Acceleration of a point at a moment in time. Determining the absolute speed of a point

Instruction

Enter the coordinate system in relation to which you will determine the direction and module. If the task already has dependencies speed from time to time, you do not need to enter a coordinate system - it is assumed that it already exists.

According to the available dependency function speed from time you can find the value speed at any time t. Let, for example, v=2t²+5t-3. If you need to find module speed at time t=1, just plug this value into and calculate v: v=2+5-3=4.

Sources:

how to find path versus time

Module numbers n is the number of unit segments from the origin to point n. And it does not matter in which direction this distance will be counted - to the right or left of zero.

Instruction

Module numbers also called absolute value this numbers. He's short vertical lines drawn to the left and right of numbers. For example, the module numbers 15 is written as follows: |15|.

Remember that the modulus can only be a positive number or . Module positive numbers is equal to the number. Module zero. That is, for any numbers n greater than or equal to zero, the following |n| = n. For example, |15| = 15, that is, the modulus numbers 15 equals 15.

Modulo negative numbers will be the same number, but with opposite sign. That is, for any numbers n, which less than zero, the formula |n| = -n. For example, |-28| = 28. Module numbers-28 is equal to 28th.

You can find not only for integers, but also for numbers. And with regard to fractional numbers the same rules apply. For example, |0.25| = 25, that is, the modulus numbers 0.25 will be equal to 0.25. A |-¾| = ¾, that is, the modulus numbers-¾ will be equal to ¾.

When working, it is useful to know that modules are always equal to each other, that is, |n| =|-n|. This is the main property. For example, |10| = |-10|. Module numbers 10 equals 10, just like a modulus numbers-ten. Moreover, |a - b| = |b - a|, since the distance from point a to point b and the distance from b to a are equal to each other. For example, |25 - 5| = |5 - 25|, i.e. |20| = |- 20|.

To find change speed determine the type of body movement. If the motion of the body is uniform, change speed equals zero. If the body is moving with acceleration, then change his speed at each moment of time can be found if we subtract from the instantaneous speed in this moment time its initial speed.

You will need

stopwatch, speedometer, radar, roulette, accelerometer.

Instruction

Definition of Change speed arbitrarily moving trajectory Using a speedometer or radar, measure the speed of the body at the beginning and end of the segment of the path. Then from end result subtract the initial, this will be change speed body.

Definition of Change speed body moving with acceleration Find the acceleration of the body. Use an accelerometer or dynamometer. If the mass of the body is known, then divide the force acting on the body by its mass (a=F/m). Then measure the time it took for the change to take place. speed. To find change speed, multiply the acceleration value by the time it took change(Δv=a t). If acceleration is measured in meters per second, and time is measured in seconds, then the speed will be in meters per second. If it is not possible to measure the time, but that the speed changed on a certain section of the road, with a speedometer or radar, measure the speed at the beginning of this segment, then use a tape measure or range finder to measure the length of this path. Using any of the above methods, measure the acceleration that acted on the body. After that, find the final speed of the body at the end of the section of the path. To do this, raise the initial speed in , add to it the product of the section by the acceleration and the number 2. Extract from the result. To find change speed, from the result, subtract the value of the initial speed.

Definition of Change speed body when turning If not only the magnitude, but also the direction speed then find it change vector difference of initial and final speed. To do this, measure the angle between the vectors. Then subtract twice their product from the sum of squared speeds, multiplied by the cosine of the angle between them: v1²+v2²-2v1v2 Cos(α). From the resulting number, extract Square root.

Trajectory, radius vector, body motion law

Now we will consider the simplest kinematics - point kinematics. Imagine that the body (material point) is moving. It doesn't matter what kind of body it is, we still consider it as a material point. Maybe it's a UFO in the sky, or maybe it's a paper plane that we launched from the window. Better yet, let it be new car on which we are traveling. Moving from point A to point B, our point describes an imaginary line, which is called the trajectory of movement. Another definition of a trajectory is the hodograph of the radius vector, that is, the line that the end of the radius vector describes. material point when moving.

Radius vector - a vector that specifies the position of a point in space .

In order to know the position of a body in space at any moment of time, you need to know the law of motion of the body - the dependence of coordinates (or the radius vector of a point) on time.

