Problem 1.6. Find graphically the displacement and the path traveled in t 1 = 5 with a material point whose movement along the axis OH described by the equation X = 6 – 4t + t 2, where all quantities are expressed in SI units.
Solution. In problem 1.5 we found (4) the projection of velocity onto the axis OH:
The speed graph corresponding to this expression is shown in Figure 1.6. Projection of movement onto the axis OH equal to the algebraic sum of the areas of the triangles AOB And BCD. Since the velocity projection in the first section is negative, the area of the triangle AOB take with a minus sign; and the projection of velocity in the second section is positive, then the area of the triangle BCD take with a plus sign:
Since the path is the length of the trajectory and cannot decrease, to find it, we add the areas of these triangles, while considering as positive the area of not only the triangle BCD, but also a triangle AOB:
Earlier (see problem 1.5) we found this path in a different way - analytically.
Problem 1.7. In Fig. 1.7, a shows a graph of the dependence of the coordinates of some body moving rectilinearly along the axis OH, from time. The curved sections of the graph are parts of parabolas. Draw graphs of speed and acceleration versus time.
Solution. To build graphs of speed and acceleration, we set according to this graph (Fig. 1.7, A) the nature of the body’s movement at different periods of time.
In the interval 0 – t 1 coordinate graph is a part of a parabola, the branches of which are directed upward. Therefore, in Eq.
expressing in general view coordinate dependence X from time t, coefficient before t 2 is positive, i.e. A x > 0. And since the parabola is shifted to the right, this means that v 0x < 0, т.е. тело имело начальную скорость, направленную противоположно направлению оси ОХ. В течение промежутка 0 – t 1 module of the body’s velocity first decreases to zero, and then the speed changes direction to the opposite and its module increases to a certain value v 1 . The speed graph in this section is a straight line segment passing at a certain angle to the axis t(Fig. 1.7, b), and the acceleration graph is a segment of a horizontal straight line lying above the time axis (Fig. 1.7, V). The vertex of the parabola in Fig. 1.7, A corresponds to the value v 0x= 0 in Fig. 1.7, b.
In the interim t 1 – t 2 the body moved uniformly with speed v 1 .
In the interim t 2 – t 3 coordinate graph is part of a parabola whose branches are directed downward. Therefore, here a x < 0, скорость тела убывает до нуля к моменту времени t 3, and in the interim t 3 – t 4 the body is at rest. Then over a period of time t 4 – t 5 a body moves uniformly with speed v 2 in reverse side. At a moment in time t 5 it reaches the origin point and stops.
Taking into account the nature of the body’s movement, we will construct corresponding graphs of projections of velocity and acceleration (Fig. 1.7, b, c).
Problem 1.8. Let the speed graph have the form shown in Fig. 1.8. Based on this graph, draw a graph of the path versus time.
Solution. Let us divide the entire time period under consideration into three sections: 1, 2, 3. In section 1, the body moves uniformly accelerated without an initial speed. The path formula for this section has the form
Where A– acceleration of the body.
Acceleration is the ratio of the change in speed to the period of time during which this change occurred. It is equal to the ratio of the segments.
In section 2 the body moves uniformly with speed v acquired by the end of section 1. The uniform movement did not begin at starting moment time, and at the moment t 1 . By this point, the body has already passed the path. The dependence of the path on time for section 2 has the following form:
In section 3 the movement is uniformly slow. The path formula for this section is as follows:
Where A 1 – acceleration in section 3. It is half the acceleration A on section 1, since section 3 is twice as long as section 1.
Let's draw conclusions. In section 1, the path graph looks like a parabola, in section 2 - a straight line, in section 3 - also a parabola, but inverted (with the convex pointing upward) (see Fig. 1.9).
The path graph should not have kinks; it is depicted as a smooth line, that is, parabolas are conjugated with a straight line. This is explained by the fact that the tangent of the angle of inclination of the tangent to the time axis determines the value of the speed at the moment of time t, i.e. By the slope of the tangents to the path graph, you can find the speed of the body at one time or another. And since the speed graph is continuous, it follows that the path graph has no breaks.
In addition, the vertex of the inverted parabola must correspond to the moment in time t 3. The vertices of the parabolas must correspond to moments 0 and t 3, since at these moments the speed of the body is zero and the paths tangent to the graph must be horizontal for these points.
The path traveled by the body in time t 2, numerically equal to area figures OABG, formed by the velocity graph on the interval From 2 .
Problem 1.9. In Fig. Figure 1.10 shows a graph of the projection of the velocity of some body moving rectilinearly along the axis OH, from time. Construct graphs of acceleration, position and path versus time. At the initial moment of time the body was at the point X 0 = –3 m. All values are given in SI units.
