Acceleration physics problem, seeking detailed solutions, a physics problem about acceleration

When the EMU is working normally, the time taken to reach station B from station A is t. The average speed of the EMU during the acceleration and deceleration phases is half of 180 km h = 50 m s, i.e. v = 25 m s. The acceleration phase took time t1=v1 a1=50 s=100s, and the deceleration phase took time t2=v1 a2=50 s=100s.

When the train is 3 minutes late for some reason, the time taken from station A to station B is (T-180s). The average speed of the EMU during the acceleration and deceleration phases is half of 216 km h = 60 m s, that is, v'=30m/s。The acceleration phase took time t3=v2 a1=60 s=120s, and the deceleration phase took time t4=v2 a2=60 s=120s.

The distance between stations A and B is s=v(t1+t2)+v(t-t1-t2)=v'(t3+t4)+v(t-180-t3-t4) solution: t=1300s

The distance between stations A and B is s=56000m=56km

Suppose the time for the first acceleration to 50m s is t1, and the time for constant speed travel is t2. The second acceleration to 60m s is t3, and the constant speed travel time is t4.

The displacement of the second exercise is the same, s = displacement of constant speed driving 50t2 + displacement of acceleration motion 2 * (50 * 50-0) (2 * displacement of uniform speed driving 60t4 + displacement of acceleration motion 2 * (60 * 60-0) (2 *).

The relationship between t2 and t4 is 6t4-5t2=220

Then according to the time difference of 3 minutes, the relationship between t1, t2, t3 and t4 is obtained, and 2t1+t2-(2t3+t4)=180 is obtained

According to the velocity formula, t1, t3, 50=, 60= can be calculated, t1, t2, t3, t4 can be calculated, and the ab distance can be calculated.

v1=180km/h=50m/s v2=216km/h=60m/st1=100s t2=120s

The displacement of the acceleration phase is equal to the deceleration phase has:

S1 = T1 * V1 = 5000m S2 = T2 * V2 = 7200m The displacement traveled by arriving at B at the same time is the distance between A and B The second departure is 180s late Let the first time be TS and the second time be (T-180)s; (t-2*t1)*v1+s1=s=(t-180-2*t2)*v2+s2 Substituting data to solve the solution: s=60000m

Multiple movement combinations, convocations and relationships.

It is clear that the acceleration is not constant.

The average acceleration between 1s and 2s is (11-5) and between 1=62s and 3s is 10

during this time period.

The trend of acceleration a is to increase.

The law of change cannot be detected.

Therefore, the initial velocity cannot be determined.

For a particle with uniform variable speed linear motion, the initial velocity is 4m more than the displacement in the 9th s than the displacement in the 5th s, and the following are obtained: (1) the acceleration of the particle; (2) The displacement of the particle through within 9s, please write the process or idea, thank you!

The displacement of the first 4 seconds is and the average velocity is and this velocity is the intermediate velocity, i.e. the velocity of the 2nd second.

The displacement of the subsequent 4s is zero, how can it be 0? Oh, because it was the top in those 4 seconds, and then it slid back to the same place. The time it takes to slide up and down is the same, 2 seconds. At the very top, it was the sixth second.

From the 2nd second to the 6th second, the velocity changes from 0, so its acceleration is minus square).

Let the half-stage displacement be s and the total time is t

Then s=1 2a*2 2

2s=1/2a*t^2

Combine the two formulas, about the s, you can get:

t = double root number two.

Let s1 be the first half of the journey, s2 be the second half of the journey, v is the initial velocity of the second half, and according to s=at 2, s1=4a 2=2a s2=vt+at 2 v=2a

and s2=s1=2a

Both 2a=2at+at 2 are simplified.

t²/2 + 2t -2 = 0

t²+ 4t + 4-4-4=0

Both (t+2) =8

Get t=2 2-2 or t=-2 2-2 (unsuitable, discarded) Answer: The remaining half of the journey takes (2 2-2) seconds, and the whole process takes 2 2 seconds.

