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Learning tasks.

1(A) Two tasks are solved:

a) the docking maneuver of two spacecraft is calculated;

b) the period of revolution of spaceships around the Earth is calculated.

In what case can spaceships be considered as material points?

1) Only in the first case.

2) Only in the second case.

3) In both cases.

4) Neither in the first nor in the second case.

2(A) The wheel rolls down a flat hill in a straight line. What trajectory does the point on the wheel rim describe with respect to the road surface?

1) Circle. 3) Spiral.

2) Cycloid. 4) Direct.

3(A) What is the displacement of a point moving along a circle of radius R when it is rotated by 60º?

1) R/2 2) R 3) 2R 4) R

indication: build a drawing, mark two positions of the body, the movement will be a chord, analyze how the triangle will turn out (all angles are 60º).

4(A) How far will the boat make a full turn with a radius of 2m?

1) 2 m 3) 6.28 m

2) 4 m 4) 12.56 m

indication: make a drawing, the path here is the length of the semicircle.

5(A) The figure shows the schedule of the bus from point A to point B and back. Point A is at the point X= 0, and point B is at the point X= 30 km. What is the maximum ground speed of the bus for the entire round trip?

6(A) The body begins to move in a straight line uniformly accelerated along the axis Ox. Indicate the correct location of the velocity and acceleration vectors at time t.

mechanical movement. Trajectory. Path. moving

Control tasks.

1 (A) The material point is:

1) a body of negligible mass;

2) the body is very small;

3) a point showing the position of the body in space;

4) a body whose dimensions can be neglected under the conditions of a given problem.

2(A) What is the change in position of one body relative to another called?

1) trajectory;

2) movement;

4) mechanical movement.

3(A) What is the displacement of a point moving along a circle of radius R when it is rotated by 180º?

1) R/2 2) R 3) 2R 4) R

4(A) The line that the body describes when moving in space is called:

1) trajectory;

2) movement;

4) mechanical movement.

5(A) The figure shows a graph of the movement of the body from point A to point B and back. Point A is located at the point x 0 = 30m, and point B is at the point x = 5 m. What is the minimum speed of the bus on the whole way there and back?

1) 5.2 m/s hm

6(A) The body starts braking rectilinearly uniformly accelerated along the axis Ox. Specify correct location velocity and acceleration vectors at time t.

7(A) A bar placed on a horizontal surface of a table was given a speed of 5 m/s. Under the action of the friction force, the bar moves with an acceleration modulo 1 m/s 2 . What is the distance traveled by the block in 6 seconds?

1) 5 m 2) 12 m 3) 12.5 m 4) 30 m

8(A) The equation for the dependence of the projection of the displacement of a moving body on time has the form: s x = 10t + 4t 2 (m). What is the equation for the coordinate of a body that started moving from a point with coordinate 5?

1) x \u003d 5 + 10t + 2t 2 (m) 3) x \u003d 5 + 10t + 4t 2 (m)

2) x \u003d 5 + 5t + 2t 2 (m) 4) x \u003d 5 + 5t + 4t 2 (m)

9(A) The crane lifts the load vertically upwards with a certain speed u 0 . When the load is at a height h = 24m, the crane cable breaks and the load falls to the ground in 3 s. With what speed will the object hit the ground?

1) 32 m/s 2) 23 m/s 3) 20 m/s 4) 21.5 m/s

10(A) A body that began to move uniformly accelerated from a state of rest with an acceleration of 2 m / s 2, then in the third second it will pass the path

1) 7 m 2) 5 m 3) 3 m 4) 2 m

11(A) The coordinates of bodies A and B moving along the same straight line change with time, as shown in the graph. What is the speed of body A relative to body B?

1) 40 m/s x.m

12(A) The escalator staircase rises with a speed u, with what speed relative to the walls, should a person go down it in order to rest relative to the people standing on the stairs going down?

