The linear motion of a particle, the kinematic equation of a particle s linear motion is, then the p

If it is a linear motion with uniform acceleration starting from rest, i.e., v0=0, then:

The instantaneous velocity at the end of each second when the particle moves uniformly and in a linear motion is (at), and the formula used is: v=v0+a*t

When the particle is uniformly accelerated and moves in a linear motion, it is within 1 second, 2 seconds, and 3 seconds. The ratio of displacements in n seconds (1:4:9:......n 2) The formula used is: s=v0*t+1 2*a*t 2

When the particle is uniformly accelerated and moves in a straight line, it passes 1 meter, 2 meters, and 3 meters. The ratio of the instantaneous velocity at n meters is ( 1: root number 2:

Root number 3: ......The formula used is: v 2-v0 2=2*a*s, v=2*a*s

When the particle is uniformly accelerated and moves in a straight line, it passes 1 meter, 2 meters, and 3 meters. The ratio of time to n meters is (1: root number 2: root number 3: ......The formula used is: s=v0*t+1 2*a*t 2 t=under the root number (2s a).

The kinematic equation of a point moving in a straight line is , then the person with the point is ().

a.Uniform acceleration in a straight line is moving, and the acceleration is in the positive direction of the x-axis.

b.Uniform acceleration in a linear motion with acceleration in the negative direction of the x-axis.

c.Linear motion with variable acceleration in which the acceleration is positive along the x-axis.

d.Variable acceleration linear motion with acceleration in the negative direction of the x-axis

Correct Answer: Variable acceleration in linear motion, acceleration in the negative direction along the x-axis

It's the first fill-in-the-blank question.

1、x=2t²-16t+5

v=dx/dt

4t-16a=4t=10s.

v=4×10-16

24m/sa=4m/s²

In the first second, the acceleration a=2m s does uniform variable speed linear motion, and in the second second, the acceleration a=-2m s does uniform variable speed linear motion, according to the conditions, the distance in each second is actually equal, and the speed at the beginning is considered to be 0, and the distance per second is obtained s=1 2*2*1*1=1m

The velocity at the end of the first second is the maximum, and the velocity at the end of the second becomes the initial velocity v=0, and so on, so that the velocity at the end of the odd number of seconds is the maximum, and the velocity at the end of the even number of seconds is the output velocity.

So when t=99s, the maximum velocity v=v+at=2m s(v=0)t=100s, total s=100s=100m

Of course, if the initial velocity v is not equal to 0.

v=v+a*1=v+2

S total = 100s = 100 (v * 1 + 1 2 * a * 1 * 1) = 100v + 100

v=0.

First, find the acceleration according to s=1 2at 2, where s=3mt=3s then a=2 3

m s 2 and then the ant segment continues to find the time it takes to carry the object through the first 4m by the formula s=1 2at2, where s=4a=2 3m s 2

then t = 2 times the root number 3

Left and right at the same time to find the derivative of time t, get.

a=2*s*v

v=1+s^2

So a=2*s*(1+s2).

4s^2+2s

b is the time point between the middle and simple times of a and c, and the velocity is equal to the average velocity of this period of ac, so there is vb=(l1+l2) 2

There is also l2-l1=at 2 (2 means 2 to the power of 2) from o to b there is vb 2=2a (s+l1).

The distance between O and A is: s=[(3L1-L2) 2] [8(L2-L1)].

Reference: Yanling Mountain Coarse.

Idea: According to the three elements of motion and known conditions, use the motion formula to list the equations and solve the equations

For the first 4 seconds, s1 = v0t + 1 2at squared, i.e.: 1 formula 24 = 4v0 + 8a, the velocity of the object at the end of 4 seconds, v4 = v0 + at meters rent banquet second noisy type preparation, for the second 4 seconds, s2 = v4t + 1 2at squared, i.e.: 2 formula 64 = v0t + at square rise and destroy + 1 2at squared, 1 and 2 formula are solved immediately.

Set the initial knowledge of Li speed v0 plus filial piety and fierce boy Qiaowang speed a

0-4 s1=(v0+v0+4a)2 4=244-8 s2=(v0+4a+v0+8a) 24=64 solve the system of equations to obtain v0=1m s a=

Let the initial velocity be v and the acceleration be a

v4+1 stool touch 2*a*4 0 5=24

v8+1/2*a*8�0�5=24+64

v=7 jujube finch talks about the year with 2m s a=5 4m s 0 5

By the formula s2-s1=a*t*t;

There is a=(s2-s1) t t=40 4 4=m s2) from the first 4 seconds of the average speed of Pega Qing v=s1 4=6m s) and average speed = the speed of the middle moment.

v0=v-a*t 2=6-5=1

m/s)