A proof question about acceleration in high school physics...

The particle first does a uniform acceleration linear motion, then a uniform deceleration linear motion, and then a reverse uniform acceleration linear motion.

In a uniform acceleration linear motion, s1=1 2a1t reaches velocity v=a1t

In the linear motion of uniform deceleration, the reverse consideration is considered from the time of stopping to the speed of v, and the elapsed time t1 is considered

S2=1 2A2*T1 is also the velocity reached, v=A2T1, then A1T=A2T1

The displacement elapsed in the linear motion of reverse uniform acceleration is the total displacement s=s1+s2Elapsed time t2

and s=1 2a2*t2 t=t1+t2

So there is the equation 1 2a2*t2 =1 2a1t +1 2a2*t1 (substituting s1, s2).

a2*(t-t1) = a1t +a2*t1 (2 sides simultaneously remove "1 2").

a2t -2*a2*tt1+a2*t1 =a1t +a2*t1 (open the brackets).

a2t -2*a2*tt1=a1t (subtract a2*t1 from both sides of the equal sign at the same time).

a2t-2*a2*t1=a1t (equal sign is divided by one t on both sides).

a2t-2*a1t=a1t (replaced by the previous a1t=a2t1).

Get a2t=3*a1t so a2=3a1

a1 a2 t1 t2 made my head big, and I wrote it on scratch paper, and I fainted, and I couldn't mark the small corner, which was really depressing.

Proof that the particle P starts from rest and returns to the starting point, and the displacement magnitude is the same in the two periods, and the direction is opposite, then.

The displacement at uniform acceleration x1 = 1 2 * a1 * t 2

The velocity at this point is v=a1*t

The displacement at uniform deceleration x2=v*t-1 2*a2*t 2=a1*t*t-1 2*a2*t 2

Because x1=x2 then.

1 2*a1*t 2=-(a1*t*t-1 2*a2*t 2). 1/2*a2*t^2=3/2*a1*t^2a2=3a1

The displacement of the whole process is 0

1/2a1t^2+a1t*t-1/2a2t^2=03/2a1t^2=1/2a2t^2

So a3=3a1

The velocity when the acceleration becomes a2: v1

v1=a1*t (1)

The displacement of the acceleration before it becomes v2: s1

s1=1/2(a1*t*t) (2)

The displacement from the time it becomes a2 to the moment when the velocity becomes zero: s2=v1t-1 2(a2*t1*t1)(3).

The displacement from the second velocity of the object to the return to the origin s3=1 2(a2*t2*t2)(4).

t1+t2=t(5)

s3=s1+s2 (6)

Substitute (1), (2), (3), (4), and (5) into (6).

a2=3a1

With the velocity image, it is proved at once.

1.Let the maximum velocity be v, then the average velocity of the descent is (0+v) 2, and the descent time is 4 (v 2)=8 v

In the same way, the time for horizontal sliding is 12 V

The total time is 8 v+12 v=20 v=10

Then v=2 slides down time is 4s

The horizontal glide time is 6s

Acceleration on slope = (2-0) 4=

Acceleration on the horizontal plane = (0-2) 6 = 1 3 (m s 2)2 The problem is to solve the time taken by the rod to descend from 5m to 20m, according to the free fall formula.

s=So when you descend 5 m, the time is 1s (let g=10) and when you descend 20m, the time is 2s

The difference between the two is 1s

3.Same as 2.

That is, the time required to solve the motion from 200m to 450m is only changed to 1, and when the front of the car just reaches the bridge, t=20s

When the rear of the car leaves the rear of the bridge, t=30s

It takes 10s to cross the bridge

1).According to the formula h=1 2gt 2, t=5s can be obtained. Then it takes 5 seconds for the ball to fall from the wellhead to the bottom, and when the sixth ball just falls from the wellhead, the first ball just reaches the bottom of the well. So there are 5 balls in the middle, so the time interval is 1s

2).The third ball falls for 3 seconds, h3 = 1 2gt 2h3 = 45m

3).The second ball falls for 4 seconds, h2 = 1 2gt 2h2 = 80m

h=h2-h3=35m.

1) Let the time interval be t seconds, and when the sixth ball falls from the wellhead, the falling time with a small ball is 5t

then t=1 second.

2) The third ball falls for 3t

then the height of the fall is meters.

3) The second ball falls for 4t

then the height of the fall is meters.

The spacing is 80-45 = 35 meters.

1）h=1/2gt^2

t=5s The first ball reaches the bottom of the well, and the sixth ball just falls from the inlet.

There are five time intervals between the 1st and the 6th.

That is, the time interval is 1s

2) The third ball falls for 3s

h1=1/2gt^2=45m

3) The second ball falls for 4s

h2=1/2gt^2=80m

The distance δH = H2 - H1 = 35m

This kind of problem can be drawn to show the intention analysis, and it is enough to figure out the actual time of the movement of a small ball.

When stopping at the top of the inclined plane, vt=0

First 30m: VO 4) 2-VO 2=2ASO=2*30A=30*(2A).

15/16)vo^2=-30*(2a)

vo^2=-32*(2a)

Full journey: 0-vo 2=(2a)s

Slope length s=-vo 2 (2a)=-32(2a)] 2a)=32m

Judging from the title, it is a uniform deceleration motion. So the acceleration can be calculated from the last segment;

1) Think of it as a uniform acceleration motion in the opposite direction, then there is, s=1 2at2 (squared).

a=2s/t2=

So the acceleration is "-1" meters per square second.

2) Calculate the initial velocity based on the first segment.

tvt+1/2at2=

vt=7m/s

So the time is vt+at=0 t=7 1=7ss=tvt+1 2at2=7*7-1 2*17*7=

Uniform deceleration linear motion, it is known that the displacement passed in the first 3 s and the last 3 s is and the acceleration is a, respectively

The first three seconds: x=1 2at2,, so, a=11 3m s2 let the total time be t

1/2at^2-1/2a(t-3)^2=

a/2*(6t-9)=

11/3)/2*(6t-9)=

t=21/11

That is, the total time is 21-11 seconds.

Total displacement x=1 2at 2=1 2*11 3*(21 11) 2=147 22m

The square of v(b) = 2a1x1

The square of v(b) = 2a2x2

x2 is in turn equal to 2x1

Summing up the above equation gives us that the square of v(b) is equal to 2a1x1

The square of v(b) is equal to 4a(2)x1

Dividing the two equations yields a1:a2 equal to 2:1

Because the velocity is equal when point B is reached.

x1=1 2a1t1 2 vb=a1t1 so vb=a1 x1 a1

x2=1 2a2t 2 vb=a2t2 so vb=a2 x2a1 again because x2=2x1

Combine the above three forms.

a1:a2= x1 a1: 2 times x1 a1a1:a2=1: 2