A bus starts from a standstill and moves in a straight line with a constant acceleration of 2m every second. After 6S, it should move in a straight line with a constant speed. When it comes to the next stop, it should turn off the engine and move in a straight line with a constant speed deceleration, and then stop after 12s Finding the speed of a car at a constant speed But what about the format? What is the positive direction, the acceleration direction or the motion direction of the object

Generally, the motion direction of the object is positive
Direct: VT = at = 2 * 6 = 12m / S

RELATED INFORMATIONS

1. When the car is in emergency braking, the acceleration is 2m / s Square, and the original speed of the car is 10m / s (1) How long does it take for the car to stop after braking? (2) What is the displacement of the car from braking to stopping?
2. When the car runs at a speed of 10m / s and gets an acceleration of 2m / S2 after braking, it will be 4 s after braking
3. When the car runs at a speed of 10 meters per second and gets an acceleration of 2m / s Square after braking, what is the displacement in 4 seconds after braking?
4. Urgency: a person can lift up to 100kg on the ground. When he rises in the elevator at a constant speed with an acceleration of a = 2m / s, he can lift up to 100kg Urgency: a person can lift up to 100kg of objects on the ground. When he rises in the elevator with a square acceleration of 2m / s, how many kg of objects can he lift? When he falls in the elevator with a square acceleration of 2m / s, how many kg of objects can he lift? (G = 10m / s Square)
5. The elevator starts to move in a straight line with a square acceleration of 2m / s from standstill, changes to a uniform motion after 3S, changes to a uniform deceleration motion after 7S, and stops after 2S 1. How fast is the elevator moving at a constant speed? 2. What is the acceleration during deceleration? 3. What is the average speed of acceleration and deceleration? Complete the formula
6. When the speed reaches 12m / s, the elevator starts to make a uniform linear motion. After 4S, the elevator stops. Calculate the height of the elevator and the average speed in the process of rising
7. A. B two cars are driving in the same direction on the straight road. A car has been moving at a constant speed of 20 m / s. when B car is 80 m in front of a car, B car's speed is 4 m / s, and it is moving at a constant deceleration with an acceleration of 2 m / S2? （　　） A. 4sB. 4.2sC. 5sD. 20s
8. There is a bicycle 120 meters ahead of the car moving at a constant speed of 6 m / s, and the car is catching up with the bicycle at a speed of 18 m / s. if two cars are moving in a straight line in the same direction on different lanes of the same road, the following questions can be found: (1) how long will it take for the two cars to meet for the first time? (2) If the car brakes immediately after catching up with the bicycle, and the acceleration during the braking process is 2m / S2, how long will it take for the two cars to meet for the second time?
9. A. Two cars B drive in the same direction on the straight road. Car a moves at a constant speed at the speed of VA = 20m / s, and car B moves at a constant deceleration straight line at the initial speed of vb0 = 24m / s, with the acceleration of a = 2m / S2. If car B is at △ s = 165m in front of car a, how long does it take for the two cars to meet? A student has the following solution: let two cars meet at the passing time t, then SA = VAT, sb = vb0t-12at2, SA = sb + △ s, The value of t can be obtained by combining the above formulas with the data. Do you think the student's solution is reasonable? If it is reasonable, please continue to calculate the time t; if not, please use your own method to calculate the correct result
10. The bus departs from the station and runs at a speed of 4 m / s along the straight road at a constant speed. Two seconds later, a motorcycle departs from the same station to catch up at a constant speed The acceleration is 3m / S & sup2;, Q 【1】 How long does it take for the motorcycle to catch up with the car? 【2】 How far is it from the starting point when the motorcycle catches up with the car
11. There is an object moving in a straight line with constant acceleration at an initial velocity of 0 and an acceleration of 10 meters every second, reaching 2M and 4m Then V1: V2 is equal to
12. When an object moves in a straight line with uniform speed change, the speed is 4m / s at a certain moment; after 1s, the speed is 10m / s, then the acceleration of the object in the 1s is 0 Why do we have two situations? Sometimes we need to set the positive direction in our current problem. How can we not use it? When can we set the positive direction? When is the same as this problem? I'm in a mess. Please help me solve my problem
13. An object moves in a straight line with an acceleration of 2m / S ^ 2 and a velocity of 4m / s at a certain time. The velocity and displacement of the object in 3S are calculated
14. When an object moves in a straight line with uniform speed change, its velocity is 4 m / s at a certain moment, and it becomes 10 m / s after 1 s. in this 1 s, the () A. The displacement may be less than 4MB. The displacement may be more than 10mc. The acceleration may be less than 4m / S2D. The acceleration may be more than 10m / S2
15. When an object moves in a straight line with uniform speed change, the velocity is 4m / s at a certain moment, and it becomes 10m / s after 1s? Let's first specify that the initial velocity direction is positive, then the velocity of 10m / s may be positive or negative If it is positive: a = 10-4 / 1 = 6m / S ^ 2 s = (4 + 10) * 1 / 2 = 7M If it is negative: a '= - 10-4 / 1 = - 14m / S ^ 2 s' = (4-10) * 1 / 2 = 3M Then the initial velocity direction should be negative, There are four answers? I thought about it for a long time
16. When the elevator is stationary, it begins to rise at a constant acceleration of A1 for 2 seconds, then decelerates at a constant acceleration of A2, and stops after 3 seconds. It is known that the elevator's rising height is 15 m. The maximum speed and the accelerations of A1 and A2 in the process of movement are calculated
17. The mass of a person is 50kg, which rises vertically by 10m with the acceleration of the elevator a = 2m / s B man's mechanical energy increased by 5000j C lift works 6000J to human D man's kinetic energy increased by 1000 J
18. A and B blocks with gravity GA = 4N and GB = 6N (a on B) are relatively stationary in the elevator with uniform deceleration, and the acceleration is 2m / s and 178; Why is the elasticity of object B just a supporting force, not a pressure The answer is support
19. When comparing the magnitude of acceleration, the sign indicates the direction, but if there is - 2m|s and#178; and - 5m|s and#178;, it is the absolute value, right
20. From - 2m / s to - 5m / s, does the acceleration increase or decrease