1) u 2) 2u 3) 3u 4) 4u

13(A) At a speed of 12 m/s, the braking time of a truck is 4 s. If, during braking, the acceleration of the car is constant and does not depend on the initial speed, then the car, during braking, will reduce the speed from 18 m/s to 15 m/s, passing

1) 12.3 m 3) 28.4 m

2) 16.5 m 4) 33.4 m

14(A) Along the roundabout highway 5 km long, a truck and a motorcyclist travel in the same direction with speeds u 1, respectively = 40 km/h and u 2 = 100 km/h. If at the initial moment of time they were in the same place, then the motorcyclist will catch up with the car, passing

1) 3.3 km 3) 8.3 km

2) 6.2 km 4) 12.5 km

15(A) A body is thrown from the Earth's surface at an angle α to the horizon with an initial velocity u 0 = 10m/s, if the range of the body is L = 10 m, then the angle α is equal to

1) 15º 2) 22.5º 3) 30º 4) 45º

16(A) The boy threw the ball horizontally from a window at a height of 20 m. The ball fell at a distance of 8 m from the wall of the house. With what initial speed was the ball thrown?

1) 0.4 m/s 2) 2.5 m/s 3) 3 m/s 4) 4 m/s

17(B) A material point moves at a constant speed along a circle of radius R. How will the physical quantities listed in the first column change if the speed of the point increases?

Physical quantities. Their change.