Considering the extreme case, that is, when the car just catches up with the bicycle, the two happen to have the same speed, this is to assume that the time t (from the car starts to slow down to just catch up) then the distance that the car has traveled in this period of time is 1 2*(v1+v2)*t, and the distance that the bicycle has traveled is v2*t, and the approximate schematic diagram can be drawn that the car has walked more than the bicycle l

Therefore, there is 1 2*(v1+v2)*t l+v2*t to solve t=2*l (v1-v2) (what I understand is that the car should be faster than the bicycle before deceleration, otherwise the car would not have to slow down).

Then its acceleration is (v1-v2) t

v1-v2)^2/2*l

If the acceleration is greater than this value, that is, the deceleration is more severe, then when the two cars are about to meet, the car is less than the bicycle and will not collide. So this value is the critical value, which is the minimum value. Therefore a is greater than or equal to (v1-v2) 2 2*l

It should be v1>v2, right?

Solution: The analysis shows that in order not to collide, when the car and the bicycle meet, the speed of the car must be below the bicycle.

Let the time it takes for the car and the bicycle to meet t, and the acceleration of the car is a, so the equation can be columned:

t=l/(v1-v2)

v1-at≤v2

Solution: A (v1-v2) 2 l

So the acceleration must be greater than or equal to (v1-v2) 2 l

Since V1 is smaller than V2, it will never collide and no brakes will be used.

The magnitude of the two speeds is wrong, it should be the car speed v1 > the bicycle speed v2

There are several ways to connect this topic, and for the sake of simplicity, I will only write one of them, forgive me.

Solution: Because the bicycle is moving at a uniform speed, the bicycle can be used as a reference, then the speed of the car relative to the bicycle is v=v1-v2, let this acceleration be a, and the distance from the reference object l when the car starts to decelerate, that is, the velocity v must be reduced to zero within the range of l.

That is, v 2-0 = 2*a*s, that is, (v1-v2) 2=2*a*s, where s<=l must be satisfied

So (v1-v2) 2=2*a*s<=2*a*l, we get a>=(v1-v2) 2 (2l).

That is to say, when the acceleration a meets the size of a>=(v1-v2) 2 (2L), it can ensure that there is no collision.

Is f equal to 3mg at the maximum?

In this case, the maximum acceleration a=(3mg-mg) m=2g

Look

Let the velocity of point A be V, then V=15-10A=Vt+ S=125Mt=10S, and the solution A=Hopefully, you can understand that remembering kinematic time and acceleration is the key! v（b）=v（a）+at，

1 From the first pole to the second pole used 5s, the average V1 = 12M S was used from the second pole to the third pole 3S, the average V2 = 20m s The car is a straight line with uniform acceleration, and the intermediate speed is the average speed, then there is a = (20-12) (

The size of the speed of the car when passing through the third pole = 20 + 2 uniform speed linear motion The speed of the first s is 2m s when it passes through 2m in the first s, and the speed of the third s is 6m s

a=vt-vo/t=2m/s^2

The average velocity in the 6th s = 6 + 3 * 2 = 12m s is generally to determine the movement of the problem first, then think about what characteristics the movement has, and finally look at the data in the problem that meet the characteristics, these data are the breakthrough.

An object with mass m is placed on a rough horizontal plane, and after receiving a thrust f at an angle to the horizontal plane, the object moves to the left with acceleration a, fcosx-fu=ma

fn=mg+fsinx

fu=ufn

If the thrust is increased to 2F, 2FCOSX-FU=MA1FN=MG+2FSINX

fu=ufn

Solution a1=

A The acceleration is constant, then the velocity v=at is changing all the time, because a is constant, but t is changing all the time, so v is also changing all the time. It's easier to analyze with a formula, but you need to be familiar with it.

B velocity to the east, then v to the east is the positive direction, acceleration can be in many ways: (1) can be east, do a uniform acceleration motion, (2) can be west, do a uniform deceleration motion, (3) can be south or north, do a curved motion. The answer is only one of these cases, and it is not comprehensive.

A: The concept of acceleration is the rate of change of velocity, when the acceleration is constant, it means that the velocity is still changing at this rate of change, B: For example, a uniform deceleration of eastward driving, acceleration of velocity to east but west, think about it slowly.