High School Physics Kinesiology 50, High School Physics Kinematics

Displacement image: The corresponding displacement at any time can be known, and the velocity can be determined according to the image, if it is an inclined straight line, the velocity can be found according to its slope, and the straight line parallel to the time axis indicates that the rest is.

Velocity image: The velocity at any time can be known, the acceleration can be calculated according to the slope of the image (a straight line parallel to the time axis represents a uniform velocity), and the displacement can be calculated according to the area enclosed by the image and the coordinate axis.

is a temporal displacement image; When the object is moving in a uniform linear motion, the image of s-t is the image of the primary function of (k>0); When an object moves in a uniform (addition, subtraction) variable speed linear motion, the image of s-t is the image of the quadratic function of (a>0).

is a temporal velocity image; When the object is moving in a uniform linear line, the image of v-t is an image of the primary function of (k=0), and the straight line of v is parallel to the t-axis; When an object moves uniformly (additionally, subtracting) in a linear motion with variable speed, the image of v-t is an image of a function of (k>0, k<0).

S-T Image:

1.The change in displacement over time, read directly from the graph.

2.The slope of a point is the instantaneous velocity of a point, and positive and negative indicate the direction.

V-T Image:

1.The change in velocity over time, read directly from the graph.

2.The slope of a point is the instantaneous acceleration of a point, and positive or negative indicates the direction.

3.The area enclosed by the image and the axis is the displacement during this time period.

Because p = fv and because p is constant, if v is the maximum, then f is the smallest.

f is the smallest then f(he)=0, that is, f=mg=500*10=5000n, so vmax=p f=10000 5000=2m s, from the kinetic energy theorem, it can be seen that 1 2mv2=-mgh+pt is substituted into the data.

So the solution is t=

Solution 1:

1) The acceleration of the A and B systems A = (mg-mgsin30°) 2m = g 4 The velocity of the A and B systems when A lands on the ground is the velocity of the A and B systems when A lands on the ground, vv 2 = 2As = 2 (g 4)*

v = 2 m sec.

The velocity of object A when it hits the ground is 2 m s.

2) After object A hits the ground, B slides along the inclined plane, and the acceleration a = -mgsin30 degrees m = -g 2

0-v^2=2a`s`=-gs`

s = v 2 g = 4 10 = m.

After object A hits the ground, the maximum distance that B slides along the inclined plane in meters.

Solution 2: From the perspective of energy.

Think in terms of energy.

1.From the beginning to the time the object A hits the ground.

mgh = (1 2) 2 mv 2 + mghsin 30 degrees (law of conservation of energy) v 2 = (1 2) gh = (1 2) * 10 * v = 2 m sec.

2.After object A hits the ground, B slides along the inclined plane.

1 2) mv 2 = mgh = mgs sin30 ° (law of conservation of energy) s = v 2 g = 4 10 = m.

Select b Parse:

The center of gravity of the small height is raised, 2as=v (t)-v (0), v (t)=0, a=10

s=upward is the positive direction), data generation formula:

2*10*(, solution:

v=3√2≈4

This question tests two knowledge points, one is how much the center of gravity rises, and the other is the relationship between displacement and velocity, the latter is the key point, use 2as=v (t)-v (0), talk about the derivation process.

1)v(t)=v(0)+at;(2)s=v(0)t+(1/2)at²

Other kinematic formulas can be derived from these two basic formulas.

t=(v(t)-v(0)) a from (1) and substituting (2) to obtain it.

s=v(0)(v(t)-v(0))/a+(1/2)a[(v(t)-v(0))/a]²

Multiply 2a on both sides at the same time

2as=2v(0)(v(t)-v(0))+v(t)-v(0))]

2as=v²(t)-v²(0) 。

Whether it's physics or mathematics, a formula illustrates a relationship. Just like this question, if you know that the distance requires speed, you need to search for the relationship between distance and speed in your memory, and this relationship is indicated.

2as=v (t)-v (0), then draw inferences, there is also the relationship between speed and time, the relationship between time and displacement, kinematics is nothing more than distance, speed, time, acceleration, these major elements, you can learn well by mastering the connection of these major elements, what is their relationship, that is, like the formula says. The same is true for mathematics.

This is the way of thinking in the formula of science, so that you can think reliably.