Task 1. Two small steel balls are thrown simultaneously from the same point from the earth's surface with initial velocities u01=5 m/s and v02 = 8 m/s, directed at angles α,=80° and a2=20° to the horizon, respectively. . What is the distance between the balls after the time / = -^ s after the throw? The trajectories of the balls lie in the same vertical plane. Ignore air resistance. Solution. The balls move in the Earth's gravitational field with a constant acceleration g (we neglect the air resistance). We choose a coordinate system as shown in Fig. 20, we place the origin at the throwing point. For radius-vectors balls Let's choose a coordinate system. desired distance. Projection of acceleration The desired distance / is equal to the modulus of the difference between the radius vectors of the balls at the moment of time / = - s. Since the balls were thrown from the same point, /*0| = r02, therefore: / = . (The remaining terms were destroyed when subtracting the radius vectors.) In turn, according to the cosine theorem (see Fig. 20): Substituting the numerical values of the quantities included in this equality, we obtain \v0l -v02\ = 7m/s. Then the desired distance between the balls at the time * Problem 2. Two bodies are thrown vertically upwards from the earth's surface from one point after each other with a time interval r, with the same initial velocities v0. Neglecting air resistance, determine how long it will take for them to “meet”? Pro- t comment on the solution for Solution. Let's direct the Oy axis vertically upwards, place the reference point at the throwing point. We will count the time starting from the moment of throwing the first body. Initial conditions for the movement of bodies: O "o \u003d 0, vy0l \u003d v0; 2) t0 \u003d g, y02 \u003d O, vy02 \u003d v0. The projections of accelerations of bodies in the absence of air resistance are equal to: avl \u003d ay2 \u003d -g. The equations of motion of bodies in projections on the Oy axis, taking into account the initial conditions, have the form: A, where the graphs yx(t) intersect. Thus, ^^ the "meeting" condition: y, (O \u003d Ug (A) "that is, \u003d v0 ft -r) 2 "2 Solving this equation for / v, we find: tx = - + - Let us analyze the obtained expression for g 2 It is known (see Example 7) that the time of flight of a body thrown vertically is equal to 2v0/g Therefore, if v0 2v0/g. This means that the first body will fall to the ground first, and only then the second will be thrown up. In other words, the bodies will "meet" at the point of throwing. Task 3. The boy, being on a flat slope of a mountain with an angle of inclination (р- 30 °), throws a stone towards the mountain, giving him an initial speed v0, directed at an angle /? = 60 ° to the horizon. At what distance from the boy will the stone fall Let's neglect the air resistance.Solution.We choose the reference frame as shown in Fig. 22, placing the reference point O at the point of throwing.In this reference frame, the initial velocity of the stone makes an angle a = ft-(p = 30°) with the Ox axis. conditions: Fig. 22 Projections of stone acceleration in the absence of air resistance are equal (see Fig. 22): is equal to the slope of the mountain (р- 30° (why?), in addition, according to the condition of the problem (р = a. Let's write down the equations of system (14) taking into account the initial conditions: t2 Г x(t) = (y0cos«)/-( gsin^>)-, y(t) = (v0sina)t-(gcosp)- We find the flight time r of the stone from the last equation, knowing that We choose the coordinate system. . Acceleration projection Namely, r = -=-. (We have discarded the value r = 0, since it is not connected with the question of the problem). Substituting the found value of g into the equation for.g(/), we determine the desired distance (in other words, the flight range): 3 g Task 4. A massive platform moves at a constant speed K0 along a horizontal floor. The ball is kicked from the rear edge of the platform. The module of the ball's initial velocity relative to the platform is y\ u = 2VQ9, and the vector u makes an angle a = 60° with the horizon (Fig. 23). What is the maximum height above the floor that the ball will rise to? At what distance from the edge of the platform will the ball be at time _ j. w_ ,0 touchdown. Ignore the platform height and air resistance. All velocities lie in the same vertical plane. (FZFTSH at MIPT, 2009.) Solution. To describe the movement of the ball and the platform, we introduce a reference system associated with the floor. We direct the Ox axis horizontally in the direction of impact, and the Oy axis vertically upwards (Fig. 23). The ball moves with a constant acceleration a, and ax = 0, aY = -g, where g is the value of the free fall acceleration. The projections of the ball's initial velocity v0 on the Ox and Oy axes are: v0,x = V0, + = -K + 2F0 cos 60° = -V0 + V0 = 0, % = K, - + =10 + sin 60° = >/ 3F0. If the ball's horizontal velocity is equal to zero, it means that it moves only vertically, and it will fall at the point of impact. The maximum lifting height (onvix) and the time of flight of the ball will be found from the laws of kinematics uniformly accelerated motion: a/ Let's choose a coordinate system. desired distance. Acceleration projection Zt Considering that at y = y^ the vertical velocity projection vanishes vY = 0 , and at the moment of the ball landing t = Гflight its coordinate along the Oy axis vanishes y = 0 , we have: ЗУ-т = 1 flight 2 g 2 g - S During the flight of the ball, the platform will move to a flight distance of 8 Y check which is the required distance between the ball and the platform at the moment the ball lands. Control questions 1. In fig. 24 shows the trajectory of the body. Its initial position is indicated by point A, the final position - by point C. What are the projections of the body's movement on the Ox and Oy axes, the displacement module and the path traveled by the body? 2. The body moves uniformly and rectilinearly on the xOy plane. Its coordinates change depending on time in accordance with the equations: (values are measured in SI). Write down the equation y = y(x) of the trajectory of the body. What are the initial coordinates of the body and its coordinates 2 s after the start of motion? 