What's the image of uniform velocity linear motion

What's the image of uniform velocity linear motion

For A-T image, it is a parameter function, and a = R, a is not equal to 0, but a can be negative, that is, opposite to the direction of motion. For S-T image, it is a quadratic function, because s = v0t + (1 / 2) at ^ 2, the positive and negative determinants of A

On another kind of uniform variable speed linear motion
In history, some scientists have called the one-way linear motion with equal velocity change in the same displacement as "uniform variable speed linear motion" (now called "alternative uniform variable speed linear motion"). The "alternative acceleration" is defined as a = (VT VO) / s, where VO and VT represent the initial velocity and final velocity in a certain displacement s respectively. A > 0 means that the object is accelerating, A0 and remains unchanged, then a gradually increases
C. If a is constant, the velocity of the object in the middle position is (VO + VT) / 2
D. If a > 0 and remains unchanged, the velocity of the object in the middle is less than (VT + VO) / 2
BCD, seek detailed explanation

It is easy to calculate the answer C. at 1 / 2S, the velocity of the object must be V0 + A * s * 1 / 2 = V0 + (vt-v0) / 2 = (V0 + VT) / 2
As for the answer A., B. and D. let V0 = 0, then the object accelerates from rest, because s is constant and V is increasing all the time, so the T required for every s is decreasing, so if a is constant and greater than zero, then a must gradually increase. Similarly, because a > 0 and is constant, a increases, so it is easy to understand that at the middle of time, the displacement of the object is less than S / 2, Because the velocity of the back t / 2 is greater than that of the front T / 2, the velocity of the object must be less than (V0 + VT) / 2 at the middle of time

Can the displacement of non-uniform linear motion be calculated by the area of V-T image?
Can the area of V-T image be used to represent the displacement of non-uniform linear motion?

Can Di, V-T image, the area represents displacement, direction!

The meaning of slope in Physics
For example, what is the meaning of slope in the st image of uniform linear motion
What is the meaning of slope in the UI image of Ohm's law
In the origin of density, what is the meaning of slope in MV image
In the origin of G, what is the meaning of slope in GM image
In a word, I don't understand what slope is

1. Speed
2.R
3. Rou (density)
4.g
In fact, the slope is Y-axis / x-axis
For example, question 1
S / T is the slope, S / T is the velocity

The relationship between the velocity and displacement of the uniform speed change linear motion
① When an object moves in a straight line at a uniform speed, the initial velocity is 10m / s, and the acceleration is 1m / S ^ 2, the average acceleration of the object in the first second before stopping is 0_____
② The acceleration of a = 4 m / S ^ 2 can be obtained when the vehicle is braked in case of obstacles, and the distance of the vehicle within 3 s after braked is_____
③ After braking, the car moves in a straight line with uniform deceleration, and stops moving after 3S. Then, the ratio of the displacement of the car passing through in the three consecutive 1s is_____
The first question is to do even deceleration exercise, oh, sorry to write wrong. trouble
The first is 0.5m/s, the second is 90m, and the third is 5:3:1

So the last second displacement x = 1 / 2at & sup2; = 1 / 2 * 1 * 1 & sup2; = 1 / 2 v = x / T = 1 / 2 / 1 = 1 / 2M2, 36km = 10m / s, t stop = 10 / 4 = 2.5s, so the car can only walk for 2.5 seconds X1 = vt-1 / 2at & sup2; = 10 * 2.5-1 / 2 * 4 * 2.5 & sup2

The problem of displacement and velocity in linear motion with uniform speed change
When the motorcycle starts to move on the straight road, the acceleration is 1.6m/s2, then it moves at a constant speed, and then decelerates. The acceleration is -6.4m/s2 until it stops. It takes 130s and the journey is 1600m. Q: what is the maximum speed of the motorcycle?

Draw a V-T diagram
The area of trapezoid is displacement 1600, 1.6 and 6.4 are the slopes of two base angles, and the time is 130
According to the geometric relation, [130 - (H / 1.6 + H / 6.4) + 130] * H / 2 = 1600
The high h in the figure is the maximum velocity v
V=12.8m/s

The relationship between the velocity and displacement of the uniform speed linear motion in the first grade physics of senior high school
What is the ratio of the displacement of an object on the inclined plane to that on the horizontal plane?

The maximum velocity of the two segments is the same, and the average velocity of the two segments is v / 2, so the displacement X1 = V / 2 × T1 = VT1 / 2, X2 = V / 2 × T2 = vt2 / 2, and the displacement ratio X1: x2 = T1: T2 = 3:9 = 1:3

The relationship between the velocity and displacement of the uniform speed change linear motion
If the car runs at a constant speed of 10m / s and moves in a straight line with uniform deceleration after braking, what is the displacement of riding within 6S after braking if the car's braking time is 6.25m [braking time exceeds 2S]?

Criticize the landlord. There are so many typos. Fortunately, I can understand them
In the second second second s = (V-A) t-at ^ 2 / 2 = 10-a-a / 2 = 6.25m, the solution a is 2.5m
2as=v^2
The total braking distance is about 20 meters
Time: T = V / a = 4 seconds
(the car stops in 4 seconds, and there is no displacement any more, so the displacement within 6 seconds is also 4 seconds.)

Physical problems (the relationship between displacement and time of uniform velocity linear motion)
On a straight road, the speed of a car is 15m / s, and it starts to brake from a certain moment. Under the action of resistance, the car moves at an acceleration of 2m / s. how far does the last car leave the starting braking point 10 s after braking?
Wrong title
2m / s to 2m / S
I don't understand why the acceleration is 2m / s instead of - 2m / s. isn't that braking? Isn't that deceleration? It should be - 2m / S! - -
Isn't that slowing down? It should be - 2m / S! - -

This is positive in the direction of acceleration
There are: under the action of resistance

The relationship between displacement and time of uniform speed linear motion in senior one physics is solved!
After walking for 12 hours, I found that there were still passengers who didn't come up, so I immediately slowed down to stop. It took 20 seconds for the car to start and stop, and it took 5 m to get the maximum speed of the car
----Our teacher told me to use V-T image to solve the problem, but after drawing it, we can see it for a long time and calculate the formula for a long time, but there are still unknown quantities, which can not be solved by displacement velocity formula, nor can we set up equations------

After the v-tl image is drawn, the area of the triangle is the displacement within 20s, the bottom of the triangle is 20s, and the height is the maximum velocity
(v/2)t=x
The result is: v = 2x / T = 0.5m/s