Test on the topic "Kinematics" Option 1.

1. The distance between the start and end points is:

A) path B) movement C) displacement D) trajectory

2. In which of the following cases the motion of a body cannot be considered as the motion of a material point?

A) The movement of the earth around the sun. B) The movement of a satellite around the Earth.

C) Airplane flight from Vladivostok to Moscow. D) Rotation of the part being machined

machine tool

3. Which of the following quantities are scalar?
A) movement B) path C) speed

4 . What does a car speedometer measure?
A) acceleration B) instantaneous speed module;
B) average speed D) moving

5. What is the basic unit of time in the International System of Units?
A) 1 hour B) 1 minute C) 1 s D) 1 day.

6. Two cars are moving along a straight highway in the same direction. If you direct the OX axis along the direction of movement of the bodies along the highway, then what will be the projections of car speeds on the OX axis?

7. The car traveled around Moscow along the ring road, the length of which is 109 km. What is the distance traveled l and displacement S of the car?
A) l = 109 km; S = 0 B) l = 218 km S = 109 kmV) l = 218 km; S = 0. D) l=109 km; S=218 km

8.

BUT ) 1 B) 2 C) 3 D) 4.

9 . Determine the path traveled by the point in 5 s. (Fig. 2).

A) 2m B) 2.5m C) 5m D) 10m.

10 .. Figure 3 shows a graph of the distance traveled by a cyclist versus time. Determine the path traveled by the cyclist in the time interval from t 1 \u003d 1c to t 2 \u003d 3s?

11 . If acceleration is 2 m/s 2 , that is:

A) uniform motion B) uniformly slow motion

C) uniformly accelerated motion D) rectilinear

12 . Acceleration characterizes the change in the velocity vector

A) in magnitude and direction B) in direction C) in magnitude

13 . A car moving in a straight line with uniform acceleration increases its speed with
3 m/s to 9 m/s in 6 seconds. With what acceleration was the car moving?
A) 0 m/s 2 B) 3 m/s 2 C) 2 m/s 2 D) 1 m/s 2

14. What speed does the car acquire when braking with an acceleration of 0.5 m / s 2 after 10 s from the start of braking, if its initial speed was 72 km / h?

A) 15m/s B) 25m/s C) 10m/s D) 20m/s.

Test on the topic "Kinematics" Option 2.

1 . A cyclist moves from point A of the cycle track to point B along the curve AB. name
the physical quantity represented by the vector AB.
A) path B) movement C) speed

2 . Why, in calculations, can the Moon be considered a material point (relative to the Earth)?

A) The moon is a ball B) The moon is a satellite of the Earth C) The mass of the Moon is less than the mass of the Earth

D) The distance from the Earth to the Moon is many times greater than the radius of the Moon.

3. . Physical quantities are vector and scalar. Which of the following physical quantities is a scalar?
A) acceleration B) time C) speed D) displacement

4. . Which of the following quantities are vector quantities:
1) path 2) movement 3) speed?
A) 1 and 2 B) 2 and 3 C) 2 D) 3 and 1.

5 . The basic units of length in SI are:
A) meter B) kilometer C) centimeter D) millimeter

6 . Two cars are driving on a straight highway in opposite directions. If you direct the OX axis along the direction of movement of the first car along the highway, then what will be the projections of the speeds of cars on the OX axis?
A) both positive B) both negative
C) the first - positive, the second - negative
D) the first - negative, the second - positive

7 . A body thrown vertically upward reaches its maximum height of 10 m and falls on
earth. What are the path l and displacement S for the entire time of its movement?
A) l = 20 m, S = 0 m B) l = 10 m, S = 0B) l = 10 m, S = 20 m D) l = 20 m, S = 10 m.

8 . Which of the graphs corresponds to uniform motion? (Fig. 1).

BUT ) 3 B) 4 C) 1 D) 2

9 . Determine the path traveled by the point in 3 s. (Fig. 2).

A) 2m B) 6m C) 5m D) 1.5m.

10. . Figure 3 shows a graph of the distance traveled by a cyclist versus time. Determine the path traveled by the cyclist in the time interval from t 1 = 2c to t 2 = 4s?

