A calculation problem of acceleration in the first year of physics urgent, urgent!!!!

Uniform acceleration motion displacement of train: s1 = 1 2 a1t2 = 1 2 1 30 m = 450 m

The speed at the end of the uniform acceleration motion of the train, that is, the constant velocity velocity v = A1T1 = 1 30 m s = 30 m s

Uniform deceleration motion displacement of train: s3 = vt3 -1 2 a2 t3 = 30 20 m - 1 2 m = 300 m

Uniform motion displacement of the train: s2 = 3000 m - 450 m - 300 m = 2250 m

The total time of the train is: t = t1 + t2 + t3 = 30s + 75s + 20s = 125s

v=a1t1 then v=30m s s1=1 2a1t1 squared then s1=450m

Let the time that t2 elapses with a uniform velocity s2=vt2

V=A2T3 then T3=20S S3=Vt3-1 2A2T3 Square then S3=300m then S2=3000-S1-S3=2250M

then s2=vt2 then t2=75s then t=t1+t2+t3=125s

You missed 2 words, the front is a uniform acceleration, and the back is a uniform deceleration.

The acceleration of a1=1m s is a uniform linear motion, and after t1=30s, it is changed to a uniform linear motion.

Then the speed that can be achieved is v=a1t1=1 30=30m, and the displacement is s1=1 2at, 2=1, 2, 1, 30, 30=450m, and the acceleration of the magnitude a2= moves in a uniform linear motion until it stops.

Then the time taken to stop t2 = v a2 = 30

The last two are concise, clear, and very good.

That's a mistake! Will there be a uniform motion if there is acceleration?

Constant motion time t2 The time of deceleration is t3

The velocity at the beginning of the deceleration v=a1t1==1-30=30ms The displacement of acceleration s1=att 2=1*30*30 2=450m.

The displacement of a constant velocity s2 = vt2 = 30t2

vt-v0=at

0-30=a2*t2 a2 deceleration is negative.

t2=30 seconds.

Displacement s3 = vt + a2tt 2 = 30 * meters.

The displacement of uniform motion s2 = 3000 - s1 - s2 = 2250 meters.

Time t2 = 2250 30 = 75 seconds.

Total time = 30 + 75 + 20 = 125 seconds.

The first 30s walked 450m

Finally, it slowed down and walked 300m

3000-450-300=2250m

The middle time distance is 2250, the speed is 30m s, and the time is 75s30+75+20=125s

This is reasoning in accordance with the process, directly with the formula.

When the truck catches up with the car, the displacement of both is equal and the travel time is equal v vapor t = at

t=2v, steam a=2*20ms, 1ms=40sv, goods=at=40ms

At the beginning, the starting speed of the truck is smaller than that of the car, and the distance between the two is getting larger and larger, and when the speed of the truck increases to be larger than the car, the distance between the two is getting smaller and smaller, and the distance is the largest when the speed of the two is the same.

The time taken by the truck to reach the speed of the car is 20m s t1 = v a = 20 1 = 20s distance = car displacement - truck displacement.

vt1 - at1²

How long does it take for a 200m truck to catch up with a car? What is the speed of the truck at this point?

Truck displacement = vehicle displacement + d

After the train starts at the intersection, how long does it take for the two cars to be furthest apart before catching up with the car? What is the distance at this point?

The conditions for reaching the maximum distance remain the same.

The third question is to do it yourself.

(1) So solution time = 40 seconds.

The speed of the truck is 40*1=40 meters per second.

2) When they are the furthest apart, the velocity of the two is equal.

Then 1*t=20

t = 20 seconds.

Distance = 20*meters.

3) The answers are: 50 seconds 50 meters seconds 20 seconds 450 meters.

A small disc rests on the tablecloth, on the horizontal table of one table. The back side of the tablecloth coincides with the ab side of the table, as shown in Figure 14. The kinetic friction factor between the disc and the tablecloth is known to be 1, and the kinetic friction factor between the disc and the tabletop is 2.

The tablecloth is suddenly pulled away from the table with a constant acceleration a, and the acceleration direction is horizontal and perpendicular to the AB edge. If the disc does not fall off the table at the end, what is the condition for acceleration a? (g for gravitational acceleration).

As shown in Figure 6, a person with mass m stands on the scale, uses a rope across the fixed pulley, lowers the object of mass m from a height, and when the object accelerates down at a (a g, the reading of the scale is (

An object moves in a uniform deceleration straight line 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 knows v0=7

After that 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=

Use the formula: vt 2-vo 2=2as

In the problem, vt=50m s, a=5m s 2, s=100m, find vo=?

Substituting the formula is: 50 2-vo 2=2*5*100vo 2=1500

VO = 10 roots 15 m s

- -Generating a maximum speed of 5m s is a bit outrageous It's acceleration, right?

You see, if you want to accelerate from initial velocity to 50m s in 100m displacement, in 5m s, wouldn't it be nice to apply the formula?

According to the velocity displacement formula. v end -vo =2axvo =2500-1000=1500

VO = 10 times the root number under 15m s

It is assumed that the muzzle velocity is VO and the take-off speed is VT

There is the formula vt 2 = vo 2 + 2as

That's 50*50=VO 2+2*5*100

It gives vo = 10 * 15 under the root number

When accelerating, the time required to accelerate to 90m s is t 90 4 = the required runway length is at 2 2 = 2*

The time from braking to stopping t1 = 90 5 = 18s

The required runway length is 90*18-(5*18 2) 2=1620--810=810m

So the runway length is designed at least.

The time of the acceleration phase is t1, then v a*t1 gives 90 4*t1 , t1 seconds.

The deceleration phase time is t2, and from v a2*t2, 90 5*t2 and t2 are 18 seconds.

The average speed in the acceleration phase is (0 90) 2 45 m s, and the average speed in the deceleration phase is (90 0) 2 45 m s, so the average speed of the whole process must be v flat 45m s, and the length sought is at least (v flat) * (t1 + t2) = 45 * (m.

Solution: With a = square constant acceleration, the required displacement from 0 to 90m s is s1 = v at the beginning of t t + at*t 2

And v initial = 0, t = 90 4 = seconds to calculate the constant acceleration of s1 = 4 * meters with a = square, the displacement required from 90 to 0 m s is s2 = v initial t + at*t 2

At this time, v = 90, a = -5, t = 90 5 = 18 seconds Calculate s2 = 90 * 18 - 5 * 18 * 18 2 = 810 meters.

So the runway is at least s=s1+s2=m.

First, calculate the displacement s1 when accelerating from rest to v=90m s, and obtain s1=from 2a1s1=v 2

The displacement s2 from v=90m s to 0 is then calculated

From 2a2s2=0-v 2, s2=810m, so the total runway length s=s1+s2=

Of course, this is a theoretical value, and the runway may be designed to 2000m when it is really designed, and we only need to answer the question now!

If you don't understand, ask me ..

The length in the acceleration phase is obtained by the square 2a of s=v.

The length of the deceleration phase is 810m from the square of 2a of s=v

So the least few total lengths are the sum of them.

It is known that the initial velocity v0 = 72 km h = 20 m s, and the acceleration a = 4 m s 2.

v=v0-at=12m/s

2.It takes time for the speed of the car to decrease to 0.

t2=v0/a=5s

So the displacement within 10s is the displacement within 5s.

Substituting the data from v0 2=2as results in s=50

3.The distance traveled in the reaction time is .

x=v0t3=10m

The total displacement is s+x=60m

Just with the formula, how simple.

vt=vo+at

s=vot-1/2at^2

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)