A high school physics question, if you have a question, ask for an answer

Ask about power, why is the answer v?

But to your problem, you delete the 2l on both sides of the formula, that is, a=gsin The acceleration here is to decompose gravity, specifically to establish a Cartesian coordinate system along the inclined plane direction, because the inclined plane is smooth, the object only has gravity and elastic force (elastic force is the supporting force of the inclined plane to the object), in this, the elastic direction has no velocity, so the acceleration of the object is all provided by the component of gravity in the inclined plane direction.

Got it

The gravity force is decomposed along the inclined plane, and the vertical component is balanced with the elastic force of the inclined plane, and the force along the inclined plane mgsin, acceleration = gsin

Because the beginning and end of the two resulting time periods are the same event. In other words, the removal of bits to minute velocity is to project the motion of changing directions onto an axis of the coordinate system, because the initial and last coordinates of the motion are the same, so no matter how you calculate, the elapsed time is the same, that is, time is considered as the fourth dimension.

This is a conceptual problem of speed and velocity, which does not require complex calculations, but it is brain-consuming. Suppose you are in a scene, you are facing a curtain, there is a parallel light source behind the curtain (to ensure that the size of the object is not affected by the distance between the object and the curtain), someone walks from your right hand side to the left hand side behind the curtain, but instead of walking in a straight line, but following a regular zigzag line, you can see the shadow, and someone walking in the same direction in a straight line, you can also see the shadow. If these two people are walking at a constant speed, starting and arriving at the same time, then the shadows you see must coincide, or the shadows you see must be traveling at the same speed.

It shows that in the dimension of the curtain, the speed of the two is the same. But because someone walks the Z-shaped line, this person must walk faster than the straight line, which is a matter of rate size. And the velocity seen on the curtain is the fractional velocity of the velocity on the displacement of the curtain.

The connection between the five-pointed star and the center is the curtain, no matter how the particle runs, it always runs to the middle in the end, and there is always speed on the line, so time is of course equal to the distance on the line divided by the speed on the line.

In fact, if you want to accurately describe the motion trajectory of the five particles, you need to use calculus, this problem only involves the initial and final, and does not describe the intermediate state, so you can understand it with the concept of average velocity, don't think about how the particles move at a certain moment, with the mathematical knowledge in middle school, you can't figure it out.

High school in the last century, don't blame me for making mistakes, I hope it can help you.

The work done by friction is equal to kinetic energy, fs=1 2mv2

50000*3 5*75=1 2*5000*v2.

Solution: When an object falls from a balloon, it has to move upwards for a certain distance because it still has an upward velocity, and the time for moving upwards for another distance is 4 10=

So this distance is 4* squared =

So the height of this weight from the ground to the highest point is h=217+, so the time t to return to the ground, can be made by.

1 2*10*tsquared=

The solution is t = so the velocity to the ground v = 10*

If you are already in the first year of high school, then I don't think there is any need for me to calculate the result, I only remind you: the initial velocity of the heavy object is 4m s upward, the acceleration g is not to mention, the displacement is 217m downward, you can calculate it if you know the formula of uniform acceleration linear motion.

Only when the speed of the two cars is exactly the same, and the two cars do not collide, can the two cars not collide to ensure that the two cars do not collide.

In other words, as long as the two cars are guaranteed to be at exactly the same speed, and the two cars do not collide, it can be guaranteed that the two cars will not collide.

The standard answer idea is wrong.

When the speed of car B drops to the same speed as car A, the elapsed time is t, and the acceleration of car A is A.

Then there is: This is the formula.

At time t, the displacement of car B minus the displacement of car A is exactly equal to 500 meters.

Then there is (30-10)t+, which is the formula.

From , the minimum acceleration A of car A can be calculated.

The equation can draw an orthogonal diagram of the speed and time of the two cars, and according to the geometric relationship, the equation is obtained).

Suppose the acceleration is a

The closest distance between the two cars should be the same speed of the two cars, because after that, car B will decelerate slower than car A, and you will never be able to catch up if you don't catch up.

So ab= m s 2

When is the speed the same?

t=(20+

At this point, the two are far apart.

500+10* t^2]>=0

Replace t, simplify to get 1 8*(15049*a-2400) (4*a+1)>=0

a>=2400/15049~

Don't think about it, assuming that car A does not accelerate, B must go 500m before reducing to 10m s if it wants to surpass A.

b acceleration a=. Calculate and you will know that B simply cannot go to 500m at 10ms.

So it is impossible for A to be caught up by B in front.

For the second question, there will be an acceleration value a after the standard answer is solved, replace a with the original equation (take t as an unknown), and then solve the equation, there will be two t corresponding to one of which is 120, and the other is what you are worried about, I didn't calculate, it is estimated that there are two cases, one is that the other t is a negative number, which can be ignored, and the second case is that t is a positive number less than 120, which is the situation you are worried about. You can do the math.

This is a pursuit problem, it is not difficult to master the method, it is best if you set a physical quantity, like an equation in mathematics to solve the problem, sometimes use inequality, take the maximum value, the minimum value, generally in this centralized form.

When the speed of car A and car B are equal, there is no collision, and then the speed of car B is not as fast as car A, and it will not collide, which is the critical condition.

My method should be shorter than the answer, you must learn this way to look at the picture:

If you think about it, you consider the formula of the constant, what is the value of time t, it is 0 120 seconds, and the critical value is 120 seconds, at this time it is the explanation on the answer, right, why don't you need to consider the situation you said, because it has been included, in fact, the principle is that the two cars take time as the x-axis, whether it is speed or distance, in this interval is a monotonic function, one increases and one decreases, and the intersection point must be 120 seconds, if you say that the situation exists, That violates the fact that for the same y-value, there are two different x-values, then it is not a function, let alone a monotonic function.

There is something wrong with the way the answer is going on. You're right.

My solution is:

Let the acceleration of car A be A, and the speed of the two cars is equal when the time is t (t is less than 120). There was no collision at this point. (After that, car A accelerates, and car B decelerates and will not collide.) So it's a critical state to analyze).

Simultaneous equations.

where t<120

You draw your own diagram, the relationship between time and distance, and you see if the two lines A and B will intersect.

Look at the picture and answer! The ab line is the tangent of the curve, that is, the acceleration a is the slope of ab, a=15-5 4-0=

Force analysis: f=mgcos, f=kv

cd is the asymptote of the curve, i.e., when v = 4, it moves at a uniform speed.

There is mgsin = kv2+ mgcos

Acceleration phase: a=mgsin -kv1- mgcos when v1=2, v2=4

Substituting Lianli can find k,

2. The acceleration of curvilinear motion is not necessarily perpendicular to velocity, such as flat throwing!

Compulsory 2 question.

The acceleration at other points is downward, because the title is talking about a linear motion with a uniform variable speed, and the acceleration is constant, and the magnitude and direction are the same.

Gravitational acceleration is related to gravity and goes all the way down.

Let the angle of the inclined plane be x and the dynamic friction factor be y, then mgsinx=ymgcosx. Therefore y=tanx. After adding F, (mg+f)sinx=(mg+f)cosxtanx=y(mg+f)cosx. i.e., A. should be chosen