The body has moved from point A to point B. In this case, the displacement of the body is a segment connecting these points directly - vector quantity. The path traveled by the body is the length of its trajectory. Obviously, movement and path should not be confused. The module of the displacement vector and the length of the path are the same only in the case of rectilinear motion.

In the SI system, displacement and path length are measured in meters.

The displacement is equal to the difference between the radius vectors at the start and end times. In other words, it is an increment of the radius vector.

Speed and acceleration

Average speed - vector physical quantity, equal to the ratio displacement vector to the time interval for which it occurred

And now imagine that the time interval decreases, decreases, and becomes very short, tends to zero. In that case about average speed I have to say, the speed becomes instantaneous. Those who remember the basics mathematical analysis, they will immediately understand that in the future we cannot do without a derivative.

Instantaneous velocity is a vector physical quantity equal to the time derivative of the radius vector. The instantaneous velocity is always directed tangentially to the trajectory.

In the SI system, speed is measured in meters per second.

If the body does not move uniformly and in a straight line, then it has not only speed, but also acceleration.

Acceleration (or instantaneous acceleration) is a vector physical quantity, the second derivative of the radius vector with respect to time, and, accordingly, the first derivative of the instantaneous speed

Acceleration shows how quickly the speed of a body changes. In the case of rectilinear motion, the directions of the velocity and acceleration vectors coincide. In the case curvilinear motion, the acceleration vector can be decomposed into two components: tangential acceleration, and acceleration is normal .

Tangential acceleration shows how quickly the speed of the body changes in absolute value and is directed tangentially to the trajectory

Normal acceleration characterizes the rate of change of speed in direction. Vectors of normal and tangential acceleration are mutually perpendicular, and the normal acceleration vector is directed to the center of the circle along which the point moves.

Here R is the radius of the circle along which the body moves

Here - x zero is the initial coordinate. v zero - initial speed. Differentiate with respect to time, and get the speed

The derivative of the speed with respect to time will give the value of the acceleration a, which is a constant.

Problem solution example

Now that we have considered physical foundations kinematics, it's time to consolidate knowledge in practice and solve some problem. And the sooner the better.

For example this one: A point moves in a circle with a radius of 4 meters. The law of its motion is expressed by the equation S=A+Bt^2. A=8m, B=-2m/s^2. At what point in time is the normal acceleration of a point equal to 9 m/s^2? Find the speed, tangential and total acceleration of the point for this moment in time.

Solution: we know that in order to find the speed, we need to take the first time derivative of the law of motion, and the normal acceleration is equal to the private square of the speed and the radius of the circle along which the point moves. Armed with this knowledge, we find the desired values.

Dear friends, congratulations! If you have read this article on the basics of kinematics, and in addition you have learned something new, you have already done a good deed! We sincerely hope that our "kinematics for dummies" will be useful to you. Dare and remember - we are always ready to help you with solving tricky puzzles with insidious cheap traps. . Good luck with your study of mechanics!

For example, a car that starts off moves faster as it increases its speed. At the starting point, the speed of the car is zero. Starting the movement, the car accelerates to a certain speed. If you need to slow down, the car will not be able to stop instantly, but for some time. That is, the speed of the car will tend to zero - the car will start to move slowly until it stops completely. But physics does not have the term "deceleration". If the body moves, decreasing speed, this process is also called acceleration, but with a "-" sign.

Average acceleration is the ratio of the change in speed to the time interval during which this change occurred. Calculate the average acceleration using the formula:

where is it . The direction of the acceleration vector is the same as the direction of the change in speed Δ = - 0

where 0 is initial speed. At the point in time t1(see figure below) the body has 0 . At the point in time t2 body has speed. Based on the vector subtraction rule, we determine the vector of speed change Δ = - 0 . From here we calculate the acceleration:

In the SI system unit of acceleration is called 1 meter per second per second (or meter per second squared):

A meter per second squared is the acceleration of a point moving in a straight line, at which the speed of this point increases by 1 m / s in 1 s. In other words, acceleration determines the degree of change in the speed of a body in 1 s. For example, if the acceleration is 5 m / s 2, then the speed of the body increases by 5 m / s every second.