Well, that's all, I hope it helps you.

Height, indicating the height of the center of gravity of the original body. In other words, the height of his actual jump is about. According to this, substituting 2gh=VO can give VO 4m s

Because his acceleration is 0 when he crosses the crossbar, and the distance he travels is , so according to v(initial)2-v(end)2=2as, since v(end)=0So v(initial)2=2as, we get v=4m s

The key is to know how high he jumped, the center of gravity is, so the starting height h=, so v 2 = 2gh calculates v = if he wants to cross the crossbar, the speed to reach the crossbar must be greater than 0, which means that the speed is the lowest, so the answer should be c

This is a strange question, because I don't remember that there is a definition of "average acceleration" in the textbook.

I have to guess based on my own understanding.

First of all, acceleration is a vector, which is difficult to average, and it is even more difficult to divide with another vector.

But you can borrow the concept of "average velocity", you can use the end velocity to subtract the starting velocity time, this subtraction is also problematic, if the velocity is used as a scalar quantity, this subtraction is equal to zero, the conclusion is 0:0, of course not equal to the ratio of centripetal acceleration.

You can also subtract the velocity as a vector, the angle at which two points on a uniform circular moving object turn at the same time is the same, and the two points subtract the velocity of the starting point and the end point by vector, imagine that the graphing method will obtain two similar isosceles triangles composed of velocity vectors, the "waist" of the two isosceles triangles represents the velocity of the two points at the beginning and end of each other, and the bottom edge is the velocity difference, and then divided by time, it becomes the "average acceleration", so the ratio of the average acceleration is equal to the ratio of velocity, Equal to the ratio of the radius, whereas the ratio of centripetal acceleration is equal to the ratio of the radius, the conclusion is equal.

There is no such thing as average acceleration at most equivalence.

And I don't know what you mean.

For example, is it a rod moving in a uniform circular motion around the ends, acceleration at different points?

a = square of angular velocity * radius.

And the angular velocity is equal so it is proportional to its distance from the center of the circle.

You can go directly to your physics teacher.

In terms of the critical state, that is, it is assumed that it happens to collide.

A: vt 2-v0 2=2as

S A 18m T A 6s

According to the assumption, then S C = S A + 10 = 28m

vt^2-v0^2=2as

It is also possible to solve a C to be equal to or add a negative sign in front of it.

ta>tb

Suppose the bend point in the path of b is d

The bend point in path A is E

When the ball B arrives at D.

The ball of A must not have reached the point at the same level as D.

Because the A to E section is slower than the B to D section.

When B arrives at D.

Closer to the end point C than A.

And its kinetic energy is even greater.

Therefore B will reach C before A

I looked at the second floor and asked.

I think there may be something wrong with the answer.

I've done the exact same thing.

The answer is ta>tb

And the answer says to use V-T image solution.

I feel a little troublesome.

The total displacement magnitude is constant, V-T image method. You try it, it's very useful. Note that the acceleration (slope) and the end velocity are equal, and the time should be greater than b, or b should return to the bottom first.

ta=tb Some time ago, we did a lot of this kind of topic.

Regardless of friction, then only gravity does the work. mgh = 1 2 mv 2v equal, acceleration a equal. Launch T equal.

I thought of doing free fall. h=1/2gt^2

Solution: When passing the first paragraph.

The average speed is s t1

That is, the velocity at the middle moment of the segment.

Similarly. The average velocity of the second segment is s t2

It is also the speed of the middle moment of the paragraph.

Then you can find the acceleration.

aδv/δt

s/t1s/t2/(t1/2

t2/22s/(t1t2)

Process. Let it take time t for A to catch up with B, where the displacement of A is 4T and the displacement of B is S=VT+

And 7+s=4t

The solution is 7 seconds.

Since the corners are both 90 degrees, the velocity accumulated in the corners is equivalent to nothing, and there is no effect on the later stage, so the movement of the two balls is exactly the same, and the time is the same.

Because the title says that the same amount of time has elapsed and the distance is the same, the average speed is the same. Let the velocity at f1 be removed as v, then (0 + v) 2 = v + 8) 2Thus, the velocity of f1 can be removed at 4m s.

You're right.

After removing f1, first do a uniform deceleration, and then a uniform acceleration.