Rectilinear uniform motion. Determining the type of movement according to the schedule

GRAPHICS

Determining the type of movement according to the schedule

1. Uniformly accelerated motion corresponds to a graph of the acceleration modulus versus time, indicated in the figure by the letter

1) A

2) B

3) IN

4) G

2. The figures show graphs of the acceleration modulus versus time for different types movements. Which graph corresponds to uniform motion?

1 4

3.
Body moving along an axis Oh rectilinearly and uniformly accelerated, over some time it reduced its speed by 2 times. Which of the graphs of the projection of acceleration versus time corresponds to such a movement?

1 4

4. The parachutist moves vertically downward at a constant speed. Which graph - 1, 2, 3 or 4 - correctly reflects the dependence of its coordinates Y from the time of movement t relative to the surface of the earth? Neglect air resistance.

1) 3 4) 4

5. Which of the graphs of the projection of velocity versus time (Fig.) corresponds to the movement of a body thrown vertically upward with a certain speed (axis Y directed vertically upward)?

13 4) 4

6.
A body is thrown vertically upward with a certain initial speed from the surface of the earth. Which of the graphs of the height of a body above the earth's surface versus time (Fig.) corresponds to this movement?

12

Determination and comparison of movement characteristics according to the schedule

7. The graph shows the dependence of the projection of the body’s velocity on time during rectilinear motion. Determine the acceleration projection of the body.

1) – 10 m/s2

2) – 8 m/s2

3) 8 m/s2

4) 10 m/s2

8. The figure shows a graph of the speed of movement of bodies versus time. What is the acceleration of the body?

1) 1 m/s2

2) 2 m/s2

3) 3 m/s2

4) 18 m/s2

9. According to the graph of the velocity projection versus timenor presentedIn the figure, determine the acceleration modulus linearlymoving body in point in time t= 2 s.

1) 2 m/s2

2) 3 m/s2

3) 10 m/s2

4) 27 m/s2

10. x = 0, and point B at point x = 30 km. What is the speed of the bus on the way from A to B?

1) 40 km/h

2) 50 km/h

3) 60 km/h

4) 75 km/h

11. The figure shows a bus schedule from point A to point B and back. Point A is at point x = 0, and point B at point x = 30 km. What is the speed of the bus on the way from B to A?

1) 40 km/h

2) 50 km/h

3) 60 km/h

4) 75 km/h

12. A car is moving along a straight street. The graph shows the dependence of the car's speed on time. The acceleration module is maximum in the time interval

1) from 0 s to 10 s

2) from 10 s to 20 s

3) from 20 s to 30 s

font-family: " times new novel>4) from 30 s to 40 s

13. Four bodies move along an axis Oh.The figure shows graphs of the dependence of velocity projectionsυx from time to time t for these bodies. Which body moves with the least absolute acceleration?

1) 3 4) 4

14. The figure shows the path dependence graphScyclist from time to timet. Determine the time interval when the cyclist was moving at a speed of 2.5 m/s.

1) from 5 s to 7 s

2) from 3 s to 5 s

3) from 1 s to 3 s

4) from 0 to 1 s

15. The figure shows a graph of the dependence of the coordinates of a body moving along the axisOX, from time to time. Compare speedsv1 , v2 Andv3 bodies at moments in time t1, t2, t3

1) v1 > v2 = v3

2) v1 > v2 > v3

3) v1 < v2 < v3

4) v 1 = v 2 > v 3

16. The figure shows a graph of the projection of the squarebody growth over time.

The projection of body acceleration in the time interval from 5 to 10 s is presented by the graph

13 4) 4

17. A material point moves rectilinearly with acceleration, the time dependence of which is shown in the figure. The initial speed of the point is 0. Which point on the graph corresponds to the maximum speed of the material point:

1) 2

2) 3

3) 4

4) 5

Drawing up kinematic dependencies (functions of the dependence of kinematic quantities on time) according to the schedule

18. In Fig. shows a graph of body coordinates versus time. Determine the kinematic law of motion of this body

1) x( t) = 2 + 2 t

2) x( t) = – 2 – 2 t

3) x( t) = 2 – 2 t

4) x ( t ) = – 2 + 2 t

19. Using the graph of the velocity of a body versus time, determine the function of the velocity of this body versus time

1) vx= – 30 + 10 t

2) vx = 30 + 10 t

3) v x = 30 – 10 t

4) vx = – 30 + 10 t

Determination of movement and path according to schedule

20. Using a graph of the velocity of a body versus time, determine the distance covered by a rectilinearly moving body in 3 s.