A) 9 m B) 6 m C) 3 m D) 12 m

11 . If acceleration is -3m/s 2 , that is:

A) uniform motion B) uniformly accelerated motion

C) uniformly slow motion D) rectilinear motion

12 . The car starts off and moves with increasing speed in a straight line.
A) acceleration is 0 B) directed against the movement of the car
B) directed in the direction of the car

13. The speed of the car decreased from 20m/s to 10m/s in 20s. What was the average acceleration of the car?

A) 0.5m/s 2 B) 5m/s 2 C) -5m/s 2 D) -0.5m/s 2

14 . Determine the speed of the body during braking with an acceleration of 0.2 m / s 2 after 30 s from the start of movement, if its initial speed was equal to 2 m / s.

A) -4m B) 4m C) -6m D) 8m.

Answers

Option 1 Option 2

1-b 1-b

2 - d 2 - d

3 - a 3 - b

4 - b 4 - c

5 - in 5 - a

6 - a 6 - in

7 - in 7 - a

8 - b 8 - d

9 - d 9 - b

10 - b 10 - b

11 - in 11 - in

12 - a 12 - in

13 - g 13 - g

14-b 14-a

1.13. The car starts off and moves with increasing speed in a straight line.
What is the direction of the acceleration vector?

1.14. The car slows down on a straight section of the road. What direction does
acceleration vector?
A) acceleration is 0; B) directed against the movement of the car;
B) is directed in the direction of movement of the car.

1.16. Physical quantities are vector and scalar. Which of the following physical quantities is a scalar?
A) acceleration B) time; B) speed D) movement.

1.18. The basic units of length in SI are:
A) kilometer B) meter; B) centimeter D) millimeter.

1.19. Which of the following quantities are vector quantities:
1) path, 2) movement, 3) speed?
A) 1 and 2; B) 2; C) 2 and 3; D) 3 and 1.

1.22. Moving in a straight line, one body travels a distance of 5 m every second, another body - 10 m every second. The movements of these bodies are: A) uniform B) uneven; C) the first is uneven, the second is uniform; D) the first uniform, the second uneven

1 25. The modulus of body speed for every second increased by 2 times. Which statement would be correct?
A) acceleration decreased by 2 times; B) the acceleration did not change;
B) the acceleration is doubled

1.26. A body thrown vertically upward reaches its maximum height of 10 m and falls on
earth. What are the path l and displacement S for the entire time of its movement?
A) l = 10 m, S = 0 m; B) l = 20 m, S = 0;
B) l = 10 m, S = 20 m; D) l = 20 m, S = 10 m.

1.35. When leaving the station, the acceleration of the train is 1 m/s2. How far does the train travel in 10 seconds?
A) 5 m; B) 10 m; C) 50 m; D) 100 m.

1.36. With uniformly accelerated movement for 5 s, the car increased its speed from 10 to
15 m/s. What is the acceleration modulus of the car?
A) 1 m/s2; B) 2 m/s2; C) 3 m/s2; D) 5 m/s2.

1.55. Which of the following functions (v(t)) describes the dependence of the velocity modulus on
time with a uniform rectilinear motion of the body along the axis ОХ with a speed of 5 m/s?
A) v = 5t; B) v = t; B) v = 5; D) v = -5.

1.65. A bar placed on a horizontal surface of a table was given a speed of 5 m/s. Under the action of friction forces, the bar moves with an acceleration of 1 m/s2. What is the distance traveled by the block in 6 seconds?
A) 48 m; B) 12 m; C) 40 m; D) 30 m.

13. Figure 3 shows a graph of the distance traveled by a cyclist versus time. Determine the path traveled by the cyclist in the time interval from t 1 = 1c to t 2 = 4s?

BUT) 15 m. B) 3m. AT) 12 m G) 9 m D) 20 m

14. Figure 3 shows a graph of the distance traveled by a cyclist versus time. Determine the speed of the cyclist at time t = 2c.

BUT) 2 m/s. B) 6 m/s. AT) 3 m/s. G) 12 m/s. D) 8 m/s.

18. The body moves in a straight line and reduces speed. Where is the acceleration directed?

BUT) Along the way. B) Normally. AT) Against movement. G) Along the radius vector to the given point of the trajectory. D) Tangent to path

BUT) The moon is a ball . B) The Moon is the Earth's satellite. AT) The mass of the Moon is less than the mass of the Earth.

G) The distance from the Earth to the Moon is many times greater than the radius of the Moon.

D) None of the suggested answers are correct.

