S, the velocity becomes V2 = 8m / s, and the acceleration a of the object is calculated
- 9. When an object moves in a variable speed linear motion, the velocity is 4 m / s at a certain moment, and the velocity after 2 s is 4 m / s. if the acceleration of the object is constant within 2 s, then the acceleration of the object is constant A / may be equal to 0 B / may be less than 4m / S C / must be 4 m / s D / may be greater than 4m / s
- 10. The initial velocity of the object is V0 = 5m / s, the initial acceleration is A1 = 4m / s # 178;, and the acceleration becomes A2 = 2m / s # 178; after moving for 2S What are the velocities at the end of 1s, 2S, 3S and 4S? The acceleration a decreases but the velocity V increases. Why?
- 11. The motorcycle starts from a standstill, runs for a certain distance with the acceleration of A1 = 1m / S2, then moves at a constant speed, and then decelerates with the acceleration of A2 = - 4 / S2 until it stops. It takes 1440m and 100s to walk. It is not necessary to solve the maximum speed VM twice in this process
- 12. The motorcycle starts from standstill, runs for a distance with the acceleration of A1 = 1m / S2, then makes a constant speed movement, and then makes a constant deceleration movement with the acceleration of A2 = - 4 / S2 until it stops. The motorcycle has walked 1440m in total and lasted for 100s. The maximum speed of the motorcycle in this process is calculated
- 13. The acceleration is A1. The motion time is T1. Then the acceleration is A2. After T2, the velocity is exactly zero. Then the average velocity of the object in the whole process can be expressed as? Why are 0.5a1t1 and 0.5a2t2 right
- 14. An object starts to move in a straight line with constant acceleration A1 from its rest, and immediately changes to a straight line with constant deceleration A2 after time T1 When an object starts to make a uniform acceleration linear motion with acceleration A1 from its rest, and immediately changes to a uniform deceleration linear motion with acceleration A2 after time T1, and stops after time T2, 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．
- 15. The object starts from the rest and moves in a straight line with uniform acceleration of A1. After a period of time, it moves in a straight line with uniform deceleration of A2 to the rest. If the total displacement is x, the maximum velocity in the motion is () A. 2a1xa2B. (a1+a2)xC. （a1+a2）xD. 2a1a2xa1+a2
- 16. The acceleration is A1 and the motion time is T1. Then the acceleration is A2 and the motion is decelerated The acceleration is A1. The motion time is T1. Then the acceleration is A2. After T2, the velocity is exactly zero. Then the average velocity of the object in the whole process can be expressed as?
- 17. When a car starts to make uniform acceleration linear motion from standstill with an acceleration of 2 meters per second, the average speed of the car in the fifth second should be explained in detail
- 18. [physics of grade one in senior high school] acceleration and force problems of trolley motion The question has a picture There is a sentence in the answer: the small plate and the weight pull the car, and the force to make the car do uniform acceleration linear motion is equal to the gravity of the small plate and the weight Let t be the one pulling the trolley and Mg be the gravity. Then, is it not mg-t = ma that we are doing the uniform acceleration motion now? How can we make the force of the trolley doing the uniform acceleration linear motion equal to the gravity of the small plate and the weight?
- 19. The second time, the car moved in the same way, but it broke down suddenly after 5 seconds. The next car stopped on the side of the road and then transported. The handover time was 20 seconds. The latter car also started with the original acceleration, Now, in order to make the car still arrive at the ambulance station at the scheduled time, what is the final uniform speed of the car?
- 20. Car a starts to move in a straight line with constant acceleration at 3 meters per second and car B lags behind for 2 seconds. At the same place, car a starts to move in a straight line with constant acceleration at 4 meters per second. The directions of the two cars are the same. What is the maximum distance between the two cars before car B catches up with car a? How long can car B catch up with car a after starting? How far are they from the starting point at this time?