3. The rod AB, oriented along the axis Ox, moves with a constant speed v = 0.1 m/s in the positive direction of the axis. The front end of the rod is point A, the rear end is point B. What is the length of the rod if at time tA \u003d 1 Oc after the start of movement, the coordinate of point A is x, \u003d 3m, and at time tB- 30s the coordinate of point B is * L \u003d 4.5 m? (MIET, 2006) 4. How is their relative speed determined when two bodies move? 5. The bus and the motorcycle are at a distance L = 20 km from each other. If they move in the same direction with some speeds r\ and v2, respectively, then the motorcycle will catch up with the bus in time / = 1 hour. What is the speed of the motorcycle relative to the bus? 6. What is called the average ground speed of the body? 7. For the first hour of the journey, the train traveled at a speed of 50 km/h, for the next 2 hours it traveled at a speed of 80 km/h. Find the average speed of the train during these 3 hours. Select correct option answer and justify your choice: 1) 60 km/h; 2) 65 km/h; 3) 70 km/h; 4) 72 km/h; 5) 75 km/h. (RGTU named after K. E. Tsiolkovsky (MATI), 2006) 8. One fifth of the way the car drove at a speed r\ = 40 km/h, and the rest of the way at a speed v2 = 60 km/h. Find the average speed of the car for the whole journey. (MEPhI, 2006) 9. A material point begins to move along the Ox axis according to the law *(/) = 5 + 4/-2r(m). At what distance from the origin of coordinates will the point's velocity be zero? (Moscow State Technical University named after N. E. Bauman, 2006) 10. The speed skater, having accelerated to a speed of v0 = 5 m/s, began to slide in a straight line and uniformly slowed down. After a time t = 20 s, the skater's speed modulus became equal to v = 3 m/s. What is the skater's acceleration? Tasks 1. A pedestrian ran at a speed of v( = 9 km / h for a third of the entire journey, a third of the entire time he walked at a speed of v2 = 4 km / h, and the rest of the time he walked at a speed equal to the average speed for the entire journey. Find this speed. (ZFTSH at the Moscow Institute of Physics and Technology, 2001) 2. The body, moving uniformly accelerated and rectilinearly from a state of rest, passed the distance S in time r. What speed did the body have at the moment when it passed the distance S / n, where n is some positive number? (MEPhI, 2006) 3. The body falls without initial velocity and reaches the ground surface in 4s. From what height did the body fall? Ignore air resistance. Choose the correct answer and justify your choice: 1) 20m; 2) 40 m; 3) 80m; 4) 120m; 5) 160 m. (Russian State Technical University named after K. E. Tsiolkovsky (MATI), 2006) 4. A stone thrown vertically upward from the surface of the earth fell to the ground after T = 2s. Determine the path 5 covered by the stone in time r = 1.5 s after the throw. Ignore air resistance. Acceleration of free fall is taken equal to g = 10m/s2. (MIET, 2006) Let's choose a coordinate system. desired distance. Projection of acceleration 5. From one point at a height h from the surface of the earth, a stone A is thrown vertically upwards and a stone B vertically downwards with the same speed. It is known that stone A reached top point its trajectory simultaneously with the fall of stone B to the ground. What is the maximum height (from the ground) that stone A reached? Air resistance is ignored. (MIPT, 1997) 6. A stone is thrown horizontally from a mountain slope forming an angle a = 45° with the horizon (Fig. 25). What is the initial speed v0 of the stone if it fell on the slope at a distance / = 50 m from the point of throwing? Ignore air resistance. 7. The body is thrown horizontally. 3 s after the throw, the angle between the direction of full velocity and the direction of full acceleration became equal to 60°. Determine the value of the total speed of the body at this moment in time. Ignore air resistance. (Russian State University of Oil and Gas named after I.M. Gubkin, 2006) Instruction. By full speed and full acceleration, understand simply the speed and acceleration of the body. 8. The projectile exploded into several fragments flying in all directions at the same speed. A fragment that flew vertically down reached the ground in time. A fragment that flew vertically upward fell to the ground after time t2. How long did the fragments fall, flying horizontally? Air resistance is ignored. (MIPT, 1997) 9. A stone thrown at an angle to the horizon has reached its maximum height of 5 m. Find the total flight time of the stone. Ignore air resistance. (Russian State University of Oil and Gas named after I.M. Gubkin, 2006) 10. A stone thrown from the surface of the earth at an angle a = 30 ° to the horizon twice visited the same height h after a time = 3s and = 5s after the start of the movement. Find the stone's initial velocity v0. Acceleration of free fall is taken equal to g = 10m/s2. Ignore air resistance. (Institute of Cryptography, Communications and Informatics of the Academy of the Federal Security Service of the Russian Federation, 2006) 11. With what speed v0 must a projectile fly out of a cannon at the moment of rocket launch in order to shoot it down? The rocket starts vertically with a constant acceleration i = 4m/s2. The distance from the cannon to the rocket launch site (they are at the same horizontal level) is / = 9 km. The gun fires at an angle α = 45° to the horizon. Ignore air resistance.

