There is a body that does a uniform acceleration of linear motion from the end of 3s

According to the displacement formula s=v0t+1 2at 2, 5v0+1 2a*5 2-3v0-1 2a*3 2=24 and 12v0+1 2a*12 2-8v0-1 2a*8 2=40, the simultaneous two-formula is easy to solve, a=-1 3m s 2, v0=40 3m s

Solution: Let the acceleration of the moving object be a and the initial velocity be v0;

From the end of 3s to the end of 5s, t1=5-3=2s 1.

Then: (v0+3a)*t1+1 2at1 2=24 2.

From the end of 8s to the end of 12s, t2=12-8=4s3.

Then: (v0+8a)*t2+1 2at2 2=40 4 formula.

Solved by the formula: a==-1 3m s 2....v0=40/3m/s ..

Use: s=(v0t+, within 5 seconds (v0*5+within seconds(v0*3+=24m.)

Within 12 seconds (v0*12+ -8 seconds (v0*8+ = 40m.)

The solution yields a=1 3m s*s, v0=32 3m s

Use the formula vt+1 2at =s

24=2v+2a

40=4v+8a

a=2 v=10 for solving the system of equations

Analysis: The object starts at rest with a muzzle velocity of 0

The displacement in the third second, the formula should be s=v2*t+1 2*a*t 2v2 is the terminal velocity of the second second, a*2

In the above formula, t is the third second, which is the third second, so t=1, s=3m are substituted into the above equation 3=2a+1 2a

The acceleration of this uniform acceleration motion is obtained: a=

The initial velocity is 0 and the displacement ratio is 1:3:5 so the total displacement is 27 5

s=v0t+1 2at.

So a=6 5

s 3 seconds before = 1 2a * 9

2 seconds before s = 1 2a * 4

s 3 seconds = s 3 seconds - s 2 seconds = 3a=

c, an object moves in a uniform accelerated linear motion, and its displacement in the 3rd s is 5m, according to this condition line, the equation for finding the displacement in the 3rd s is listed first.

The displacement of the previous 3 seconds minus the displacement of the previous 2 seconds: v*3+

Simplify. To the end V+

Then look at option C.

The displacement in the first five seconds is s=v*5+

Isn't it the above equation*5?

25m correct.

According to the inscription, the velocity is 5m s

5 This is the maximum acceleration, i.e., the initial velocity is 0

When the initial velocity is not zero, the acceleration decreases.

The speed at the end of 3s is 6m s

The displacement is 25m in 5s

The displacement within the 5th is 9m

So all right.

From the meaning of the title. x/（x+6）=3/7

Get x=9 2

Total displacement s=25 2 (the calculation process is very cumbersome, omitted).

C. Question analysis: Since the initial velocity of the uniform acceleration linear motion is unknown, knowing the displacement in the 3rd s, it is impossible to find the acceleration of the object, the displacement in the first 3 seconds and the displacement in the 5th s; However, the average velocity over the first 5 seconds can be found; The average velocity within the 3s is v=<>

m/s=5m/s；The speed is v1

5m/s；The time is the middle moment within 5s, so the average velocity in the first 5s is 5m s; Therefore, the displacement in the first 5s must be 5 5m=25m;

s2=v0*2+

v0+s3=v0*3+

v0+4a=

A = the 4S car balance.

s = m. Displacement of the 5th s.

s = m. si:sii:siii:siiii...1:3:5:7...

siiii=siii*5 Closed mind to do 7=4*5 high modification 7=20 7mRevise.