High 1 physics urgency After the car brakes, it does a uniform deceleration and linear movement

v0 = 12m/s a=2m/s^2

Let the displacement within t=8s be x

After the car brakes, pass the T'=v0 a=6s stops moving, and the displacement in t=8s is t'= displacement within 6 s, at t'Average velocity in =6s v=v0 2=6m s, x=vt'=36m

Let the speed of the car be reduced to V'=6m s, the distance it has traveled is s,v0 2-v'^2=2as，s=27m

From the formula of uniform acceleration linear motion, it can be seen that x=v0t+1 2at is substituted into v0=12m s, a=-2m s, t=8s (note that a is reversed to velocity, so the v0 direction is regarded as the positive direction, then a=-2m s).

The solution is x=32m

From the formula of uniform acceleration linear motion, it can be seen that v end -v beginning = 2ax is substituted by v beginning = v0 = 12m s, v end = 6m s, a = -2m s (note that a is reversed to velocity, so the v0 direction is regarded as the positive direction, then a = -2m s).

Solution: x=27m

It takes 12 seconds to slow down from 12m s to stop, not 8 seconds.

2as=0^2-12^2

4s=-144

s=36(m)

In the same way: 2as = 6 2-12 2

s=27(m)

The velocity formula vt=vo-at shows that the car stops moving at the end of 6s, so the displacement within 8s can only be substituted into the displacement formula for 6s, and the displacement is 36m. 27m based on VT2=VO2-2AS.

1.The time to decelerate to 0 is v0 a=6s, and it doesn't take 8 seconds, so s=m.

2.The time it takes to decelerate to 6m s is (12-6) 2 = 3s, so s = v0*

The speed of the half-way sock is set to v

The average velocity of the first half of the attack v1=(vo+v) 2 and the average velocity of the second half v2=v 2

The ratio of the average Zen velocity of the first half to the second half v1 v2=[(vo+v) 2)] v 2)=(vo v)+1

vo^2=2as

v^2=2a(s/2)

vo v = root number 2

v1 v2 = 1 + root number 2

Hehe, draw the velocity, the time axis can clearly see the answer, the ratio of the small triangle row of 3 seconds and the medium triangle of 2,3 seconds and the large triangle of 1,2,3 seconds is 1 4 9, so the displacement ratio of 3,2,1 seconds is 1 4-1 9-4 = 1 3 5, then the displacement ratio of 3 consecutive one-second is 5 3 1

5:3:1 is a uniform acceleration motion with an initial velocity of 0, and according to s=1 2at 2, it can be judged to be 5:3:1

Uniform deceleration linear motion, which can be reversed, is a uniform acceleration linear motion.

The first half of the distance, 1 2a*t 2, the end speed a*t, the average speed 1 2at

The distance in the second half of the time is 1 2*a*(2t) 2=4at 2, the end velocity is a*2t, and the average speed is (1 2at+2at) 2=5 4at

The average speed ratio is 5 to 2

After the car brakes, it does a uniform deceleration linear motion until it comes to a standstill, which can be seen as a uniform acceleration linear motion with zero initial velocity, and according to the inference, it can be seen that the ratio of displacements in adjacent equal time is 1:3:

Shirt disturbance...., i.e.: x3x5

After the braking starts, the ratio of the line collapse or displacement in the 1st s and the 3rd s is 9:5

So the answer is: 9:5

Chuzai Lee has a speed of +

Acceleration - vt=0

vt-v0=at

0-10=-5t

t = 2 seconds. It stopped.

The displacement within 3s after braking is good, and the displacement of 2 seconds is s=v0t+att 2=10*2-5*2*2 2=10 meters = the second second post-braking displacement.

I think we can set t to be a second, divide the second into 7 time intervals, and then look at it in reverse, because the orange is a uniform acceleration motion with a velocity of 0, the ratio of the displacement in the same time interval is 1:3:5:

13, so the answer is (11+13):( missing band 7+9):(3+5)=3:

2:1, I hope to give a reply, I don't think it's right.

Reverse wax with observation, the car is from the stationary start to do uniform wheel reed acceleration linear motion, then s=a*t*t 2

In units, the displacement ratio is 1:4:9:

The displacement ratio of 49 is 1:3:5 per second

13 The flip side is 13:11:9:

The result of 1 is (13+11):(9+7):(5+3)=3:2:1