§1.18. graphs of the dependence of the module and projection of acceleration and the module and projection of velocity on time when moving with constant acceleration. Uniform linear motion Projection of the speed of movement of a body how to find
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.
This is a movement in which a body makes equal movements at any equal intervals 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:
vcp = vSpeed of uniform rectilinear motion is a physical vector quantity equal to the ratio of the movement of a body over any period of time to the value of this interval t:
=/t
Thus, the speed of uniform rectilinear motion shows how much movement a material point makes per unit time.
Moving with uniform linear motion is determined by the formula:
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:
vx = v, that is v > 0The projection of displacement onto the OX axis is equal to:
s = vt = x - x0where 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 = x0 + vtIf 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 = x0 - vtUniform 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 modulus of the body’s velocity increases over time, and 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.
vcp = s/tThis 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 interval 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:
vx = 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).
This 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 interval 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:
= " = " Considering that 0 is the speed of the body at the initial moment of time (initial speed), is the speed of the body at a given moment of time (final speed), t is the period of time during which the change in speed occurred, will be as follows:
From here uniform speed formula at any time:
0 + t 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:
vx = v0x ± axtThe “-” (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 = v0 bc = vThe 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 acceleration values.
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 moment in 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 graph of displacement), but the vertex of the parabola in the general case does not coincide with the origin. When a x< 0 и х 0 = 0 ветви параболы направлены вниз (рис. 1.18).
To perform calculations of velocities and accelerations, it is necessary to move from writing equations in vector form to writing equations in algebraic form.
Initial velocity and acceleration vectors 
may have different directions, so the transition from vector to algebraic writing of equations can be very labor-intensive.
It is known that the projection of the sum of two vectors onto any coordinate axis is equal to the sum of the projections of the summands of the vectors onto the same axis.
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Therefore, to find the projection
The projection of a vector onto an axis is considered positive if it is necessary to go from the projection of the beginning to the projection of the end of the vector in the direction of the axis, and negative in the opposite case. |
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velocity vector 
on an arbitrary axis OX you need to find the algebraic sum of projections of vectors 
And 
on the same axis.
Speed graph
From Eq. 
it follows that the graph of the projection of the speed of uniformly accelerated motion versus time is a straight line. If the projection of the initial velocity onto the OX axis is zero, then the straight line passes through the origin.
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Main types of movement
A n = 0, a = 0 – rectilinear uniform motion;
A n = 0, a = const– rectilinear uniform motion;
A n = 0, a 0 – rectilinear with variable acceleration;
A n = const, a = 0 – uniform around the circumference
A n = const, a = const– uniformly variable around the circumference
A n const, a const– curvilinear with variable acceleration.
Rotational motion of a rigid body.
Rotational motion of a rigid body relative to a fixed axis - a movement in which all points of a rigid body describe circles whose centers lie on the same straight line, called axis of rotation.
Uniform movement around a circle
Let's consider the simplest type of rotational motion, and pay special attention to centripetal acceleration.
With uniform motion in a circle, the speed value remains constant, and the direction of the speed vector 
changes during movement.
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Over time interval ∆
t the body goes through the journey |
. This path is equal to the arc length AB. Velocity vectors 
And 
at points A And B are directed tangent to the circle at these points, and the angle
between vectors 
And 
equal to the angle between the radii O.A. And O.B. Let's find the vector difference 
and determine the ratio of the change in speed to ∆
t:
From the similarity of triangles OAB and BCD it follows
If the time interval ∆t is small, then the angle is also small. At small values of the angle , the length of the chord AB is approximately equal to the length of the arc AB, i.e. 
. Because 
,
, then we get
.
Because the 
, then we get
Period and frequency
The period of time during which a body makes a complete revolution when moving in a circle is called circulation periods (T). Because circumference is equal to 2 R, period of revolution for uniform motion of a body with speed v in a circle of radius R equals:
The reciprocal of the period of revolution is called frequency. Frequency shows how many revolutions a body makes in a circle per unit time:
(s -1)
Speed is one of the main characteristics. It expresses the very essence of the movement, i.e. determines the difference that exists between a stationary body and a moving body.
The SI unit of speed is m/s.
It is important to remember that speed is a vector quantity. The direction of the velocity vector is determined by the movement. The velocity vector is always directed tangentially to the trajectory at the point through which the moving body passes (Fig. 1).
