Physics acceleration questions! Please, everybody!

I don't know formulas, so I'll use kanji instead!

In this 1s hour, the square of the final velocity minus the square of the initial velocity 2as, a 2m s, s=15m, which is Equation 1; Furthermore, the average velocity is used to form the second equation: the average velocity displacement s is at time 1s, and the average velocity is equal to one-half of the sum of the initial velocity and the final velocity. Solving this binary equation gives that the initial velocity is equal to 14m s, and the final velocity is equal to 16m s, so all problems can be easily solved, using the basic formula, the velocity of the object at the beginning of this 1s is 14m s, the object has been moving for 7s before this 1s, and the object has passed 49m before this 1s!

How's that, plus points!

To put it simply, in this second, the average velocity is 15m s, because of the uniform acceleration of linear motion, and the Gaussian fourth grade algorithm, the average velocity is the velocity in the middle of time in the acceleration process, then the initial velocity is 15m s-2m s*1 2=14

From stationary to 14m s, the velocity is increased by 14m s per second by 2m s, then it takes 7 seconds.

The distance is s=vt, this is constant, but the speed is not necessarily, then use the average speed * time, time knows, the key is the average speed. Same as above, the velocity is 2m s*7 2s=7m s

I just went to high school 1

Let the velocity at t seconds be 2t at t seconds and 2 (t+1) at t+1 second According to the V-T image, (2t+2t+2) 2=15 t=7, then the velocity at the beginning of 1 second is 2t=15m s 1 second ago and 7 seconds after 7 seconds. The distance that has been moved before 1 second is measured by the v-t image, s=(7+14)2=

1.By the formula v0t+1 2at 2=15

v0=(15-1/2*2*1)*1=14m/s2.Because at the beginning of the stationary initial velocity is 0

So t=v0 a=14 2=7s

Let the velocity at the beginning of this second be va, and the velocity at the end of this second is vbvb 2-va 2) 2as=1

vb=va+at

a=2 t=1 s=15

vb=16m/s

2.The previous exercise time was:

to=14/a=14/2=7s

If the initial velocity of 1s is V, then the velocity after 1s is V+A

v=14m s from (v+v+a)*1*(1 2)=15 of the image method

And because v0=0 v=at --t=7s

by s = (1 2)at 2 = 49m

An object moves in a uniform deceleration linear motion under the action of frictional force, the initial velocity is 12m s, the acceleration is 2 meters per square second, the total motion time t=(v-v0) a=(0-12) (-2)s=6s, and the total displacement x=((vo+v) 2)*t=6*6m=36m, which is a uniform acceleration motion with an initial velocity of 0.

The displacement ratio is 1:2:3....Namely.

36/(1+2+3)=6m

Illustrates that the displacement of the last 1s is 6m, so after that it is no longer in motion and the displacement is 0

Time t, velocity v(t), displacement s(t), t=0 at t=12, s(0)=0, acceleration a=2m s 2.

There are v(t) = 12-2t, s(t) = v(0)·t-1 2·a·t 2=12t-t 2.

Let the displacement in t (t+1) be 6 at a certain time, then s(t+1)-s(t)=12(t+1)-(t+1) 2-(12t-t 2)=11-2t=6,t=,t+1=

s(t=6)=36,s(t=

The displacement that can also be moved is s(t=6)-s(t=

The drawing method is also very simple, multi-hands-on, and it is super simple after combining a lot of physics and mathematical geometry.

s=v0t+1/2 at*t

The displacement in a second is 6m: t=1, and a=-2 is known as v0=7

After this second, the velocity v=v0-at=7-2=5

Consists of: v(t)*v(t)-v*v=2as

Because at the end of the rest, v(t)=0

That is: 0-25=-2*2s

It is possible to solve s=25 4=

1.Before a train enters the station, close the air valve and slow down. When the train has taxied for 300 meters, the speed is reduced to half of what it was when the valve was closed, and then it continues to slide for another 20 seconds and stops at the station.

Assuming that the acceleration of the car remains constant during the coasting process, we find: (1) the total displacement of the coasting from the time the car closes the valve to the time when the car stops coasting? (2) How fast is the car coasting?

3) How fast is the car when the valve is closed?

