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Is velocity and velocity a ratio of displacement to time and a ratio of distance to time
not a quarter
Velocity + velocity direction = velocity direction
In other words, the speed and the speed are the same, but the speed has a direction, but the speed does not
The ratio of displacement to time = average velocity, directional
Distance to time ratio = average speed, no direction
Attention is "average"
Why use the ratio of displacement to time instead of distance when calculating velocity Isn't the displacement of the object a directed line segment from the initial position of the particle to the final position? That is to say, the distance is greater than or equal to the displacement. When calculating the velocity, the ratio of displacement to time is used instead of the distance?
After entering high school, there is a new definition of speed. You should not take the previous feeling of speed directly to high school. You should take a closer look at the definition of speed, because the research problems have changed, and the definition has changed accordingly
Application of ratio: is the ratio of distance to time calculated as speed
No, it's velocity. Displacement divided by time is velocity
The ratio of distance to speed represents time This is a judgment question
Not necessarily. If it's uniform, it's right. If it's not, it's wrong
Is speed the ratio of distance to time
Yes
On the slope 10 m away from the bottom of the slope, a car runs up at a constant speed of 4 m / s, and then goes down at a speed of 2 m / s after 5 s. let the car move in a straight line with the downward direction along the slope as the positive direction. The starting point of the car is the origin of the sitting mark, and the starting time of the car is the starting point of timing. The x-t image and V-T of the car in 20s are made The position of the car at the end of 20s is determined by the image
In the first five seconds, the displacement of the trolley is: X1 = v1t1 = (- 4) × 5m = - 20m, and the direction is upward along the slope;
In the last 15 seconds, the displacement of the trolley is: x2 = v2t2 = 2 × 15m = 30m, and the direction is downward along the slope;
Taking the starting point of the car as the coordinate origin, the x-t image and V-T image of the car within 20 seconds are shown in the figure
It can be seen from the figure that the displacement of the trolley in the 20s is x = 10m, and the direction is downward along the slope, that is, the trolley is at the bottom of the slope at the end of 20s
A: the x-t and V-T images of the car in 20s are shown in the figure. It can be seen from the figure that the displacement of the car in the 20s is x = 10m, and the direction is downward along the slope, that is, the car is at the bottom of the slope at the end of the 20s
How is v ^ 2-V ^ 2 = 2aX derived? Is there any difference or connection between displacement and distance? Can you explain in detail how V ^ 2-V ^ 2 = 2aX is derived?
The formula is based on S = 1 / 2T (V1-V2) and then according to v = at, displacement and distance are two different concepts. Displacement refers to the linear distance from the starting position to the end position, and the distance is the length you pass. For example, if you go from home to school and then home, your starting position is your home, and your ending position is also your home
When an athlete runs along a circular track with a radius of 200m, when his displacement is 400m, his distance may be______ m. When his displacement is 200 At 2m, his distance may be______ .
When the athlete's displacement is 400m, it is exactly the diameter of the circular track, so the distance is half circle, that is, s = π, r = 3.14 × 200m = 628m
When his displacement is 200
At 2m, it can be seen from the geometric relationship that the athlete's movement may be 1
Four circles, maybe three
4 circumference, the distance may be: 1
4 × 2 π r = 314m or: 3
4×2πR＝942m
So the answer is: 628m, 314m or 942m
At the speed of u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u
The correspondent arrived at the head of the line at the speed u from the end of the team. The relative speed of the team was (U-V), time T1 = L / (U-V),
The speed of the relative team is (U + V), time T2 = L / (U + V),
During this period, the distance that the team moves forward s = VT = V (T1 + T2) = 2uvl (u ^ 2-V ^ 2)
When a particle moves 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 When the particle starts to move, (1) Maximum displacement in a few seconds (2) Maximum distance in a few seconds I answer: (1) the second second (2) Second second Why not? Wrong number. The form should be: When a particle moves 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 In addition, please write the maximum displacement and the maximum distance,
What second or the first few seconds should be confirmed?
Your answer is correct for the displacement distance in seconds
But the problem here is a few seconds,
The magnitude here is independent of the positive and negative half axis, only the value of displacement can be concerned
So it should be
The maximum displacement in the first four seconds is 7m,
The maximum distance within 5 seconds is 26m
RELATED INFORMATIONS
- 1. Vehicle a and vehicle B pass the same displacement along the straight road. Vehicle a moves at the speed of 40 km / h in the first half of displacement and 60 km / h in the second half of displacement; vehicle B moves at the speed of 40 km / h and 60 km / h in the second half of the displacement （ ） A. . V A= . V b B. . V a ＞ . V b C. . V a ＜ . V b D. Can't be sure
- 2. Is the average velocity equal to the instantaneous velocity in any period of time of uniform linear motion? Is that correct? There is another question—— Is the displacement equal to the distance in any time of uniform linear motion? Is that correct?
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