Instantaneous acceleration of a body (material point) at a given moment of time - this is a physical quantity, which is equal to the limit to which the average acceleration tends when the time interval tends to 0. In other words, this is the acceleration developed by the body for very small segment time:

The acceleration has the same direction as the change in speed Δ in extremely small time intervals during which the speed changes. The acceleration vector can be set using projections on the corresponding coordinate axes in a given reference system (projections a X, a Y , a Z).

With accelerated rectilinear motion the speed of the body increases modulo, i.e. v 2 > v 1 , and the acceleration vector has the same direction as the velocity vector 2 .

If the modulo velocity of the body decreases (v 2< v 1), значит, у вектора ускорения направление противоположно направлению вектора скорости 2 . Другими словами, в таком случае наблюдаем deceleration(acceleration is negative, and< 0). На рисунке ниже изображено направление векторов ускорения при прямолинейном движении тела для случая ускорения и замедления.

If there is movement in curvilinear trajectory, then the modulus and direction of the velocity change. This means that the acceleration vector is represented as 2 components.

Tangential (tangential) acceleration call that component of the acceleration vector, which is directed tangentially to the trajectory at a given point of the trajectory of motion. Tangential acceleration describes the degree of change in speed modulo when making a curvilinear motion.

At tangential acceleration vectorsτ (see figure above) the direction is the same as that of linear speed or opposite to it. Those. the vector of tangential acceleration is in the same axis as the tangent circle, which is the trajectory of the body.

Velocity is a vector quantity that characterizes not only the speed of movement of a particle along a trajectory, but also the direction in which the particle moves at each moment of time.

Average speed over time from t1 before t2 is equal to the ratio of the movement during this time to the time interval for which this movement took place:

The fact that this is precisely the average speed we will note, concluding average value in angle brackets:<...>, as done above.

The above formula for the mean velocity vector is a direct consequence of the general mathematical definition mean value<f(x)> arbitrary function f(x) on the interval [ a,b]:

Really

The average speed may be too rough characteristic of the movement. For example, the average speed over a period of oscillations is always zero, regardless of the nature of these oscillations, for the simple reason that over a period - by the definition of a period - an oscillating body will return to starting point and hence the displacement per period is always zero. For this and a number of other reasons, instantaneous speed is introduced - the speed at a given moment in time. In the future, implying instantaneous speed, we will write simply: “speed”, omitting the words “instantaneous” or “at a given moment of time” whenever this cannot lead to misunderstandings. To obtain speed at a moment of time t gotta do the obvious thing: Calculate the ratio limit when aiming for a span of time t2 – t1 to zero. Let's rename: t1 = t and t 2 \u003d t + and rewrite the upper relation as:

Speed at time t is equal to the limit of the ratio of movement in time to the time interval during which this movement took place, when the latter tends to zero

Rice. 2.5. To the definition of instantaneous speed.

At the moment, we do not consider the question of the existence of this limit, assuming that it exists. Note that if there is a finite displacement and a finite interval of time, then and are their limit values: an infinitesimal displacement and an infinitesimal interval of time. So that right part speed detection

is nothing more than a fraction - a quotient of division by , so the last ratio can be rewritten and is quite often used in the form

By geometric sense derivative, the velocity vector at each point of the trajectory is directed tangentially to the trajectory at this point in its direction of motion.

Video 2.1. The velocity vector is directed tangentially to the trajectory. Sharpener experiment.

Any vector can be expanded in a basis (for unit vectors of the basis, in other words, unit vectors that determine the positive directions of the axes OX,OY,oz we use the notation , , or , respectively). The coefficients of this expansion are the projections of the vector onto the corresponding axes. The following is important: in the algebra of vectors, it is proved that the expansion in terms of the basis is unique. Let us expand the radius vector of some moving material point in terms of the basis

Taking into account the constancy of the Cartesian unit vectors , , , we will differentiate this expression with respect to time

On the other hand, the expansion in terms of the basis of the velocity vector has the form

comparing the last two expressions, taking into account the uniqueness of the expansion of any vector in terms of the basis, gives the following result: the projections of the velocity vector on the Cartesian axes are equal to the time derivatives of the corresponding coordinates, that is

The modulus of the velocity vector is

Let's get one more, important, expression for the modulus of the velocity vector.

It has already been noted that for the value || less and less different from the corresponding path (see Fig. 2). That's why

and in the limit (>0)

In other words, the modulus of speed is the derivative of the distance traveled with respect to time.