Vehicle speed for 20 s decreased from 20 m/s before 10 m/s . What was the average acceleration of the car? [−0.5 m/s 2 ]

Example 1 According to the given law of motion S= 10 + 20t - 5t 2 ([S]= m; [t]= with ) determine the type of movement, the initial speed and tangential acceleration of the point, the time to stop.

Decision

1. Type of movement: equally variable

2. When comparing the equations, it is obvious that

• the initial path traveled before the reference point is 10 m;
• initial speed 20 m/s;
• constant tangential acceleration a t/2 = 5 m/s; a t= - 10 m/s.
• the acceleration is negative, therefore, the movement is slow (equally slow), the acceleration is directed in the direction opposite to the direction of the speed of movement.

3. You can determine the time at which the speed of the point will be equal to zero:

v=S"= 20 - 25t; v= 20 – 10t = 0;t= 20/10 = 2 s.

Note. If the speed increases during uniformly variable motion, then the acceleration is a positive value, the path graph is a concave parabola. When braking, the speed drops, acceleration (deceleration) is a negative value, the path graph is a convex parabola (Fig. 10.4).

Example 2 The point moves along the chute from the point BUT exactly D(Fig. 10.5).

How will the tangent and normal accelerations change when a point passes through AT and With?

Decision

1. Consider the site AB. Tangent acceleration is zero (v= const).

Normal acceleration ( a p = v2/r) when passing through a point AT increases by 2 times, it changes direction, because the center of the arc AB does not coincide with the center of the arc BC.

2. On site Sun:

The tangential acceleration is zero: a t = 0;

Normal acceleration when passing through a point With changes: to the point With the movement is rotational, after point C the movement becomes rectilinear, the normal stress on the rectilinear section is zero.

3. On site CD total acceleration is zero.

Example 3 According to a given speed graph, find the path traveled during the movement (Fig. 10.6).

Decision

1. According to the schedule, three traffic sections should be considered. The first section is acceleration from a state of rest (uniformly accelerated motion).

The second section is uniform motion: v= 8 m/s; a 2 = 0.

The third section is braking to a stop (equally slow motion).

2. The path traveled during the movement will be equal to:

Example 4 A body with an initial speed of 36 km/h travels 50 m before stopping. Assuming the motion is uniformly slowed down, determine the deceleration time.

Decision

1. We write the equation of speed for uniformly slow motion:

v \u003d v o + at \u003d 0.

Determine the initial speed in m/s: v about\u003d 36 * 1000/3600 \u003d 10 m / s.

We express the acceleration (deceleration) from the velocity equation: a = - v 0 /t

2. Write down the path equation: S \u003d v o t / 2 + at 2 / 2. After substitution, we get: S = v o t/2

3. Determine the time to a complete stop (braking time):

Example 5 The point moves in a straight line according to the equation s = 20t – 5t2 (s- m, t- with). Plot graphs of distances, speeds and accelerations for the first 4 seconds of movement. Determine the path traveled by the point in 4 s and describe the movement of the point.

Decision

1. The point moves in a straight line according to the equation s = 20t – 5t2 hence the speed of the point u = ds/d/t = 20 - 10t and acceleration a = a t = dv/dt =-10 m/s 2 . This means that the motion of the point is uniform (a = a t = - 10 m/s 2 = const) with initial speed v0= 20 m/s.

2. Compose the dependence of numerical values s and v for the first 4 s of motion

3. Based on the given numerical values, we construct distance graphs (Fig. a), speed (Fig. b) and acceleration (Fig. in), selecting scales for the image along the ordinates of distances s, speed v and acceleration a, as well as the same time scale for all graphs along the x-axis. For example, if the distance s \u003d 5 m is plotted on a graph with a segment length l s \u003d 10 mm, then 5m \u003d μ s * 10 mm, where the proportionality factor μ s is the scale along the axis Os: μ s \u003d 5/10 \u003d 0.5 m / mm (0.5 m in 1 mm); if the speed module v= 10 m/s depicted on a graph with a length lv\u003d 10 mm, then 10 m / s \u003d μ v * 10 mm and scale along the axis Ovμ v = 1 m/(s-mm) (1 m/s in 1 mm); if the acceleration module a\u003d 10 m / s 2 represent a segment l a \u003d 10 mm, then, similarly to the previous one, the scale along the axis Oaμ a \u003d 1 m / (s 2 -mm) (1 m / s 2 in 1 mm); and finally, depicting the time interval Δt= 1 with a segment μ t = 10 mm, we get on all graphs the scale along the axes Ot μ t= 0.1 s/mm (0.1 s in 1 mm).