1) 2 m

2) 4 m

3) 18 m

4) 36 m

21. A stone is thrown vertically upward. The projection of its velocity onto the vertical direction changes over time according to the graph in the figure. What is the distance traveled by the stone in the first 3 s?

1) 30 m

2) 45 m

3) 60 m

4) 90 m

22. A stone is thrown vertically upward. The projection of its speed onto the vertical direction changes over time according to the graph in the figure for section 21. What is the distance traveled by the stone during the entire flight?

1) 30 m

2) 45 m

3) 60 m

4) 90 m

23. A stone is thrown vertically upward. The projection of its speed onto the vertical direction changes over time according to the graph in the figure for section 21. What is the movement of the stone in the first 3 s?

1) 0 m

2) 30 m

3) 45 m

4) 60 m

24. A stone is thrown vertically upward. The projection of its speed onto the vertical direction changes over time according to the graph in the figure for section 21. What is the displacement of the stone during the entire flight?

1) 0 m

2) 30 m

3) 60 m

4) 90 m

25. The figure shows a graph of the projection of the velocity of a body moving along the Ox axis as a function of time. What is the distance traveled by the body at time t = 10 s?

1) 1m

2) 6 m

3) 7 m

4) 13 m

26. position:relative; z-index:24">The cart begins to move from rest along the paper strip. There is a dropper on the cart, which leaves paint spots on the tape at regular intervals.

Choose a graph of velocity versus time that correctly describes the motion of the cart.

1 4

EQUATIONS

27. The movement of a trolleybus during emergency braking is given by the equation: x = 30 + 15t – 2.5 t2, m What is the initial coordinate of the trolleybus?

1) 2.5 m

2) 5 m

3) 15 m

4) 30 m

28. The movement of the aircraft during the take-off run is given by the equation: x = 100 + 0.85t2, m What is the acceleration of the plane?

1) 0 m/s2

2) 0.85 m/s2

3) 1.7 m/s2

4) 100 m/s2

29. The movement of a passenger car is given by the equation: x = 150 + 30t + 0.7t2, m. What is the initial speed of the car?

1) 0.7 m/s

2) 1.4 m/s

3) 30 m/s

4) 150 m/s

30. Equation for the dependence of the projection of the speed of a moving body on time:vx= 2 +3t(m/s). What is the corresponding projection equation for the displacement of the body?

1) Sx = 2 t + 3 t2 2) Sx = 4 t + 3 t2 3) Sx = t + 6 t2 4) Sx = 2 t + 1,5 t 2

31. The dependence of the coordinate on time for a certain body is described by the equation x = 8t – t2. At what point in time is the speed of the body equal to zero?

1) 8 s

2) 4 s

3) 3 s

4) 0 s

TABLES

32. X uniform motion of a body as a function of time t:

t, With
X , m

At what speed did the body move from time 0 s to motime ment 4 s?

1) 0.5 m/s

2) 1.5 m/s

3) 2 m/s

4) 3 m/s

33. The table shows the dependence of the coordinates X body movements over time t:

t, With
X, m

Determine the average speed of the body in the time interval from 1s to 3s.

1) 0 m/s

2) ≈0.33 m/s

3) 0.5 m/s

4) 1 m/s

t, With	0	1	2	3	4	5
x1 m
x2, m
x3, m
x4, m

Which body could have a constant velocity and different from zero?

1) 1

35. Four bodies moved along the Ox axis. The table shows the dependence of their coordinates on time.

t, With	0	1	2	3	4	5
x1 m
x2, m
x3, m
x4, m

Which body could have constant acceleration and be different from zero?

3.1. Uniform motion in a straight line.

3.1.1. Uniform motion in a straight line- movement in a straight line with acceleration constant in magnitude and direction:

3.1.2. Acceleration()- a physical vector quantity showing how much the speed will change in 1 s.

In vector form:

where is the initial speed of the body, is the speed of the body at the moment of time t.

In projection onto the axis Ox:

where is the projection of the initial velocity onto the axis Ox, - projection of the body velocity onto the axis Ox at a point in time t.

The signs of the projections depend on the direction of the vectors and the axis Ox.