Finally we have:

Middle module velocity vector, is defined as follows:

The average value of the module of the velocity vector is equal to the ratio of the distance traveled to the time during which this path was traveled:

Here s(t1,t2)- path in time from t1 before t2 and correspondingly, s(t0,t2)- path in time from t0 before t2 and s(t0,t2)- path in time from t0 before t1.

Average vector speed, or simply average speed, as above, is

Note that, first of all, this is a vector, its module - the module of the average velocity vector should not be confused with the average value of the module of the velocity vector. AT general case they are not equal: the modulus of the average vector is not at all equal to the average modulus of this vector. Two operations: the calculation of the module and the calculation of the average, in the general case, cannot be swapped.

Consider an example. Let the point move in one direction. On fig. 2.6. shows a graph of the path she traveled s at the time (for the time from 0 before t). Using physical meaning speed, use this graph to find the point in time at which the instantaneous speed is equal to the average ground speed for the first seconds of the movement of the point.

Rice. 2.6. Determination of the instantaneous and average speed of the body

Velocity modulus at a given time

being the derivative of the path with respect to time, it is equal to the angular coefficient of the rocking to the dependence graph to the point corresponding to the moment of time t*. The average module of speed for a period of time from 0 before t* is the slope of the secant passing through the points of the same graph corresponding to the beginning t = 0 and end t = t* time interval. We need to find such a moment in time t* when both slope match. To do this, we draw a straight line through the origin of coordinates, tangent to the trajectory. As can be seen from the figure, the point of contact of this straight line s(t) and gives t*. In our example, we get

An example of solving a problem with a complex motion of a point is considered. The point moves in a straight line along the plate. The plate revolves around fixed axle. The absolute speed is determined and absolute acceleration points.

The theory used to solve the problem below is described on the page “Complex motion of a point, Coriolis theorem”.

The task

A rectangular plate rotates around a fixed axis according to the law φ = 6 t 2 - 3 t 3. The positive direction of reading the angle φ is shown in the figures by an arc arrow. Rotation axis OO 1 lies in the plane of the plate (the plate rotates in space).

The point M moves along the straight line BD along the plate. The law of its relative motion is given, i.e., the dependence s = AM = 40(t - 2 t 3) - 40(s - in centimeters, t - in seconds). Distance b = 20 cm. In the figure, point M is shown in the position where s = AM > 0 (for s< 0 point M is on the other side of point A).

Find the absolute speed and absolute acceleration of point M at time t 1 = 1 s.

Directions. This task is for a complex movement of a point. To solve it, it is necessary to use the theorems on the addition of velocities and on the addition of accelerations (the Coriolis theorem). Before performing all calculations, it is necessary to determine, according to the conditions of the problem, where the point M is located on the plate at the time t 1 = 1 s, and draw a point exactly in this position (and not in an arbitrary one shown in the figure for the problem).

The solution of the problem

Given: b= 20 cm, φ = 6 t 2 - 3 t 3, s = |AM| = 40(t - 2 t 3) - 40, t 1 = 1 s.

Find: v abs , a abs

Determining the position of a point

Determine the position of the point at time t = t 1 = 1 s.
s= 40(t 1 - 2 t 1 3) - 40 = 40 (1 - 2 1 3) - 40 \u003d -80 cm.
Because s< 0 , then point M is closer to point B than to D.
|AM| = |-80| = 80 cm.
We make a drawing.

According to the velocity addition theorem, the absolute velocity of a point is equal to the vector sum of the relative and translational velocities:
.

Determining the relative speed of a point

Determine the relative speed. To do this, we assume that the plate is stationary, and the point M makes a given movement. That is, the point M moves along the straight line BD. Differentiating s with respect to time t , we find the projection of the velocity onto the direction BD :
.
At time t = t 1 = 1 s,
cm/s.
Since , then the vector is directed in the direction opposite to BD . That is, from point M to point B. Relative velocity module
v from = 200 cm/s.

