How to understand the meaning of time in the following scene (1) A lesson took 45 minutes (2) The students are on time for class at 7:00 (3) It takes three hours to climb the mountain (4) The train leaves at 13:55 From this we can get the definition of time and moment

How to understand the meaning of time in the following scene (1) A lesson took 45 minutes (2) The students are on time for class at 7:00 (3) It takes three hours to climb the mountain (4) The train leaves at 13:55 From this we can get the definition of time and moment

A class takes 45 minutes. There is a period of time in his class, not short, so it is time
The class starts at 7:00 on time. There is no interval between them. It's just a moment, so it's a moment
Mountaineering takes 3 hours. It takes a while from the beginning to the end, so it's time
The train starts at 13:55. The train starts at 13:55, and there is no interval, so it is the time. In short, the time is on the time axis, corresponding to a point, such as 8:05, which is the time;
Time is the time interval between two moments. On the time axis, it corresponds to a period
Time and position correspond, time and displacement correspond

A particle is on the x-axis. The position coordinates of each moment are as follows
t/s∶ 0 1 2 3 4 5
x/m：0 5 -4 -1 -7 1
(1) Maximum displacement in the first few seconds
(2) The displacement is the largest in seconds
(3) It's the longest distance in the first few seconds
Can you explain it clearly
I think you're wrong, 1L
I think the displacement is the largest in the first two seconds, because in two seconds, the position is 5 and the end position is - 4, so - 4-5 = - 9, because the absolute value is 9

The displacement refers to the change to the position of the initial displacement, with size and direction, while the distance has no direction. The longer the walking time is, the longer the distance is. The displacement in the first N seconds refers to the distance of the position when t = 0 and T = n, so the maximum position in the first 4 seconds is 7 m; the position in the nth second refers to the distance when t = n-1 and T = n, so the maximum displacement in the second second second second is 9 m; as for the distance, the maximum displacement in the first 5 seconds is 26 M

How to get rid of the two answers of calculating time by physical displacement formula in grade one of senior high school
The formula is x = VOT + 1 / 2at ^ 2, because t is the square, always calculate two numbers, how to exclude the other

Time cannot be negative

Several calculation problems about the relationship between displacement and time of uniform linear motion in senior one physics
1. According to the regulations of a certain city, the speed of a truck in the urban area shall not exceed 40km / h. one truck at a time stops on the urban road after emergency braking for 1.5s, and the length of the brake mark is 9m. Assuming that the truck decelerates evenly after braking, how much km / h is the speed? Is the truck against the regulations?
2. The car is moving along the straight road at the speed of V1 = 10m / s, and suddenly finds a bicycle moving in the same direction at the speed of V2 = 4m / s at the speed of S0 = 6m in front of it. The car immediately brakes and decelerates at the acceleration of a = - 5m / S & # 178; after t = 3S, how far is the distance between the car and the bicycle?
3. On a smooth horizontal plane, an object is moving at a speed of 4 m / s from the right at a constant speed. From a certain moment t = 0, it suddenly receives a horizontal force to the left, which makes the object move in a straight line at a constant speed with an acceleration of 2 m / s. find the displacement, velocity and average velocity of the object passing through t = 5 s. is the average velocity equal to (V0 + VT) / 2?
4. An object decelerates along a smooth slope at a speed of 20 m / s, and the acceleration is a = 5 m / S & #. If the slope is long enough, what is the distance the object may pass when the speed becomes 10 m / S?
5. When a car brakes, it can produce a maximum acceleration of 10m / s, when the driver finds that there is danger ahead, he can react in 0.7s and brake immediately. This time is called reaction time. If the car runs at a speed of 20m / s, what is the minimum distance between cars?
6. The bus starts from the station and runs at a speed of 4 m / s along the straight road at a constant speed. After 2 s, a motorcycle starts from the same station and accelerates to catch up with the bus at an acceleration of 3 M / s?

1. After stopping for 1.5 s, the average speed of braking process is 6 m / s if the length of brake mark is 9 M
The final velocity is zero, the average velocity is 6m / s, so the initial velocity is 12m / s
2. Notice that the car is stationary after 2 seconds of braking, so the moving distance of the car within 3 seconds is the total braking distance, the average speed of braking process is 5 MGS, the time is 2 seconds, and the displacement is S1 = 10 m
When bicycles move at a constant speed in 3S, the distance they travel is S2 = VT = 12m. At the beginning, they are 6m apart, so at last they are d = 2m + 6m = 8m apart
3. Take the right direction as the positive direction
v0=4m/s,a=-2m/s²
So VT = - 6m / S
At this time, the average velocity can still be used, and the average velocity vping = (4-6) / 2 = - 1m / s
So the displacement is - 5 m, that is, 5 m to the left
4. The time required for uniform deceleration to decrease from 20m / s to 10m / s along the inclined plane is t = △ V / a = 2S. The average velocity in this section is 15m / s, so the displacement is 30m
If it moves to 10m / S (i.e. - 10m / s) downward along the slope, the required time t '= 30 / 5 S = 6S
The average velocity is equal to + 5m / s, so the displacement is also 30m above the slope
(in fact, the positions of these two times are the same, which can be seen from the perspective of energy conservation.)
5. Before braking, the car keeps moving at a constant speed for 0.7s, and the displacement is S1 = VT = 14m
The braking time is t = △ V / a = 2S, and the average speed is vping = 20 / 2m / S = 10m / s
So the braking distance S2 = 20m
The total distance is S1 + S2 = 34m
6. Set the required time as t
Displacement of motorcycle S1 = at & # / 2
The displacement of the car S2 = V (T + 2) - (the car moved for 2S first)
When catching up, S1 = S2, we can solve t = 4S or T = - 4 / 3 s (rounding off)
So the distance from the starting point can be substituted into S1 or S2 to calculate S1 = S2 = 24m
That's all,