A car that does a linear motion with variable speed will travel a certain distance forward after bra

The state in which the object remains in a straight line at a constant speed or at rest without external force is the state of speed when he brakes.

Let's say that the car turns off the engine for 10 m seconds, and if all resistance is removed, the car will keep moving forever.

It's inertia! Constant speed movement is sufficient as long as the power of the car and the resistance (friction force) are equal.

Driving forward some distance after braking is because of inertia.

When a car is driving at a constant speed, it can be seen as a balance between drag (including rolling resistance, air resistance) and driving force (provided by the engine). When the car brakes, the linear velocity of the wheels is lower than the driving speed, resulting in relative sliding against the road surface. At this point, the road surface creates a large resistance to the wheels.

That's the braking power of a car. The braking force forces the car to slow down. (Law of Inertia:.)

Until an external force changes this state, the braking force is the external force that forces the car to slow down.

Don't know if you agree or not?

I've done so much and found that I didn't understand your question. But your question is weird, I'll do it all over again.

First of all, any object has inertia.

Secondly, if it is considered impossible for a car to move in a uniform linear velocity, then the motion of any object cannot be regarded as a uniform linear motion.

On a well-leveled road surface, the car maintains the same gear and the same throttle opening, which can be regarded as a uniform linear movement.

In order to drive at a constant speed, it is necessary to balance the driving force and driving resistance (including rolling resistance, hill resistance, air resistance, and acceleration resistance). Once this equilibrium is disrupted, the car has to change speeds.

How is this equilibrium out of balance? The possibilities are endless. For example, the driving force is related to the throttle opening.

When braking, the clutch is stepped on, there is no driving force, and a large braking force is generated, and the uneven, rough road surface, slope and other changes in resistance. The car is about to shift gear.

To put it simply, in a bustling city and poor road conditions, it is impossible for the car to move at a uniform speed. On a well-leveled road surface, you can do a uniform movement.

In fact, the car has been doing uniform linear motion, which should be different road conditions, different tires, and different friction, so the car is a variable speed, because every moving object has inertia, so it will remain stationary or uniform linear motion, provided that (not subject to any external force [including friction]).

Summary. Hello, professional analysis of automobiles: the car will always move forward again when braking for a period of time, the reason is that the brake fluid is not enough, resulting in the brake not being completely braked.

Hello, professional analysis of automobiles: the car will always move forward again when braking for a period of time, the reason is that the brake fluid is not enough, resulting in the brake not being completely braked.

1. The brake fluid is not enough, resulting in the brake not being completely hail brake. Workaround: Check and add brake fluid. 2. The side circle destruction is that the brake pads are worn too badly, resulting in the vehicle not being able to stop. Solution: Replace the brake pads.

3.When operating the lead brakes, you don't fully step on the bottom. Workaround:

Apply the brakes hard again. Braking is a necessary factor for safe driving, and it is a sharp weapon when you need to stop in an emergency, playing a role in slowing down, stopping, and ensuring safety. In case of failure, insensitive tremors, etc.

Timely maintenance is required to avoid accidents.

1.Quadratic function y= +

The solution yields x=150 or x=-155

So the speed is 150kmh overspeeding.

You draw the picture and you get it ...

Through mathematical calculations, overspeeding is a certainty, but it is necessary to take into account the friction and resistance to learn and analyze!

2ay=v0^2

The standard equation goes like this, v0 is the speed of the vehicle and a is the acceleration.

Is this a physics problem?

The average velocity of the first half v1=(vo+v) 2 and the average velocity of the second half v2=v 2

The joint solution yields the ratio of the average velocity of the first half to the second half v1 v2=[(vo+v) 2)] (v 2)=(vo v)+1

Again: the whole process vo 2 = 2as

First half v 2=2a(s 2).

The joint solution obtains vo v = root number 2 substitution

Get v1 v2 = 1 + root number 2

Get the answer. The following two steps are redundant.

vo^2=2as

v^2=2a(s/2)

A usable time displacement x = 1 2at2 resulting acceleration a =

The rest of the requirements can not.

If Ting Uranium Zheng Barrel Ying embroidered father, then.

Burn the beggars and hang them on a broomstick, and then.

Dumplings make news to assist Tibetan children, yes.

Li Zheng concealed his clothes.

October 18, 2012.

11:24:21 min.

The front and back halves are doing even deceleration movements.

You have to figure out the braking distance and inertia. Inertia is the ability to maintain the original state of motion. The greater the mass, the stronger the maintenance ability, an object has the same mass, the same velocity, that is to say, the ability to maintain the original state of motion is the same, for example, two small trucks with the same mass, one 10 meters per second, one 20 meters per second, the inertia is the same, and the force required by the outside world to change its motion state is the same.

With the same force, it takes longer for a moving object with a fast speed to stop. The long time and fast speed will make the braking distance longer.

Set the car speed is v (m s).

So there is a meaning from the title.

Solution v1=30

v2 = - 40 (the speed cannot be negative, it is not suitable to be discarded) so the speed of the car is 30m s

It can be seen as a uniform acceleration motion with an initial velocity of 0.

The first half of the journey is 2=

The second half of the distance is 2=(a*t1)*t2+

The two equations give =(a*t1)*t2+

Eliminate a to get t1 2-t2 2-2t1*t2=0 Solve the relationship between t1 t2 t2: t1 is the ratio of the average speed of the car in the first half of the distance to the average speed in the second half.

According to known conditions, it is available:

Find the ratio of the average velocity of the "first half" and the "second half", then the displacement of the two halves is equal, let it be s

Since it is a linear motion with uniform variable speed, the acceleration of the whole process is equal, let a the initial velocity be set to v1, the initial velocity of the second half of the journey, that is, the end velocity of the first half of the journey is set to v2, and the final velocity of the whole journey is 0

The following equation can be obtained:

v12-v22=2as

v22=2as

Solve the system of equations and obtain:

v1=2∨2*v2=

The initial velocity of the whole journey can be written as, the initial velocity of the second half, that is, the speed at the end of the first half, is v2, and the final velocity of the whole journey is 0

And because it is a linear motion with a uniform variable speed, it can be known:

Average velocity = (muzzle velocity + terminal velocity) 2

Average speed of the first half = (v1+v2) 2=(

Average velocity in the second half = (v2+0) 2=v2 2 Average velocity in the first half Average velocity in the second half = ((

Note: It should be written as root number 2!

For the convenience of expression in the computer, it is written as a decimal!

Answer: No speeding.

Substituting the vehicle speed and braking distance into the relationship yields:

100a+b=21

150a+b=

Calculated: a= b=-489

When the braking distance is, the substitution relation: ax+b= i.e. x=i.e., when the braking distance is the speed of the vehicle, it is not more than 120km h, so there is no speeding.

Answer: Speeding.

The title is a quadratic function!

Not to be confused with once.

If you have more squares, the difference is very large, and it is very cumbersome to calculate.

Solution: Obtained from known conditions.

21=10000a+100b

The solution is a=1 500 b=1 100

So the quadratic function expression is y=1 500 x 2+1 100 x when y=.

Substituting according to the root-finding formula.

The available x1 is negative (rounded) and x2 is 140

So the car is speeding.

Uniform deceleration linear motion, which can be reversed, is a uniform acceleration linear motion.

The first half of the distance, 1 2a*t 2, the end speed a*t, the average speed 1 2at

The distance in the second half of the time is 1 2*a*(2t) 2=4at 2, the end velocity is a*2t, and the average speed is (1 2at+2at) 2=5 4at

The average speed ratio is 5 to 2