4. From the consideration of the graphs, it follows that during the time from 0 to 2 s the point moves uniformly slow (velocity v and acceleration during this period of time have different signs, which means that their vectors are directed in opposite directions); in a period of time from 2 to 4 s, the point moves uniformly accelerated (velocity v and acceleration have the same signs, i.e., their vectors are directed in the same direction).

For 4 s, the point traveled the path s o _ 4 = 40 m. Starting to move at a speed v 0 \u003d 20 m / s, the point traveled 20 m in a straight line, and then returned to its original position, having the same speed, but directed in the opposite direction.

If we conditionally accept the free fall acceleration g = 10 ms 2 and neglect air resistance, then we can say that the graphs describe the movement of a point thrown vertically upwards with a speed a 0 = 20 m/s.

Example 6 The point moves along the trajectory shown in Fig. 1.44, but, according to the equation s = 0.2t4 (s- in meters, t- in seconds). Determine the speed and acceleration of the point in positions 1 and 2.

Decision

The time required to move a point from position 0 (reference point) to position 1 is determined from the equation of motion by substituting the partial values ​​of distance and time:

Rate change equation

Point speed at position 1

Tangential acceleration of point at position 1

The normal acceleration of a point on a straight section of the trajectory is zero. The speed and acceleration of the point at the end of this section of the trajectory are shown in Fig. 1.44, b.

Let us determine the speed and acceleration of the point at the beginning of the curved section of the trajectory. It's obvious that v1\u003d 11.5 m / s, and t1 \u003d 14.2 m / s 2.

Normal acceleration of a point at the beginning of a curved section

The speed and acceleration at the beginning of the curved section are shown in fig. 1.44 in(vectors a t 1 and a a 1 shown not to scale).

Position 2 moving point is determined by the path traveled, consisting of a straight section 0 - 1 and circular arcs 1 - 2, corresponding to the central angle of 90°:

The time required to move the point from position 0 to position 2,

Point speed in position 2

Tangential acceleration of a point at a position 2

Normal acceleration of a point at a position 2

Acceleration of a point in a position 2

Velocity and acceleration of a point in a position 2 shown in fig. 1.44 in(vectors at" and a Pg shown not to scale).

Example 7 The point moves along a given trajectory (Fig. 1.45, a) according to the equation s = 5t3(s - in meters, t - in seconds). Determine point acceleration and angle α between acceleration and speed at the moment t1 when the speed of the point v 1 \u003d 135 m / s.

Decision

Rate change equation

Time t1 we determine from the equation for changing the speed by substituting the partial values ​​of speed and time:

Let us determine the position of the point on the trajectory at the moment 3 s:

An arc of a circle with a length of 135 m corresponds to the central angle

The equation for changing the tangential acceleration

Tangential acceleration of a point at a moment t t

Normal acceleration of a point at a moment t t

Acceleration of a point at the moment t x

Velocity and acceleration of a point at a moment in time t1 shown in fig. 1.45, b.

As can be seen from fig. 1.45, b

Example 8 An object is thrown into a mine with a depth of H = 3000 m from the surface of the earth without an initial velocity. Determine after how many seconds the sound that occurs when an object hits the bottom of the mine reaches the surface of the earth. The speed of sound is 333 m/s.

Decision

The equation of motion of a freely falling body

The time required to move an object from the surface of the earth to the bottom of the mine, we determine from the equation of motion.

Problem 1.6. Find in a graphical way the displacement and the path traveled for t 1 \u003d 5 with a material point, the movement of which along the axis OH is described by the equation X = 6 – 4t + t 2 , where all quantities are expressed in SI units.

Decision. In problem 1.5 we found (4) the projection of the velocity on the axis OH:

The velocity graph corresponding to this expression is shown in Figure 1.6. Projection of displacement onto the axis OH is equal to the algebraic sum of the areas of 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 velocity projection 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, in order to find it, we add the areas of these triangles, considering that the area of ​​not only the triangle is positive BCD, but also triangles AOB:

Earlier (see problem 1.5) we found this way in a different way - analytically.

Problem 1.7. On fig. 1.7, a shows a graph of the dependence of the coordinates of some body moving rectilinearly along the axis OH, from time. Curvilinear sections of the graph are parts of parabolas. Plot graphs of speed and acceleration versus time.

