Uniform acceleration problem, uniform acceleration, uniform variable speed, how to understand variab

Solution: Let the distance between Oa be s, the acceleration of the object is a, the velocity of the object at point A is v, and the time taken to pass AB, BC, CD is t

then there is v 2=2as———

v+at)^2=2a(s+2)——

v+2at)^2=2a(s+2+3)——

v+3at) 2=2a(s+2+3+4)——The system of equations composed of the above four equations is solved.

s = 9 8 m.

A: The distance between the OAs is 9 8 meters.

Here's how to solve the system of equations:

v^2=2as———

v+at)^2=2a(s+2)——

v+2at)^2=2a(s+2+3)——

v+3at) 2=2a(s+2+3+4)——from , get:

2vt+at^2=4———

By , get:

2vt+2at^2=5———

By , get:

2vt+3at^2=5———

v^2=9a/4

Substitution can be solved:

s = 9 8 m.

You should have learned a theorem, right?

The ratio of the displacement of a uniformly accelerated motion with a stationary start in an equal time is .

x1:x2:x3...xn=1：2：3...n so oa = x:2:3:4

oa=1

The ratio of the displacement of a uniformly accelerated motion with a stationary start in an equal time is .

Uniform acceleration means that the acceleration a in the acceleration motion remains unchanged, that is, the acceleration remains unchanged and the velocity changes uniformly.

Uniform speed change means that the acceleration a in the acceleration motion is also changing uniformly, that is, the acceleration changes uniformly, and the change in velocity is not uniform.

Variable acceleration is an uneven change in acceleration a, that is, an uneven change in acceleration, a change in velocity, and of course even more uneven.

To distinguish between uniform acceleration and uniform deceleration, it depends on the acceleration.

direction, independent of speed.

If the direction of acceleration is the same as the specified positive direction, then it is a uniform acceleration, a>0. Conversely, for uniform deceleration, a<0.

In this problem, it is specified that the direction of the starting motion is the positive direction. The initial velocity is 5m s and the final velocity is -12m s, indicating that the acceleration direction is negative from 5m s to rest. From stationary to -12m s, although the velocity increases, the direction of acceleration remains the same and remains negative.

So the motion is a uniform deceleration motion because the direction of acceleration is always negative.

If the speed of "uniform acceleration" reaches the speed of "uniform speed", and the "uniform speed" has not caught up with the "uniform acceleration", then it is impossible to catch up. (Because after that, the speed of "uniform acceleration" will be greater than that of Patrick's "uniform speed").

Therefore, let the initial two objects be separated from each other, and the velocity of the object a moving at a uniform speed is V; The initial velocity of the uniformly accelerated moving object b is v0 and the acceleration is a

The time it takes for b to accelerate to a velocity v is t=(v-v0) a, and the distance traveled by a is s1=vt

b distance of movement s2 = (at 2) 2

Subtract the distance of B from the distance of A to get the distance of A more than B and the distance of the cluster of motion, if this distance is greater than or equal to their initial distance, it means that A has caught up with B; If it is less than, it means that it has not caught up, and after that, B's speed will be greater than A, and A will never be able to catch up with B. Namely:

If s1-s2>=s you can catch up.

s1-s2

Assuming that the two cars do not collide in each direction, the speed of A to B will be smaller than that of B. That's why we're all about inequality.

VA2=2AAAAA. VB2=2ABBClose grip SB.

Critical conditions are assumed to be sA-dSB.

sA = v A 2 car dust qing 2a A, s B = v B 2 2a B.

v A2A-D VB2A2AB.

So you have to meet such a relationship in order not to bump into it.

According to V=AT, T=V A=, according to V1=VO+AT=, the speed after the collision is halved as, after the collision, the velocity of ball A is greater than B, and the acceleration is smaller than B, so the velocity of B first becomes zero, and then B moves in the opposite direction, and the ball A first decelerates and then moves in the opposite direction, so when the velocity of ball B becomes zero, the distance between AB is the largest, according to V square-V0 square = 2ax, the distance of ball B is , when the velocity of ball A is , the distance is , so the maximum distance is.