3.1.3. Projection graph of acceleration versus time.

With uniformly alternating motion, the acceleration is constant, therefore it will appear as straight lines parallel to the time axis (see figure):

3.1.4. Speed ​​during uniform motion.

In vector form:

In projection onto the axis Ox:

For uniformly accelerated motion:

For uniform slow motion:

3.1.5. Projection graph of speed versus time.

The graph of the projection of speed versus time is a straight line.

Direction of movement: if the graph (or part of it) is above the time axis, then the body is moving in the positive direction of the axis Ox.

Acceleration value: the greater the tangent of the angle of inclination (the steeper it rises up or down), the greater the acceleration module; where is the change in speed over time

Intersection with the time axis: if the graph intersects the time axis, then before the intersection point the body slowed down (uniformly slow motion), and after the intersection point it began to accelerate in the opposite direction (uniformly accelerated motion).

3.1.6. Geometric meaning area under the graph in axes

Area under the graph when on the axis Oy the speed is delayed, and on the axis Ox- time is the path traveled by the body.

In Fig. 3.5 shows the case of uniformly accelerated motion. The path in this case will be equal to the area of ​​the trapezoid: (3.9)

3.1.7. Formulas for calculating path

Uniformly accelerated motion	Equal slow motion
(3.10)	(3.12)
(3.11)	(3.13)
(3.14)

All formulas presented in the table work only when the direction of movement is maintained, that is, until the straight line intersects the time axis on the graph of the velocity projection versus time.

If the intersection has occurred, then the movement is easier to divide into two stages:

before crossing (braking):

After the intersection (acceleration, movement in reverse side)

In the formulas above - the time from the beginning of movement to the intersection with the time axis (time before stopping), - the path that the body has traveled from the beginning of movement to the intersection with the time axis, - the time elapsed from the moment of crossing the time axis to this moment t, - the path that the body has traveled in the opposite direction during the time elapsed from the moment of crossing the time axis to this moment t, - the module of the displacement vector for the entire time of movement, L- the path traveled by the body during the entire movement.

3.1.8. Movement in the th second.

During this time the body will travel the following distance:

During this time the body will travel the following distance:

Then during the th interval the body will travel the following distance:

Any period of time can be taken as an interval. Most often with.

Then in 1 second the body travels the following distance:

In 2 seconds:

In 3 seconds:

If we look carefully, we will see that, etc.

Thus, we arrive at the formula:

In words: the paths traversed by a body over successive periods of time are related to each other as a series of odd numbers, and this does not depend on the acceleration with which the body moves. We emphasize that this relation is valid for

3.1.9. Equation of body coordinates for uniform motion

Coordinate equation

The signs of the projections of the initial velocity and acceleration depend on the relative position of the corresponding vectors and the axis Ox.

To solve problems, it is necessary to add to the equation the equation for changing the velocity projection onto the axis:

3.2. Graphs of kinematic quantities for rectilinear motion

3.3. Free fall body

By free fall we mean the following physical model:

1) The fall occurs under the influence of gravity:

2) There is no air resistance (in problems they sometimes write “neglect air resistance”);

3) All bodies, regardless of mass, fall with the same acceleration (sometimes they add “regardless of the shape of the body,” but we are considering the movement of only a material point, so the shape of the body is no longer taken into account);

4) The acceleration of gravity is directed strictly downward and is equal on the surface of the Earth (in problems we often assume for convenience of calculations);

3.3.1. Equations of motion in projection onto the axis Oy

Unlike movement along a horizontal straight line, when not all tasks involve a change in direction of movement, in free fall it is best to immediately use the equations written in projections onto the axis Oy.

Body coordinate equation:

Velocity projection equation:

As a rule, in problems it is convenient to select the axis Oy as follows:

Axis Oy directed vertically upward;

The origin coincides with the level of the Earth or the lowest point of the trajectory.

With this choice, the equations will be rewritten in the following form:

3.4. Movement in a plane Oxy.

We considered the motion of a body with acceleration along a straight line. However, the uniformly variable motion is not limited to this. For example, a body thrown at an angle to the horizontal. In such problems, it is necessary to take into account movement along two axes at once:

Or in vector form:

And changing the projection of speed on both axes:

3.5. Application of the concept of derivative and integral

We will not provide a detailed definition of the derivative and integral here. To solve problems we need only a small set of formulas.

Derivative:

Where A, B and that is, constant values.

Integral:

Now let's see how the concepts of derivative and integral apply to physical quantities. In mathematics, the derivative is denoted by """, in physics, the derivative with respect to time is denoted by "∙" above the function.