Decision. To build graphs of speed and acceleration, we set according to this graph (Fig. 1.7, a) the nature of the movement of the body at different intervals of time.

Between 0 - t 1, the coordinate graph is a part of a parabola, the branches of which are directed upwards. Therefore, in the equation

expressing in general terms the dependence of the coordinate 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 the modulus of the body's velocity first decreases to zero, and then the velocity reverses direction and its modulus increases to a certain value v one . The speed graph in this section is a straight line segment passing at some 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, in). The top of the parabola in Fig. 1.7, a corresponds to the value v 0x= 0 in fig. 1.7, b.

In the span of time t 1 – t 2 the body is moving uniformly with a speed v 1 .

In the interim t 2 – t 3 coordinate graph - part of the parabola, the branches of which are directed downwards. Hence, here a x < 0, скорость тела убывает до нуля к моменту времени t 3 , and in the time interval t 3 – t 4 the body is at rest. Then for a period of time t 4 – t 5 a body is moving at a uniform speed v 2 in reverse. At the point in time t 5 it reaches the point of origin of the coordinates and stops.

Considering the nature of the motion of the body, we will construct the corresponding graphs of the 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 path vs. time graph.

Decision. Let us divide the entire time interval under consideration into three sections: 1, 2, 3. In section 1, the body moves uniformly accelerated without initial velocity. The path formula for this segment is

where a is the acceleration of the body.

Acceleration is the ratio of a change in speed to the time it takes for that change to occur. It is equal to the ratio of the segments.

In section 2, the body moves uniformly with a speed v, acquired by the end of section 1. Uniform motion began not at the initial moment of time, but at the moment t one . By this point, the body has already passed the way. The dependence of the path on time for section 2 has the following form:

In section 3, the movement is equally slow. The path formula for this section is as follows:

where a 1 - acceleration in section 3. It is half the acceleration a in 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 a bulge facing upwards) (see Fig. 1.9).

The path graph should not have kinks, it is depicted as a smooth line, i.e. parabolas mate 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 velocity 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 from this that the path graph has no breaks.

In addition, the vertex of the inverted parabola must correspond to the time t 3 . The vertices of the parabolas must correspond to the 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 the area of ​​the figure OABG, formed by the velocity graph on the interval From 2 .

Problem 1.9. On fig. 1.10 shows a graph of the projection of the velocity of a body moving rectilinearly along the axis OH, from time. Plot graphs of acceleration, coordinates 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.

Decision. To plot the acceleration curve a x(t), we will determine according to the schedule v x(t) the nature of the movement of the body at different intervals of time. Recall 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 motion is uniformly slowed down.

In section c, the projection of velocity , i.e. movement is in the positive direction of the axis OH.

In section c, the projection of velocity is that the body is at rest (and ).

In 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.

EN 01 MATHEMATICS

Collection of assignments for extracurricular independent work on the topic: "Application of a definite integral for solving physical problems."

for the specialty:

100126 Household and communal services

Vologda 2013

Mathematics: Collection of assignments for extracurricular independent work on the topic: "The use of a definite integral to solve physical problems" for the specialty: 100126 Household and communal services

This collection of assignments for extracurricular independent work on the topic: "Application of a definite integral for solving physical problems" is a teaching aid for organizing independent extracurricular work of students.

Contains tasks for independent extracurricular work for six options and criteria for evaluating the performance of independent work.

The set is designed to help students systematize and consolidate the theoretical material received in the classroom in mathematics, to form practical skills.

Compiled by: E. A. Sevaleva - teacher of mathematics of the highest category, BEI SPO VO "Vologda Construction College"

1. Explanatory note.

2. Independent work.

3. Evaluation criteria.

4. Literature.

Explanatory note

This work is a teaching aid on the organization of independent extracurricular work of students in the discipline EN 01 "Mathematics" for the specialty 100126 Household and communal services.

The purpose of the guidelines is to ensure the effectiveness of independent work, determine its content, establish requirements for the design and results of independent work.

The goals of independent work of students in the discipline EN 01 "Mathematics" are:

systematization and consolidation of the received theoretical knowledge and practical skills;

deepening and expansion of theoretical knowledge;

formation of skills to use reference and additional literature;

development of cognitive abilities and activity of students, creative initiative, independence and self-organization;

· activation of educational and cognitive activity of future specialists.

Independent work is carried out individually in their free time.

