A higher displacement physics problem, a higher level physics problem about displacement,

Let the acceleration be a meter second 2, then the velocity at the end of t seconds is at meter second.

1).3a) 2-(2a) 2=2as where s=15 m solution gives a=6 m sec 2

So displacement in the 6th s = [(6a) 2-(5a) 2] 2a = 33 m2) displacement in the first 6 seconds = 1 2at 2 = 108 m.

3) by (1), a = 6 m sec 2

4) v 2=2as where s=12m is found v=12m sec.

1: First find the velocity at the end of the 5th second and the velocity at the end of the 6th second, and use Formula 3 (see step 3 for details of how to find the acceleration).

2: According to the initial velocity is 0, the last velocity is the velocity at the end of 6 seconds (the first question is found), and formula 3 is used.

3: Formula 3 deformation: a=(vt2-v02) 2s, replace vt=v0+at in the formula, where v0=0 (because a particle starts from rest to do uniform acceleration linear motion, the initial velocity is 0), you can find a.

4: Or formula three: vt2-v02=2as, replace vt=v0+at in the formula, where v0=0, it becomes at2=2s, bring in s=12, and the acceleration just calculated, you will know t Then use formula one:

vt=v0+at, you know the speed.

This kind of problem can be done by drawing pictures, or you can use the formula given to you by your teacher, since you have written out the formula, then use the formula to explain (v0=0).

1) The 6th second is the displacement of the particle between the 5th and 6th seconds.

Conditionally bringing in the second formula, we can see that a=6, s=1 2*a(6 2-5 2)=33;

2) The displacement in the first 6 seconds s=1 2*a*6 2=1083)a=64) is brought into the data by the third formula, and vt=12 can be obtained

Just remember a proportional formula: the displacement in the nth second, as long as the time is evenly distributed, always follows 1:3:

11...Such a regularity. Then in the first problem, the displacement in the third second is 15m, then the corresponding sixth second is 33m

2。According to the second formula, v0t is 0 and can not be counted. The result is 108 meters.

3。Using the proportional formula I gave you, the displacement of the first three seconds is 3 + 9 + 15 = 27 meters, and according to the second formula, we get a = 6 (the unit is difficult to hit).

4。According to the third formula, the initial velocity v0 is 0, can be ignored: algebra, resulting in 12 (units).

Poor expression skills...

The displacement of the particle in the third second can be obtained by subtracting the displacement at the end of 2s from the displacement at the end of 3s

1 2*a*3 2-1 2*a*2 2=15 solution: a=6m s 2

1) Displacement in the sixth second s1 = 1 2 * 6 * 6 2-1 2 * 6 * 5 2 = 33m

2) Displacement in the first six seconds s2 = 1 2 * 6 * 6 2 = 108m3) acceleration a = 6 m s 2

4) vt²-v0²=2as=2*6*12=144vo=0vt=12m/s

1) 33m s1 = 1 2 * 6 * 6 2-1 2 * 6 * 5 2 = 33 m2) 108m s2 = 1 2 * 6 * 6 2 = 108 m3) The acceleration is 6 meters per two seconds.

4) 12 meters per second vt -v0 =2as=2*6*12=144vo=0vt=12m s

From a standstill, the linear motion is uniformly accelerated, and the proportion of distance traveled per second is 1:3:5:7....

Very useful.

Find the acceleration first, because the end velocity of the first 2s is the initial velocity of the 3s, so the final velocity of the first 2s can be obtained by vt=v0 at is 2a, so from s=vot 1 2a*t 2 can get 2a a 2=15m, so a=6m s 2, so the displacement of the first 5s can be obtained from s=v0 1 2a*t 2 is 75mThe displacement of the first 6s is 108m, so the displacement of the 6th is 33m. From s=v0 1 2a*t 2, the displacement of the first 6s is 108m.

From s=v0 1 2a*t 2, a=6m s 2. From vt 2-v0 2=2as, the velocity of 12m is 12m s! Look....

In terms of displacement, you can articulate an axis. All displacements are independent of the movement process, but related to the coordinates of the start and end points. In other words, the displacement is the line segment formed by the start and end points!

In the first void, the particle returns to the origin after one circle, so the displacement is 0 without change. The distance is related to the trajectory of the particle, and to count all the paths, the motion of the particle, which is the circumference of the circle, that is, the circumference of the circle. So it's 2 r.

When the particle moves for 7/4 of a week, the angle is calculated, that is, the difference of a quarter of a circle is two turns, and the angle is 45°, so the displacement is the root number 2r, and the distance can be directly multiplied by seven-quarters of the circumference.

The maximum displacement, in a circle, the maximum distance between two points is the diameter, so the maximum displacement is 2r. The maximum path is seven-quarters of the circumference of the circle.

In this type of problem, all displacements are considered to be the length of the line segment between two points, and the distance is considered to be the curve of the motion of the particle.

When I was in high school, I thought about this kind of problem, and I was afraid that I would not be able to figure out the movement of the particles on the paper, so you should do the same! It will soon be understood

Analysis: 1. The displacement is the straight-line distance from the starting point to the end point, which has nothing to do with the actual path. In one circle, the starting point is at the same point as the end point, so the displacement is 0.