Solution. To plot the acceleration dependence a x(t), we will determine according to the schedule v x(t) the nature of the body’s movement at different periods of time. Let us remember that by definition
where is the projection of velocity , .
In the time interval c:
In this section and (the signs are the same), i.e. the body moves with uniform acceleration.
In the time interval c:
those. and (projection signs are opposite) – the movement is equally slow.
In section c the velocity projection, i.e. movement occurs in the positive direction of the axis OH.
In section c, the velocity projection is that the body is at rest (and ).
On section c:
And (the signs are the same) – the movement is uniformly accelerated, but since , then the body moves against the axis OH.
After the sixth second, the body moves uniformly () against the axis OH. looks like shown in Fig. 1.11,.
G
Let's consider solving the following problems.
1. A current pulse passes through an area of the animal’s body, which changes over time according to the mA law. Pulse duration 0.1 s. Determine the work done by the current during this time if the resistance of the section is 20 kOhm. t Over a short time interval d , when the current practically does not change, across the resistance R
.
work is done. During the entire pulse, work will be done
Substituting the current value into the resulting expression, we obtain. 
2. The speed of the point is (m/s). Find the way S t traversed by a point in time
=4s elapsed from the beginning of the movement.
Let's find the path traveled by a point in an infinitesimal period of time. Since during this time the speed can be considered constant, then . Integrating, we have a 3. Find the force of fluid pressure on a vertical triangular plate with a base and height h
immersed in a liquid so that its top lies on the surface.
We will place the coordinate system as shown in Fig. 5. x Consider a horizontal infinitesimal strip of thickness d x, located at an arbitrary depth . Taking this strip as a rectangle, we find its base EF . From the similarity of triangles And ABC AEF
we get
Then the area of the strip is Since the strength P (m/s). Find the way fluid pressure on the platform , the immersion depth of which r
, according to Pascal's law is equal to where r is the density of the liquid, g (m/s). Find the way- acceleration of gravity, then the desired pressure force on the considered area d
.
Therefore, the pressure force Since the strength liquids on the platform . From the similarity of triangles
.
Solve problems.
5.41 The speed of a point is determined by the equation 
cm/s. Find the path traveled by a point in time t=5s elapsed from the start of movement.
5.42 The speed of a body is expressed by the formula m/s. Find the path traveled by the body in the first three seconds after the start of movement.
5.43 The speed of a body is determined by the equation 
cm/s. Which way the body will pass in the third second of movement?
5.44 Two bodies begin to move simultaneously from the same point: one with speed (m/min), and the other with speed (m/min). At what distance from each other will they be after 10 minutes if they move along the same line in the same direction?
5.45 A body of mass 5 g moving in a straight line is acted upon by a force (dyne). Find the distance traveled by the body during the third second of movement.
5.46 The speed of an oscillating point changes according to the law 
(sm/s). Determine the displacement of the point 0.1 s after the start of the movement.
5.47 How much work must be done to stretch a spring by 0.06 m if a force of 1 N stretches it by 0.01 m?
5.48 The speed of an oscillating point changes according to the law 
(m/s). Determine the distance traveled by the point for s from the beginning of the movement.
5.49 Nitrogen, whose mass is 7 g, expands at a constant temperature of 300°K so that its volume doubles. Determine the work done by the gas. Universal gas constant 
J/kmol.
5.50 How much work must be done to stretch a spring 25 cm long to a length of 35 cm, if it is known that the spring stiffness coefficient is 400 N/m?
5.51 A current pulse passes through the body of an animal, which changes over time according to the law (mA). The pulse duration is 0.1s. Determine the charge flowing through the animal's body.
5.52 What work is done when a muscle is stretched? l mm, if it is known that under load Since the strength 0 the muscle is stretched by l 0 mm? Assume that the force required to stretch a muscle is proportional to its lengthening.
5.53 A body moves in a certain medium rectilinearly according to the law. The resistance of the medium is proportional to the square of the speed. Find the work done by the resistance force of the medium when the body moves from (m/s). Find the way=0 to (m/s). Find the way=a meters.
Where x And y– in cm, a t- in the village Determine the trajectory of a point, speed and acceleration at moments of time t 0 =0 s, t 1 =1 s And t 2 =5 s, as well as the path traveled by the point in 5 s.
Solution
Trajectory calculation
We determine the trajectory of the point. We multiply the first given equation by 3, the second by (-4), and then add their left and right sides:
3x=6t 2 +6
-4y=-6t 2 -4
————
3x-4y=2
The result is an equation of the first degree - the equation of a straight line, which means the movement of the point is rectilinear (Figure 1.5).