Topic number 1. Kinematics.

mechanical movement - change in the position of the body in space over time relative to other bodies.

Progressive movement -movement in which all points of the body pass the same trajectory.

Material point - a body whose dimensions under given conditions can be neglected, because its dimensions are negligibly small compared to the considered distances.

Trajectory – line of motion of the body.(Trajectory equation - dependence y(x))

Path l (m) – path length.Properties: l ≥ 0, does not decrease!

moving s(m) – a vector connecting the initial and final position of the body.

s x \u003d x - x 0- displacement vector projection length

Properties: s≤ l, s = 0 in a closed area. l

Speed u (m/s)– 1) average track u = ; average displacement = ; ;

2) instantaneous - speed at a given point, can only be found according to the speed equation u x \u003d u 0x + a x t or on schedule u(t)

Acceleration a (m / s 2) - change in speed per unit of time.

; = if - accelerated rectilinear movement

( )if ↓ - slow straight motion

if ^ - circular motion

Relativity of motion- dependence on the choice of reference system: trajectories, displacements, speeds, accelerations of mechanical movement.

Galileo's principle of relativity– all laws of mechanics are equally valid in all inertial frames of reference.

The transition from one reference system to another is carried out according to the rule:

And = -

Where u 1 - the speed of a body relative to a fixed frame of reference,

u 2 - the speed of the moving frame of reference,

u rel (υ 12) – the speed of the 1st body relative to the 2nd.

Types of movement.

Rectilinear motion.

Rectilinear uniform motion.	Rectilinear uniformly accelerated motion.
x o = const x s x	x o x x o x s x s x accelerated slow
x \u003d x 0 + u x t x along the x axis ~ t x 0 t against the axis	x = x 0 + u 0 x t + x x x ~ t 2 x o x o t t accelerated slow
s x = u x t	s x =u 0 x t + or s x = without t!
u x \u003d const u x along the Ox axis t against the Ox axis	ux = uox + a x t u x along the x-axis u x u o u o slow motion by oh υ = 0 t t accelerated accelerated against axis Ox
a = 0 a x t	x = const ahah t t

Curvilinear motion.

Movement along a circle with a constant modulo speed	Parabolic motion with free fall acceleration.
\u003d 2πRn (m / s) - linear speed \u003d 2πn (rad / s) - angular velocity i.e. u = ω R (m / s 2) - centripetal acceleration T \u003d - period (s), T \u003d n= - frequency (Hz=1/s), n =	x = x o + u ox t + ; y = y o + u oy t + u x = u ox + g x t ; u y = u oy + g y t u o x = u 0 cosa u o y = u 0 sina g x = 0 g y = - g y u x u y s x

Special cases of uniformly accelerated motion under the action of gravity.

Additional Information

for special cases of problem solving.

1. Decomposition of a vector into projections. The modulus of the vector can be found using the Pythagorean theorem: S =	2. Average speed. 1) by definition 2) for 2 x S; if 3) , if t 1 \u003d t 2 \u003d ... \u003d t n u 1 u 2
3. Method of areas. On the chart u x (t) figure area numerically equal to the displacement or distance traveled. S \u003d S 1 - S 2 ℓ \u003d S 1 + S 2	4. Physical meaning of the derivative. For coordinate equations x(t) and y(t) → u x = x΄, u y = y΄, and a x = u΄ x = x΄΄, a y = u΄ y = y΄΄,
5. Wheel movement without slippage. u post = u rotation (if there is no slippage) The speed of a point on the wheel rim relative to the ground.	6. Flight range. Flight range is maximum at a throw angle of 45˚ υ 0 = const

S 1: S 2: S 3: ...: S n = 1: 3: 5: 7: ....: (2n-1)

S n \u003d S 1 (2n - 1) \u003d (2n - 1)

2) The ratio of movements made for time from countdown, at u o =0 equals:

S 1: S 2: S 3: …: S n = 1 2: 2 2: 3 2: 4 2: ….: n 2

S n \u003d S 1 n 2 \u003d n 2

indication: with rectilinear motion, the vectors v and a are directed along one straight line, with an increase in speed, they are co-directed.