2. The distance is the trajectory of the actual motion of the object, that is, the circumference of the circumference of the circle is the circumference of the circle, then the circumference is 2 r.

4 weeks, the size of the journey is 7 The circumference of 4 circumferences is 7 r 2

4 weeks, the displacement magnitude is the chord length of 1 4 circles, and the starting point and the end point are connected to the center of the circle, and an isosceles right triangle can be obtained. So the displacement magnitude s= 2r

5. The maximum displacement during motion is 2r because the maximum distance from one point on the circle to another point on the circle is the diameter.

6. The maximum distance, that is, the actual trajectory of the movement, is 7 r 2

If a particle moves one time from point A around a circle of radius r, its displacement magnitude is 0 and the distance is 2 r. If the particle moves for 7/4 cycles, then the displacement magnitude is 2r and the distance is 14/4 r, and the maximum displacement during this motion is 2r and the maximum distance is 14/4 r

Analysis: 7/4 weeks is exercise, and 3/4 is 2 weeks. So the displacement is the hypotenuse of an isosceles right triangle and the right angle is r. The maximum displacement is the movement to the opposite of the starting point, and the displacement is the diameter 2r.

The displacement size is always 0, but it does not mean that the displacement is unchanged, the displacement direction changes all the time, and the distance is seven-quarters of its circumference.

2r is equal to 3/4 of the time of displacement, 7 r 2 7 4*2 r

2r The maximum distance from one point on the circle to another point on the circle is 7 r 2 in diameter

(1) Displacement in the 2nd s = 30-20 = 10m

2) Displacement in the third second = 30-30 = 0m

3) Total distance in the first 5 seconds = (30-10) + 30 = 50m Total displacement in the first 5 seconds = 0-10 = -10m

If you find xab with an arrow pointing right, a at 1 and b at 2, why use xb-xa=2-1. Whereas, xba is found using xa-xb=1-2=-1.

You can solve this problem by subtracting the initial state from the end state.

Because XAB is from A to B, and XBA is from B to A

The displacement in the first 2 seconds is 20m, there is no displacement in the 3rd second, and the distance in the first 5 seconds is 30+20=50m

In the first second, in the second second. The displacement ratio in the fifth second is x1:x2:x3:x4:x5=1:3:5:7:9, so x2:x5=3:9

i.e. 3:9=2:x5

So x5=6

The time from the beginning to the meeting of the two cars is the time for the bird to fly, that is, (6 30x2 = that is, hours) 6 minutes. Multiply by 2 because two cars running at the same time need to be multiplied by 2.

The two cars must have met in the middle of the journey, so the bird's displacement is 6 2 = 3km. The distance the bird flies is equal to the speed multiplied by the time, that is, the first question is over!

The second question: you must first know that the displacement is the distance from the starting point to the destination point, so the displacement of the bird after the round trip is the same as the displacement of car A, and how much car A goes, how much car B goes.

Now step by step, the bird gets T1 from A-B, if this time is t1, that is, 60t1 + 30t1 = 6 (the distance of the bird plus the distance of the car b is exactly 6km).

Now look at b-a, at this time, a, b have gone t1 time, so a, b car distance is still 6-30x2t1 = 2, and if the bird returns to t2 time, then 60t2 + 30t2 = 2 (the same as the first step, but the total length is only 2km), get t2The rest is simple, the bird's round-trip time is t1+t2, the distance is equal to 60x(t1+t2), and the displacement is equal to 30x(t1+t2)See if you can figure it out!

I don't know if there was a miscalculation, but the idea is this.

Because it is a uniform deceleration motion, the acceleration is constant, and the velocity is exactly half of the initial velocity, so the time for the velocity to be reduced to half is also 20 seconds, and the total t=40s

20=30 gives v0 2

x＝（v0/2）t＝（v0/2）×40＝40

The distance will be 12km.

Solution: The time that the dog travels = the time when A and B meet = 12 (5+3) = hours.

So, the trip the dog travels is s=vt=6*

The displacement of the dog x = the displacement of A = 5*

Solution: We can regard this process as the sum of a uniform linear motion and a uniform acceleration linear motion with an initial velocity of 0, and the ratio of the displacement in the 1s, 2s, 3s, and 4s of the uniform acceleration linear motion with an initial velocity of 0 is 1:3:5:7, and from the data in the question, we can find that the displacement of the uniform acceleration linear motion with an initial velocity of 0 in the first second of motion = 3m, and according to s=at 2 2, t=1, it can be solved that a=6m s 2, and the displacement of the 3s of the uniform acceleration linear motion with an initial velocity of 0 = 6 (9-4) 2 = 15m, so the velocity of the uniform linear motion = (21-15) = 6m s

It's obvious. It's the wrong answer.

PS: There is another situation, which is the one inside"3" is the question number.

You're not mistaken. The answer is wrong.

From S 1 2AT2, V at, it can be seen that the displacement is proportional to the square of time, and the velocity is proportional to time. So through s, 2s, 3s ......The ns time ratio, i.e., the velocity ratio, is 1: root number 2:

Root number 3: root number n, through the time of each segment S with the difference between two adjacent items.