The student must:

• before performing independent work, repeat the theoretical material covered in the classroom;
• perform work according to the task;
• for each independent work, submit a report to the teacher in the form of a written work.

Independent work on the topic:

"Application of a definite integral for solving physical problems"

Target: learn to apply a definite integral to solve physical problems.

Theory.

Computing the path traveled by a point.

The path traveled by a point during non-uniform movement in a straight line with a 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.

Decision: According to the condition , , .

According to formula (1) we find:

Answer: .

Example 2 The speed of the point changes according to the law m/s. Find the path traveled by the point in the 4th second.

Decision: According to the condition , ,

Hence:

Answer: .

Example 3 The speed of the point changes according to the law m/s. Find the path traveled by the point from the beginning of the movement to its stop.

Decision:

· The speed of the point is 0 at the moment of the beginning of the movement and at the moment of the stop.

Determine at what point in time the point will stop, for this we will solve the equation:

I.e , .

By formula (1) we find:

Answer: .

Calculation of 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 for calculating the work of a force, it is often used Hooke's law: ……(3), where

Force ( H);

X is the absolute elongation (compression) of the spring caused by the force ( m);

Proportionality coefficient ( N/m).

Example 4 Calculate the work done when the spring is compressed by 0.04 m, if to compress it by 0.01 m need strength 10 H.

Decision:

· As x = 0,01 m with force =10 H

, we find , i.e. .

Answer:J.

Example 5 The spring at rest has a length of 0.2 m. Strength at 50 H stretches the spring by 0.01 m. What work must be done to stretch the spring from 0.22 m up to 0.32 m?

Decision:

· As x = 0.01 at force =50 H, then, substituting these values ​​into equality (3): , we get:

Substituting now in the same equality the found value , we find , i.e. .

We find the limits of integration: m, m.

Find the desired job by the formula (2):

Consider the solution of the following problems.

1. A current pulse passes through a part of the animal's body, which changes with time according to the mA law. The pulse duration is 0.1 s. Determine the work done by the current during this time if the resistance of the section is 20 kOhm.

For a small time interval d t, when the current practically does not change, on the resistance R work is being done. During the entire impulse, work will be done

.

Substituting the value of the current into the resulting expression, we obtain.

2. The speed of the point is (m/s). Find a way S, passed by the point in time t\u003d 4s, elapsed from the beginning of the movement.

Let's find the path traveled by the point in an infinitesimal time interval. Since during this time the speed can be considered constant, then . Integrating, we have

3. Find the fluid pressure force on a vertical triangular plate with a base a and height h immersed in a liquid so that its vertex lies on the surface.

Let's place the coordinate system as shown in Fig. 5.

Consider a horizontal infinitesimal strip of thickness d x located at an arbitrary depth x. Taking this strip as a rectangle, find its base EF. From the similarity of triangles ABC and AEF we get

Then the area of ​​the strip is

Since the strength P fluid pressure on the pad S, the immersion depth of which r, according to Pascal's law is equal to

where r is the density of the liquid, g is the acceleration of gravity, then the desired pressure force on the area under consideration d S calculated by the formula

.

Therefore, the pressure force P liquids on the pad ABC

.

solve problems.

5.41 The speed of a point is given by the equation cm/s. Find the path traveled by a point in time t\u003d 5 s, which has elapsed since the beginning of the 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. What is the distance traveled by the body in the third second of motion?

5.44 Two bodies start moving simultaneously from the same point: one with a speed (m/min) and the other with a speed (m/min). How far apart will they be in 10 minutes if they move in the same line in the same direction?

5.45 A force (dyn) acts on a body of mass 5 g moving in a straight line. Find the distance traveled by the body during the third second of motion.

5.46 The speed of an oscillating point varies according to the law (cm/s). Determine the displacement of the point 0.1 s after the start of the movement.

5.47 What work must be done to stretch the spring 0.06 m if a force of 1N stretches it 0.01 m?

5.48 The speed of an oscillating point varies according to the law (m/s). Determine the path traveled by the point in s from the start 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 What work must be done to stretch a spring 25 cm long to a length of 35 cm if the spring constant is known to be 400 N/m?

5.51 A current pulse passes through the body of an animal, which changes with time according to the law (mA). The pulse duration is 0.1 s. 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 P 0 the muscle is stretched l 0 mm? Assume that the force required to stretch the muscle is proportional to its lengthening.

5.53 The body moves in a certain medium in a straight line 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 moving the body from S=0 to S=a meters.