7(A) The car travels half way at a speed u 1 , and the second half of the way with the speed u 2 ,

indication: this problem is a special case of finding the average speed. The derivation of the formula comes from the definition

, where s 1 \u003d s 2, and t 1 \u003d and t 2 \u003d

8(A) The equation for the dependence of the projection of the speed of a moving body on time has the form: u x = 3-2t(m/s). What is the equation for the projection of the body displacement?

1) s x =2t 2 (m) 3) s x =2t-3t 2 (m)

2) s x \u003d 3t-2t 2 (m) 4) s x \u003d 3t-t 2 (m)

indication: write the equation for the speed of uniformly accelerated motion in general view and, comparing it with the data in the problem, find what u 0 and a are equal to, insert these data into the displacement equation written in general form.

9(A) What is the distance traveled by a freely falling body from rest in 5 seconds? Acceleration of free fall is taken as 10m/s 2 .

1) 45 m 2) 55 m 3) 125 m 4) 250 m

indication: write down the expression h for the case u o \u003d 0, the desired h \u003d h 5 - h 4, where, respectively, h for 5 s and 4 s.

10(A) If a body that started to move uniformly accelerated from a state of rest covers the path S in the first second, then in the first three seconds it will cover the path

1) 3S 2) 4S 3) 8S 4) 9S

indication: use the displacement properties of uniformly accelerated motion for u 0 =0

11(A) Two cars are moving towards each other with speeds of 20 m/s and 90 km/h, respectively. What is the absolute speed of the first relative to the second?

1) 110 m/s 2) 60 m/s 3) 45 m/s 4) 5 m/s

indication: Relative speed is the difference of vectors, because the velocity vectors are directed oppositely, it is equal to the sum of their modules.

12(A) An observer from the bank sees that the swimmer is crossing a river of width h=189 m perpendicular to the bank. At the same time, the speed of the river flow is u=1.2 m/s, and the speed of the swimmer relative to the water is u=1.5 m/s. The swimmer will cross the river for ....

1) 70 s 2) 98 s 3) 126 s 4) 210 s

indication: construct a velocity triangle based on = + , go to the Pythagorean theorem, express from it the speed of the swimmer relative to the shore, and with it find the time.

13(A) At a speed of 10 m/s, the braking time of a truck is 3 s. If, during braking, the acceleration of the car is constant and does not depend on the initial speed, then when braking, the car will reduce its speed from 16 m/s to 9 m/s in ...

1) 1.5 s 2) 2.1 s 3) 3.5 s 4) 4.5 s

indication: from the consideration of the first situation, find the acceleration and substitute it into the velocity equation for the second situation, from which you can express the desired time.

14(A) A ship departs from the pier, moving at a constant speed of 18 km / h, after 40 s a boat departs from the same pier with an acceleration of 0.5 m / s 2. How long will it take him to overtake the ship, moving with constant acceleration?

1) 20 s 2) 30 s 3) 40 s 4) 50 s

indication: take the time of movement of the boat as t, then the time of movement of the ship is t + 40, write down the expressions for the movement of the ship (uniform movement) and the boat (uniformly accelerated movement) and equate them. Solve the resulting square quadratic equation relative to t. Don't forget to convert units 18 km/h = 5 m/s.

15(A) Two people play a ball, throwing it at an angle α=60º to the horizon. The ball is in flight t =2 s. In this case, the distance at which the players are located is equal to

1) 9.5 m 2) 10 m 3) 10.5 m 4) 11.5 m

indication: make a drawing - in axes x,y- parabola trajectory, the point of intersection of the parabola with the x-axis corresponds to the flight range, at this point the equation x(t) has the form s=u o cos60º t. To find u 0 use the equation y(t), which at the same point is 0=u o sin60º t- . Express u o from this equation and substitute it into the first equation. The calculation formula has the form

16(A) The aircraft flies with cargo to its destination at an altitude of 405m above sandy terrain with a horizontal profile at a speed of 130 m/s. In order for the load to hit the intended place on the ground (neglect the force of resistance to movement), the pilot must release it from the fasteners before reaching the target

1) 0.53 km 3) 0.95 km

2) 0.81 km 4) 1.17 km

indication: consider in theory the example "Movement of a body thrown horizontally." From the flight altitude expression, express the fall time and substitute it into the flight range formula.