In order to determine the coordinates of the initial position of point A 0, we substitute the values into the given equations t 0 =0; from the first equation we get x 0 =2 cm, from the second y 0 =1 cm. For any other value of t, the x and y coordinates of the moving point only increase, so the trajectory of the point is a half-line 3x-4y=2 with the beginning at point A 0 (2; 1).
Figure 1.5
Speed calculation
We determine by first finding its projections on the coordinate axes:
At t 0 =0s point speed v 0 =0, at t 1 =1s – v 1 =5 cm/s, at t 2 =5s – v 2 =25cm/s.
Acceleration calculation
Determine the acceleration of the point. Its projections on the coordinate axes:
Acceleration projections do not depend on the time of movement,
those. the motion of the point is uniformly accelerated, the velocity and acceleration vectors coincide with the trajectory of the point and are directed along it.
On the other hand, since the motion of a point is rectilinear, the acceleration modulus can be determined by directly differentiating the velocity equation.
EN 01 MATHEMATICS
A collection of assignments for extracurricular independent work on the topic: “Application of a definite integral to solve physical problems.”
for specialty:
100126 Home and utilities
Vologda 2013
Mathematics: Collection of assignments for extracurricular independent work on the topic: “Application of a definite integral to solve physical problems” for the specialty: 100126 Household and utility services
This collection of assignments for extracurricular independent work on the topic: “Application of a definite integral to solve physical problems” is teaching aid on organizing independent extracurricular work students.
Contains tasks for independent extracurricular work for six options and criteria for assessing the completion of independent work.
The kit is designed to help students systematize and consolidate theoretical material acquired in classroom mathematics and develop practical skills.
Compiled by: E. A. Sevaleva – mathematics teacher highest category BOU SPO VO "Vologda Construction College»
1. Explanatory note.
2. Independent work.
3. Evaluation criteria.
4. Literature.
Explanatory note
this work is an educational and methodological manual for organizing independent extracurricular work of students in the discipline EN 01 “Mathematics” for specialty 100126 Household and utility services.
Target methodological instructions consists of ensuring the effectiveness of independent work, determining its content, establishing requirements for the design and results of independent work.
The goals of students’ independent work in the discipline EN 01 “Mathematics” are:
· systematization and consolidation of acquired theoretical knowledge and practical skills;
· deepening and expanding theoretical knowledge;
· developing the ability to use reference and additional literature;
· development of students’ cognitive abilities and activity, creative initiative, independence and self-organization;
· activation of educational and cognitive activities of future specialists.
Independent work is performed individually in free time from classes.
The student is obliged:
- before performing independent work, repeat the theoretical material covered in classroom lessons;
- perform the work according to the assignment;
- for each independent work Submit a report to the teacher in the form of written work.
Independent work on the topic:
“Application of a definite integral to solve physical problems”
Target: learn to apply definite integral for solving physical problems.
Theory.
Calculation of the path traveled by a point.
The path traveled by the point at uneven movement in a straight line with variable speed and the time interval from to is calculated by the formula
…… (1)
Example 1. 
m/s. Find the path traveled by a point in 10 With from the start of the movement.
Solution: According to the condition 
, , .
Using formula (1) we find:
Answer: .
Example 2. The speed of a point varies according to the law 
m/s. Find the path traveled by the point in 4 seconds.
Solution: According to the condition 
, ,
Hence:
Answer: .
Example 3. The speed of a point varies according to the law 
m/s. Find the path traveled by a point from the start of its movement to its stop.
Solution:
· The speed of the point is 0 at the moment it starts moving and at the moment it stops.
· Let’s determine at what point in time the point will stop; to do this, solve the equation: 
That is , .
· Using formula (1) we find:
Answer: .
Calculation of the work of force.
Work done by a variable force when moving along an axis Oh material point from x = a before x =, is found by the formula:
…… (2)
When solving problems involving calculating the work of force, it is often used Hooke's law: ……(3), where
Force ( N);
X– absolute elongation (compression) of the spring caused by force ( m);
Proportionality factor ( N/m).
Example 4. Calculate the work done by the force when the spring is compressed by 0.04 m, if to compress it by 0.01 m need strength 10 N.
Solution:
· Because x = 0,01 m at strength =10 N
, we find, i.e. 
.
Answer:J.
Example 5. Spring in calm state has a length of 0.2 m. Strength at 50 N stretches the spring by 0.01 m. How much work must be done to stretch the spring from 0.22 m up to 0.32 m?
Solution:
· Because x = 0.01 at force =50 N, then, substituting these values into equality (3): , we get:
· Now substituting the found value into the same equality 
, we find, i.e. 
.
· Finding the limits of integration: m, m.
· We will find the job you are looking for using formula (2):