Speed:

that is, the speed is a derivative of the radius vector.

For velocity projection:

Acceleration:

that is, acceleration is a derivative of speed.

For acceleration projection:

Thus, if the law of motion is known, then we can easily find both the speed and acceleration of the body.

Now let's use the concept of integral.

Speed:

that is, the speed can be found as the time integral of the acceleration.

Radius vector:

that is, the radius vector can be found by taking the integral of the velocity function.

Thus, if the function is known, we can easily find both the speed and the law of motion of the body.

The constants in the formulas are determined from initial conditions- values ​​and at time

3.6. Velocity triangle and displacement triangle

3.6.1. Speed ​​triangle

In vector form with constant acceleration, the law of speed change has the form (3.5):

This formula means that a vector is equal to the vector sum of vectors and the vector sum can always be depicted in a figure (see figure).

In each problem, depending on the conditions, the velocity triangle will have its own form. This representation allows the use of geometric considerations in the solution, which often simplifies the solution of the problem.

3.6.2. Triangle of movements

In vector form, the law of motion with constant acceleration has the form:

When solving a problem, you can choose the reference system in the most convenient way, therefore, without loss of generality, we can choose the reference system in such a way that, that is, we place the origin of the coordinate system at the point where the body is located at the initial moment. Then

that is, the vector is equal to the vector sum of the vectors and Let us depict it in the figure (see figure).

As in the previous case, depending on the conditions, the displacement triangle will have its own shape. This representation allows the use of geometric considerations in the solution, which often simplifies the solution of the problem.

Uniform movement– this is movement at a constant speed, that is, when the speed does not change (v = const) and acceleration or deceleration does not occur (a = 0).

Straight-line movement- this is movement in a straight line, that is, the trajectory of rectilinear movement is a straight line.

Uniform linear movement- this is a movement in which a body makes equal movements over any equal periods of time. For example, if we divide a certain time interval into one-second intervals, then with uniform motion the body will move the same distance for each of these time intervals.

The speed of uniform rectilinear motion does not depend on time and at each point of the trajectory is directed in the same way as the movement of the body. That is, the displacement vector coincides in direction with the velocity vector. In this case, the average speed for any period of time is equal to the instantaneous speed:

V cp = v

Distance traveled in linear motion is equal to the displacement module. If the positive direction of the OX axis coincides with the direction of movement, then the projection of the velocity onto the OX axis is equal to the magnitude of the velocity and is positive:

V x = v, that is v > 0

The projection of displacement onto the OX axis is equal to:

S = vt = x – x 0

where x 0 is the initial coordinate of the body, x is the final coordinate of the body (or the coordinate of the body at any time)

Equation of motion, that is, the dependence of the body coordinates on time x = x(t), takes the form:

X = x 0 + vt

If the positive direction of the OX axis is opposite to the direction of motion of the body, then the projection of the body’s velocity onto the OX axis is negative, the speed is less than zero (v< 0), и тогда уравнение движения принимает вид:

X = x 0 - vt

Dependence of speed, coordinates and path on time

The dependence of the projection of the body velocity on time is shown in Fig. 1.11. Since the speed is constant (v = const), the speed graph is a straight line parallel to the time axis Ot.

Rice. 1.11. Dependence of the projection of body velocity on time for uniform rectilinear motion.

The projection of movement onto the coordinate axis is numerically equal to the area of ​​the rectangle OABC (Fig. 1.12), since the magnitude of the movement vector is equal to the product of the velocity vector and the time during which the movement was made.

Rice. 1.12. Dependence of the projection of body displacement on time for uniform rectilinear motion.

A graph of displacement versus time is shown in Fig. 1.13. The graph shows that the projection of the velocity is equal to

V = s 1 / t 1 = tan α

where α is the angle of inclination of the graph to the time axis. The greater the angle α, the faster the body moves, that is, the greater its speed (the longer the body travels in less time). The tangent of the tangent to the graph of the coordinate versus time is equal to the speed:

Tg α = v

Rice. 1.13. Dependence of the projection of body displacement on time for uniform rectilinear motion.

The dependence of the coordinate on time is shown in Fig. 1.14. From the figure it is clear that

Tg α 1 > tg α 2

therefore, the speed of body 1 is higher than the speed of body 2 (v 1 > v 2).

