An object is uniformly accelerating with zero speed.

(1) It takes 2s to pass the 1 4m, so how long does it take to pass the 8th 4m?

(2) The displacement ratio of the first three consecutive actions is 1: 2: 3, so what is the time ratio required to pass these three actions? What is the ratio of average speed?

The second question answered upstairs is the segmented displacement from the initial position, not the continuous three-segment displacement.

Uniformly accelerated linear motion with initial velocity of 0, s = 1/2at 2,

(1)t=2, s=2a=4, so a = 2m/s 2,

Therefore, when s=7*4m, t1= 28 = 5.3s; Under the root symbol; When s=8*4m, t2= 32=5.6568s under the root sign,

It takes 5.6568-5.3 = 0.3568 s for an object to pass the 8th 4m;

(2) Let the time of three displacements be t 1, t2, t3, and the three displacements be s 1, s2, s3, respectively, and then there are the following three equations.

s 1= 1/2at 1^2

s2=at 1t2+ 1/2at2^2

s3=a(t 1+t2)t3+ 1/2at3^2

Assuming that s 1, s2 and s3 are 1.2.3 respectively, we can calculate the square root of t 1= 1, t2= root number 3- 1, T3 = 6- root number 3; T 1:t2:t3= 1: (root number 3- 1):(6- square root of root number 3)

Average speed v1:v2: v3 = (1/2at1): (at1+kloc-0//2at2): (a (t1+t2)+/kl.