When an object moves in a straight line with constant speed change, its speed when it passes through the middle of the whole journey is V1, and its speed at the middle of the whole journey is v2?

When an object moves in a straight line with constant speed change, its speed when it passes through the middle of the whole journey is V1, and its speed at the middle of the whole journey is v2?

Let V0 be the initial velocity, VT be the final velocity, and the whole journey time be t
V1=（V0+Vt）T/2
V2=(V0+Vt)/2
In general, V1 is greater than v2

When an object moves from point a to point B in a straight line with constant velocity, if the velocity at the midpoint of a and B is V1 and the velocity at the middle of the process is V2, then the object ()
When an object moves from point a to point B in a straight line with constant velocity, if the velocity at the midpoint of a and B is V1 and the velocity at the middle of the process is V2, then the object ()
(A) V1 > V2 (b) for uniform acceleration, V1 > V2 (b) for uniform deceleration

In the case of uniform acceleration: V1 > V2
In uniform deceleration: V1 > V2
In uniform acceleration, the average speed of the first half of the journey is slow, and the second half is fast. Therefore, the first half of the journey takes longer than the second half of the journey, that is, 1 / 2 of the time should be before the midpoint, at this time v2

For an object moving in a straight line with uniform speed change, the velocity at the middle position is V2 in a certain period of time, and the velocity at the middle time is v2. It is proved that the greater the acceleration is, the greater the absolute value v2-v1 is
V1 = (VO + at + V0) / 2, V2 = root sign V0 ^ 2 + (at + VO) ^, which can be approximately regarded as root sign (at + VO) ^ 2 / 2, then the absolute value v2-v1 is equivalent to (v2-v1) ^ 2 = root sign 2 multiplied by (at + VO) / 2 - (VO + at + V0) / 2, and the function is monotonic increasing without looking at the constant, so it is proved! Can we do this?

Look at the functional relationship between the absolute value v2-v1 and a, a is regarded as an independent variable
It's unnecessary to approximate that step. You can complete the whole formula and do it by derivation

A hare was playing on the grass at a distance of 200m from the cave S1, and was chased by the hound at the maximum speed. When the rabbit found the hound, it was 60m away from the hound, and the rabbit immediately ran to the cave, Then at least what is the acceleration of the hare to ensure safe return to the cave?
My question is that the hound time is 260 / 10 = 26S, the rabbit is 200m away from the hole, the speed is 8m / s, and the time is 200 / 8 = 25s. Why accelerate?

It takes time for a rabbit to start at 8 m / s. at this time, the Hound is already chasing it. If the acceleration is 1, it takes 8 seconds to accelerate to 8 m / s, and the hound can't escape being caught by the hound. If the acceleration is 8, it takes only one second to accelerate to 8 m / s, and the hound can't catch up with it