Determining the transfer speed of a point

Determining the carry speed. To do this, we assume that the point M is rigidly connected to the plate, and the plate performs a given movement. That is, the plate rotates around the OO 1 axis. Differentiating φ with respect to time t, we find the angular velocity of rotation of the plate:
.
At time t = t 1 = 1 s,
.
Since , then the angular velocity vector is directed towards the positive angle of rotation φ, that is, from the point O to the point O 1 . Angular velocity module:
ω = 3 s -1.
We depict the vector of the angular velocity of the plate in the figure.

From the point M we lower the perpendicular HM to the axis OO 1 .
During translational motion, the point M moves along a circle of radius |HM| centered at point H.
|HM| = |HK| + |KM| = 3b + |AM| sin 30° = 60 + 80 0.5 = 100 cm;
Carrying speed:
v lane = ω|HM| = 3 100 = 300 cm/s.

The vector is directed tangentially to the circle in the direction of rotation.

Determining the absolute speed of a point

Determine the absolute speed. The absolute speed of a point is equal to the vector sum of the relative and translational speeds:
.
Draw the axes of the fixed coordinate system Oxyz . Let us direct the z axis along the axis of rotation of the plate. Let the x-axis be perpendicular to the plate at the considered moment of time, the y-axis lies in the plane of the plate. Then the relative velocity vector lies in the yz plane. The translational velocity vector is directed opposite to the x axis. Since the vector is perpendicular to the vector, then, according to the Pythagorean theorem, the absolute velocity modulus:
.

Determining the absolute acceleration of a point

According to the acceleration addition theorem (Coriolis theorem), the absolute acceleration of a point is equal to the vector sum of the relative, translational and Coriolis accelerations:
,
where
- Coriolis acceleration.

Definition of relative acceleration

Determine relative acceleration. To do this, we assume that the plate is stationary, and the point M makes a given movement. That is, the point M moves along the straight line BD. Differentiating s twice with respect to time t , we find the projection of the acceleration onto the direction BD :
.
At time t = t 1 = 1 s,
cm/s 2 .
Since , then the vector is directed in the direction opposite to BD . That is, from point M to point B. Relative acceleration module
a from = 480 cm/s 2.
We represent the vector in the figure.

Translational Acceleration Definition

Define portable acceleration. During translational motion, the point M is rigidly connected to the plate, that is, it moves along a circle of radius |HM| centered at point H. Let us decompose the portable acceleration into the tangent to the circle and the normal acceleration:
.
Differentiating φ twice with respect to time t, we find the projection of the angular acceleration of the plate onto the axis OO 1 :
.
At time t = t 1 = 1 s,
with -2 .
Since , then the angular acceleration vector is directed in the direction opposite to the positive angle of rotation φ, that is, from point O 1 to point O. Angular acceleration module:
ε = 6 s -2.
We depict the vector of the angular acceleration of the plate in the figure.

Portable tangential acceleration:
a τ lane = ε |HM| \u003d 6 100 \u003d 600 cm / s 2.
The vector is tangent to the circle. Since the angular acceleration vector is directed in the direction opposite to the positive angle of rotation φ , it is directed in the direction opposite to the positive direction of rotation φ . That is, it is directed towards the x-axis.

Portable normal acceleration:
a n lane = ω 2 |HM| = 3 2 100 = 900 cm/s 2.
The vector is directed towards the center of the circle. That is, in the direction opposite to the y axis.

Definition of Coriolis acceleration

Coriolis (rotary) acceleration:
.
The angular velocity vector is directed along the z axis. The relative velocity vector is directed along the straight line |DB| . The angle between these vectors is 150°. By property vector product,
.
The direction of the vector is determined by the gimlet rule. If the gimlet handle is turned from position to position , then the gimlet screw will move in the direction opposite to the x axis.

Definition of absolute acceleration

Absolute acceleration:
.
Design it vector equation on the xyz axis of the coordinate system.

;

;

.
Absolute acceleration module:

.

Answer

Absolute speed ;
absolute acceleration.

Portal for the student. Self-training

Trajectory, radius vector, body motion law

Speed ​​and acceleration

Problem solution example

The task

The solution of the problem

Determining the position of a point

Determining the relative speed of a point

Determining the transfer speed of a point

Determining the absolute speed of a point

Determining the absolute acceleration of a point

Definition of relative acceleration

Translational Acceleration Definition

Definition of Coriolis acceleration

Definition of absolute acceleration

Answer

RELATED ARTICLES

Speed and acceleration