Seeking, the classic exercises of linear motion with uniform variable speed with answers are best an

The first problem is in the variant problem of uniform variable speed motion, if it is a uniform variable speed motion, satisfying the velocity relationship of the problem, the displacement of the latter will be greater than the displacement of the former; However, if the displacement of the problem is equal and the velocity is equal, the speed of the latter changes faster, the acceleration is greater, and the time taken is shorter, which is the empirical judgment method. Another rigorous method is to use the velocity image to represent the relationship between the two physical quantities, because the area of the image represents the displacement, the slope represents the acceleration, and the relationship between the velocity can also be intuitively reflected by the image, so it can quickly and intuitively reflect the relationship between the physical quantities before and after.

The second difficulty lies in the fact that the car emits, reflects, and **ultrasonic waves in different positions, and the relationship between displacements needs to be very clear here, so that the correct formula can be listed. This is also what experts often refer to as how to establish relationships through physical situations and list the correct formulas.

The third question is the problem of pursuit, for the object in motion of the former, it is necessary to clarify the conditions for collision, that is, the displacement of the pursuing object, that is, when the speed of the latter is reduced to the speed of the former, if the position is in front of the former, it means that it has been hit, if it is behind or just touches the former, it means that it has not hit, which is used as a foreshadowing of knowledge, and it is also the key to listing the correct formula. Then it is necessary to find their displacement relation, usually the displacement of the latter minus the displacement of the former, if it is greater than the distance before them, it means that they have collided, otherwise they have not collided.

Of course, for complex calculations, students must have solid basic skills in calculations, not only accurate, but also fast.

12. A and B trains travel in the same direction on the same track, car A is in front, and its speed is V A 10 m s, and car B is behind, and its speed is V B 30 m sDue to the fog and low visibility, car B only noticed that car A was in front of car A when it was 700 m away from car A, and car B braked immediately, but car B could not stop until 1 800 m Question: If car A moves forward at the original speed, will the two cars collide? Explain why

In short, if students can analyze the test questions every week, think about the mistakes clearly, understand the reasons for the mistakes, so as not to make mistakes in the same place next time, and understand the methods used to do the right thing, and then expand it to think of the method of the same fight chain type questions. Then, the improvement is very effective and fast.

Assuming that the maximum speed is v, the acceleration time is v a1 and the deceleration time is v a2.

The acceleration distance is S1 and the deceleration distance is S2. From 2as=v 2, we get s1=v 2 2a1, s2=v 2 2a2.

So the constant velocity distance s3=d-s1-s2=d-v 2 2a1-v 2 2a2.

then the constant velocity time is s3 V.

The total time is t = v a1 + v a2 + s3 v = d v+ is a hook function about v, so t is greater than or equal to [(d v)*2d(1 a1+1 a2)] under the root number

An object starts from rest to do uniform acceleration linear motion, the acceleration magnitude is a1 after the time t to do uniform deceleration linear motion, the acceleration magnitude is a2, if the time t can return to the starting point, then a1:a2 should be.

Let the velocity at the end of acceleration be: v1, and the velocity back to the starting point as: —v2 The average velocity represents the displacement.

x=[(v1+0)/2]*t

x=[(v1-v2)/2]*t

solve, v2=2v1

Acceleration: a1=(v1-0) t

a2=-(-v2-v1)/t

a1/a2=1/3

g is the gravitational plus or balanced velocity, which is downward, so it is a negative number in the group. g=-10s(t)=v*t+ 4=11t+(

t= This is the second before being picked up (t is still filial piety to do a solution, this is the time to fly to the top and then fall back to 4 meters).

So the real time is.

t=s(t)=vt+

I want to give a detailed explanation, but I want you to learn to use the middle image of speed to solve the problem of uniform speed and straight line movement with clever movements, you look at the diagram first.

The time is equal.

For vertical rod AD:

d (note: diameter) = 1 2gt1 .Type 1.

For inclined rods (angle a to vertical direction):

The length is d*cosa

The acceleration component in the direction of the oblique rod is a=g*cosa

So d*cosa=1 2*gcosa*t2 i.e. d=1 2gt2

Comparing the above equation with equation 1 gives t1=t2

In the same way, t3=t1

The initial velocity at the 5th second: v=v0+at=1+5a

The displacement in the 5th s = vt + (1 2) at2

That is: (1+5a)+(1 2)a(1 1)=8a=

Displacement in the first 5 s: s=1 5+1 2 (14 11) (5 5)=

Unit conversion: 60km h=(50 3)m s distance: s1=[(50 3)m s] 2 * 6s=50m (this is the formula:

Average velocity = (initial velocity + final velocity) 2, average velocity * time = distance) The distance that has been moved for another 20s is s2 = (50 3) m s * 20s = (1000 3) m

Average speed: v=(s1+s2) t= 50m+=(1000 3)m 26s

Answer: 50;

First Empty:

We know that it is accelerating at a uniform speed (question), 60 kmh = 16Another two-thirds m ss = one-half at = one-half vt (this v is the last speed after acceleration), = 50m

v average = (s acceleration + s constant velocity) total time = (50m + 16 and two-thirds m s x20s) 26s =