Tg α 3 = v 3< 0

If the body is at rest, then the coordinate graph is a straight line parallel to the time axis, that is

X = x 0

Rice. 1.14. Dependence of body coordinates on time for uniform rectilinear motion.

Uniform linear movement- This is a special case of uneven movement.

Uneven movement- this is a movement in which a body (material point) makes unequal movements over equal periods of time. For example, a city bus moves unevenly, since its movement consists mainly of acceleration and deceleration.

Equally alternating motion- this is a movement in which the speed of a body (material point) changes equally over any equal periods of time.

Acceleration of a body during uniform motion remains constant in magnitude and direction (a = const).

Uniform motion can be uniformly accelerated or uniformly decelerated.

Uniformly accelerated motion- this is the movement of a body (material point) with positive acceleration, that is, with such movement the body accelerates with constant acceleration. In the case of uniformly accelerated motion, the velocity module of the body increases over time, the direction of acceleration coincides with the direction of the speed of movement.

Equal slow motion- this is the movement of a body (material point) with negative acceleration, that is, with such movement the body uniformly slows down. In uniformly slow motion, the velocity and acceleration vectors are opposite, and the velocity modulus decreases over time.

In mechanics, any rectilinear motion is accelerated, therefore slow motion differs from accelerated motion only in the sign of the projection of the acceleration vector onto the selected axis of the coordinate system.

Average variable speed is determined by dividing the movement of the body by the time during which this movement was made. The unit of average speed is m/s.

V cp = s/t

is the speed of a body (material point) at a given moment of time or at a given point of the trajectory, that is, the limit to which the average speed tends with an infinite decrease in the time period Δt:

Instantaneous velocity vector uniformly alternating motion can be found as the first derivative of the displacement vector with respect to time:

Velocity vector projection on the OX axis:

V x = x’

this is the derivative of the coordinate with respect to time (the projections of the velocity vector onto other coordinate axes are similarly obtained).

is a quantity that determines the rate of change in the speed of a body, that is, the limit to which the change in speed tends with an infinite decrease in the time period Δt:

Acceleration vector of uniformly alternating motion can be found as the first derivative of the velocity vector with respect to time or as the second derivative of the displacement vector with respect to time:

If a body moves rectilinearly along the OX axis of a rectilinear Cartesian coordinate system, coinciding in direction with the body’s trajectory, then the projection of the velocity vector onto this axis is determined by the formula:

V x = v 0x ± a x t

The “-” (minus) sign in front of the projection of the acceleration vector refers to uniformly slow motion. The equations for projections of the velocity vector onto other coordinate axes are written similarly.

Since in uniform motion the acceleration is constant (a = const), the acceleration graph is a straight line parallel to the 0t axis (time axis, Fig. 1.15).

Rice. 1.15. Dependence of body acceleration on time.

Dependence of speed on time is a linear function, the graph of which is a straight line (Fig. 1.16).

Rice. 1.16. Dependence of body speed on time.

Speed ​​versus time graph(Fig. 1.16) shows that

In this case, the displacement is numerically equal to the area of ​​the figure 0abc (Fig. 1.16).

The area of ​​a trapezoid is equal to the product of half the sum of the lengths of its bases and its height. The bases of the trapezoid 0abc are numerically equal:

0a = v 0 bc = v

The height of the trapezoid is t. Thus, the area of ​​the trapezoid, and therefore the projection of displacement onto the OX axis is equal to:

In the case of uniformly slow motion, the acceleration projection is negative and in the formula for the displacement projection a “–” (minus) sign is placed before the acceleration.

A graph of the velocity of a body versus time at various accelerations is shown in Fig. 1.17. The graph of displacement versus time for v0 = 0 is shown in Fig. 1.18.

Rice. 1.17. Dependence of body speed on time for different meanings acceleration.

Rice. 1.18. Dependence of body movement on time.

The speed of the body at a given time t 1 is equal to the tangent of the angle of inclination between the tangent to the graph and the time axis v = tg α, and the displacement is determined by the formula:

If the time of movement of the body is unknown, you can use another displacement formula by solving a system of two equations:

It will help us derive the formula for displacement projection:

Since the coordinate of the body at any time is determined by the sum of the initial coordinate and the displacement projection, it will look like this:

The graph of the coordinate x(t) is also a parabola (like the displacement graph), but the vertex of the parabola is at general case does not coincide with the origin. When a x< 0 и х 0 = 0 ветви параболы направлены вниз (рис. 1.18).