For example, consider the wheel of a moving car. The wheel rotates and all points of the wheel move in circles. The splashes flying from the wheel will fly along tangents to these circles, indicating the directions of the velocity vectors of individual points of the wheel.
Thus, speed characterizes the direction of movement of a body (direction of the velocity vector) and the speed of its movement (modulus of the velocity vector).
Negative speed
Can the speed of a body be negative? Yes maybe. If the speed of a body is negative, this means that the body is moving in the direction opposite to the direction of the coordinate axis in the chosen reference system. Figure 2 shows the movement of a bus and a car. The speed of the car is negative and the speed of the bus is positive. It should be remembered that when we talk about the sign of velocity, we mean the projection of the velocity vector onto the coordinate axis.
Uniform and uneven movement
In general, speed depends on time. According to the nature of the dependence of speed on time, movement can be uniform or uneven.
DEFINITION
Uniform movement– this is movement with a constant modulus speed.
In case of uneven movement we speak of:
Examples of solving problems on the topic “Speed”
EXAMPLE 1
| Exercise | The car covered the first half of the journey between two settlements at a speed of 90 km/h, and the second half at a speed of 54 km/h. Determine the average speed of the car. |
| Solution | It would be incorrect to calculate the average speed of a car as the arithmetic mean of the two indicated speeds. Let's use the definition of average speed: Since rectilinear uniform motion is assumed, the signs of the vectors can be omitted. Time spent by the car to cover the entire distance: where is the time spent on completing the first half of the path, and is the time spent on completing the second half of the path.
The total movement is equal to the distance between populated areas, i.e. . Substituting these ratios into the formula for average speed, we get: Let's convert the speeds in individual sections to the SI system: Then the average speed of the car is:
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| Answer | The average speed of the car is 18.8 m/s |
(m/s)EXAMPLE 2
| Exercise | A car travels for 10 seconds at a speed of 10 m/s and then drives for another 2 minutes at a speed of 25 m/s. Determine the average speed of the car. |
| Solution | Let's make a drawing. |
The projections of the velocities of two points of a rigid body onto an axis passing through these points are equal to each other.
vA cos α = v B cos β.
Proof
Let us choose a rectangular fixed coordinate system Oxyz. Let's take two arbitrary points of a rigid body A and B. Let (x A , y A , z A ) And (x B , y B , z B )- coordinates of these points. When a rigid body moves, they are functions of time t. Differentiating with respect to time, we obtain projections of the velocities of the points.
,
.
Let us take advantage of the fact that when a rigid body moves, the distance | AB| between points remains constant, that is, does not depend on time t. Also constant is the square of the distance
.
Let's differentiate this equation with respect to time t, applying the rule for differentiating a complex function.
Let's shorten it by 2
.
(1)
Let's introduce the vector
.
Then the equation (1)
can be represented as a scalar product of vectors.
(2)
We carry out transformations.
;
(3)
.
By the scalar product property
,
.
Substitute in (3)
and reduce by | AB|.
;
Q.E.D.
Relative speed
Consider the movement of point B relative to point A. Let us introduce the relative speed of point B relative to A.
Then the equation (2)
can be rewritten in the form
.
That is, the relative speed is perpendicular to the vector drawn from point A to point B. Since point B is taken arbitrarily, the relative speed of any point on a rigid body is perpendicular to the radius vector drawn from point A. That is, relative to point A, the body undergoes rotational motion. The relative speed of body points is determined by the formula for rotational motion
.
Point A, relative to which motion is considered, is often called pole.
The absolute speed of point B relative to a fixed coordinate system can be written in the following form:
.
It is equal to the sum of the speed of translational motion of an arbitrary point A (pole) and the speed of rotational motion relative to pole A.
Example of problem solution
The task
Wheels 1 and 2 with radii R 1 = 0.15 m and R 2 = 0.3 m, respectively, are connected by hinges to a rod of 3 lengths | AB| = 0.5 m. Wheel 1 rotates with angular velocity ω 1 = 1 rad/s. For the position of the mechanism shown in the figure, determine the angular velocity ω 2 wheels 2. Take L = 0.3 m.
The solution of the problem
Point A moves in a circle radius R 1
around the center of rotation O 1
. The speed of point A is determined by the formula
V A = ω 1 R 1.
The vector is directed vertically (perpendicular to O 1 A).