From the definition of "uniform (decreasing) velocity", it can be seen that the variable of velocity is equal at equal time intervals. As a result, the time taken to reduce the speed to half is equal to that of halving to zero, so the total time for the train to coast is 40 seconds. Uniform acceleration motion From the start of the motion timing, the ratio of the distance passed in each adjacent same time interval is 1:

3:5：……The odd ratio of consecutive numbers), and the uniform deceleration motion is timed forward from the last stop.

It is known that the distance of the last part of the train is 100 meters, and the total distance of the taxi is 400 meters. The second question is to find the average speed of the slide, right? From the above, it can be seen that the average gliding speed is 400 meters 40 seconds = 10 meters seconds.

The uniform acceleration motion should start at half the average speed from start to stop, so the speed when closing the valve should be 20 m seconds.

2.On a horizontal straight track there are two trains A and B with a distance of x. It is known that car A does a uniform deceleration linear motion with an initial velocity of v0 and an acceleration of 2a behind it; At the same time, car B moves in a uniform linear motion with an initial velocity of 0 and an acceleration of a.

The two cars move in the same direction, in order to make the two cars not collide, what conditions should be met to find the initial velocity v0 of car A?

Equation at 2 2+x=v0t-(2a)t 2 2 without a solution, there will be no collision, sort out the equation, use the condition that the discriminant formula of the root is less than zero, that is, v0 2-6ax<0, and solve: - root number (6ax) 3The highway brings convenience to people, but because of the high speed of vehicles driving on the highway, there are often car accidents in which more than a dozen vehicles collide in a row on foggy days. The normal speed of the car on the Shanghai-Nanjing Expressway is 120km h, and the maximum acceleration generated by braking is 8m s2.

If there is fog on a given day and the visibility is about 37 meters, what is the speed limit for the car to drive safely? (Assuming the driver's reaction time is seconds).

The maximum acceleration generated by braking should be -8m s2). vt 2-v0 2=2as, here the last stop, vt=0, so there is s=-v0 2 (2a), that is, the gliding distance of the initial velocity v0 is -v0 2 (2a), let the maximum velocity be v, solve the equation: <37 (a=-8m/s2)

A pair The velocity of the object that does high-speed linear motion is very large, and the acceleration is 0b to the locomotive when it is just started, the velocity is 0Acceleration is not 0c wrong When the velocity is reversed, the velocity keeps increasing.

dError As long as the acceleration is in the same direction as the velocity, the velocity keeps increasing.

a.Correct, it can meet the velocity is very large, but the resultant external force is 0, and the acceleration is 0bTrue, the resultant force on an object with a velocity of 0 is not necessarily 0, so the acceleration may not be 0

c.Correct, the acceleration is negative, and the velocity cannot increase, but only decreases continuously (note that the velocity is a vector) dError, Acceleration》0, Acceleration.

If the acceleration < 0, it will be decelerated.

For the first three seconds, s = 1 2at squared, and the displacement is 9The displacement of the last two seconds becomes 6, so the displacement of the last two seconds is -3, and the acceleration is -3 2 by the formula just now, and the velocity at the end of the four seconds is v=at, so the velocity is -3 2.

The first 3s displacement can be calculated as 9

If the displacement is six after two seconds of returning, then the displacement is 3 or 12 in two seconds, so the acceleration is or 6

A velocity of or 6 is the opposite of the initial velocity.

c is absolutely correct If the direction of flight is positive, then the initial velocity is +12m s and the final velocity is -24m s, so the acceleration is |Time = —360m s square in the opposite direction of the initial velocity.

When the ball comes to rest when it makes contact with the coach, it is kicked out in the same direction of acceleration as the kick is in the same direction as the whole process: (v2-v1) t=a

So: (-24-12) choose C

mv=fs=ma*t

If 12m s is positive, the velocity of 24m s is reversed, and the combined velocity is -24-12=-36m s

36m/s*m=ma*

a=-360

It is opposite to the direction of 12m s, and the same direction as the kick out.

If the original direction is positive, the final velocity is -24m s, and the velocity change is -24-12=-36m s

So a = velocity change t = -360m s 2

The direction is the same as the kick out direction, select C