Point B moves in a circle radius R 2
around the center of rotation O 2
. The speed of point B is determined by the formula
V B = ω 2 R 2.
From here
.
The vector is directed horizontally (perpendicular to O 2 B).
We are building right triangle ABC. We apply the Pythagorean theorem.
(m)
.
The cosine of the angle between the velocity vector and the straight line AB, in the direction of the vector, is equal to
.
By velocity projection theorem two points of a rigid body on a straight line we have:
V A cos α = V B cos β.
From here
.
Finding the angular velocity of wheel 2.
rad/s .
Definition
Uniform rectilinear motion is motion at a constant speed, in which there is no acceleration, and the trajectory of motion is a straight line.
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: $\left\langle v\right\rangle =v$
Definition
The speed of uniform rectilinear motion is a physical vector quantity equal to the ratio of the movement of the body $\overrightarrow(S)$ for any period of time to the value of this interval t:
$$\overrightarrow(v)=\frac(\overrightarrow(S))(t)$$
Thus, the speed of uniform rectilinear motion shows how much movement a material point makes per unit time.
Displacement during uniform linear motion is determined by the formula:
$$ \overrightarrow(S) = \overrightarrow(v) \cdot t $$
The distance traveled during rectilinear 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: $s = v_t = x - x0$
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)
The equation of motion, that is, the dependence of the body coordinates on time $x = x(t)$, takes the form: $x = x_0 + v_t$
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 $
The dependence of the projection of the body velocity on time is shown in Fig. 1. Since the speed is constant ($v = const$), the speed graph is a straight line parallel to the time axis Ot.
Rice. 1. Dependence of the projection of the velocity of a body 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. 2), 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. 2. Dependence of the projection of body displacement on time for uniform rectilinear motion.
A graph of displacement versus time is shown in Fig. 3. From the graph it is clear that the projection of the velocity onto the Ot axis is numerically equal to the tangent of the angle of inclination of the graph to the time axis:
Rice. 3. Dependence of the projection of body displacement on time for uniform rectilinear motion.
The dependence of the coordinate on time is shown in Fig. 4. From the figure it is clear that
tg $\alpha $1 $>$ tg $\alpha $2, therefore, the speed of body 1 is higher than the speed of body 2 (v1 $>$ v2).
tg $\alpha $3 = v3 $
Rice. 4. Dependence of body coordinates on time for uniform rectilinear motion.
If the body is at rest, then the coordinate graph is a straight line parallel to the time axis, that is, x = x0
Problem 1
Two trains are moving towards each other on parallel rails. The speed of the first train is 10 meters per second, the length of the first train is 500 meters. The speed of the second train is 30 meters per second, the length of the second train is 300 meters. Determine how long it will take for the second train to pass the first.
Given: $v_1$=10 m/s; $v_2$=30 m/s; $L_1$=500 m; $L_2$=300 m
Find: t --- ?
The time it will take the trains to pass each other can be determined by dividing the total length of the trains by their relative speed. The speed of the first train relative to the second is determined by the formula v= v1+v2 Then the formula for determining time takes the form: $t=\frac(L_1+L_2)(v_1+v_2)=\frac(500+300)(10+30)= 20\c$
Answer: The second train will pass the first one within 20 seconds.
Problem 2
Determine the speed of the river flow and the speed of the boat in still water, if it is known that the boat travels a distance of 300 kilometers downstream in 4 hours, and against the current in 6 hours.
Given: $L$=300000 m; $t_1$=14400 s; $t_2$=21600 s
Find: $v_p$ - ?; $v_k$ - ?
The speed of the boat along the river relative to the shore is $v_1=v_k+v_p$, and against the current $v_2=v_k-v_p$. Let us write down the law of motion for both cases:
Having solved the equations for vp and vk, we obtain formulas for calculating the speed of the river flow and the speed of the boat.
River flow speed: $v_p=\frac(L\left(t_2-t_1\right))(2t_1t_2)=\frac(300000\left(21600-14400\right))(2\times 14400\times 21600)=3 .47\ m/s$
Boat speed: $v_к=\frac(L\left(t_2+t_1\right))(2t_1t_2)=\frac(300000\left(21600+14400\right))(2\times 14400\times 21600)=17, 36\ m/s$
Answer: the speed of the river is 3.47 meters per second, the speed of the boat is 17.36 meters per second.