17(B) A material point moves at a constant speed along a circle of radius R, making one revolution in time T. How will the physical quantities listed in the first column change if the radius of the circle increases and the period of revolution remains the same

Physical quantities. Their change.

A) Speed 1) will increase

B) Angular velocity 2) will decrease

B) centripetal 3) will not change

acceleration

BUT	B	AT

indication: write down the defining formulas of the proposed values through R and analyze their mathematical dependence, taking into account the constancy of the period, The numbers in the right column can be repeated.

18(B) What is the linear velocity of a surface point the globe corresponding to 60º north latitude? The radius of the Earth is 6400 km. Give your answer in m/s, rounded to the nearest integer.

indication: make a drawing and note that a point at the indicated latitude rotates relative to the earth's axis in a circle with a radius of r = R earth cos60º.

19(B) υ, m/s

indication: the easiest way to find a path through the area of the figure below the graph. A complex figure can be represented as the sum of two trapezoids and one rectangle.

20(С) = 2 m/s at an angle β=60º to the straight line AB. During the movement, the puck moves down to the straight line AB at point B. Neglecting the friction between the puck and the inclined plane, find the distance AB.

indication: to solve the problem, one should consider the trajectory of the washer - a parabola lying on an inclined plane and choose the coordinate axes, see fig.

In t.B x=s and the equation x(t) has the form s=u o cos60º t

You can find t from the equation y(t), at this point it will look like 0=u o sin60ºt - . Solving this system of equations together, find s.

Answers to learning tasks.

1A	2A	3A	4A	5A	6A	7A	8A	9A	10A

11A	12A	13A	14A	15A	16A	17V	18V	19V	20C
									69 cm

Training tasks.

1(A) In which case can a projectile be taken as a material point:

a) calculation of the range of the projectile;

b) calculation of the shape of the projectile, providing a decrease in air resistance.

1) Only in the first case. 2) Only in the second case.

3) In both cases. 4) Neither in the first nor in the second case.

2(A) The wheel rolls down a flat hill in a straight line. What trajectory

describes the center of the wheel relative to the road surface?

1) Circle. 3) Spiral.

2) Cycloid. 4) Direct.

3(A) What is the displacement of a point moving along a circle of radius R when it is rotated 90º?

1) R/2 2) R 3) 2R 4) R

4(A) Which of the graphs can be a graph of the path traveled by the body?

5(A) The figure shows a graph of the projection of the speed of the body. What is the modulus of the minimum acceleration of the body along the entire route?

1) 2.4 m/s2 u x, m/s

6(A) The body moves uniformly in a circle. Indicate the correct location of the linear velocity and acceleration vectors in t.A.

2) 4)

7(A) The car travels half the time at a speed u 1 , and the second half of the time with the speed u 2 , moving in the same direction. What is the average speed of the car?

8(A) The equation for the dependence of the coordinate of a moving body on time has the form:

X = 4 - 5t + 3t 2 (m). What is the equation for the projection of the body's velocity?

1) u x = - 5 + 6t (m/s) 3) u x = - 5t + 3t 2 (m/s)

2) u x = 4 - 5t (m/s) 4) u x = - 5t + 3t (m/s)

9(A) The parachutist descends vertically downwards with a constant speed u = 7 m/s. When he is at a height h = 160 m, a lighter falls out of his pocket. The time for the lighter to fall to the ground is

1) 4 s 2) 5 s 3) 8 s 4) 10 s

10(A) If a body that started to move uniformly accelerated from a state of rest covers the path S in the first second, then in the fourth second it will cover the path

1) 3S 2) 5S 3) 7S 4) 9S

11(A) With what speed do two cars move away from each other, moving away from the intersection along mutually perpendicular roads with speeds of 40 km/h and 30 km/h?

1) 50km/h 2) 70km/h 3) 10km/h 4) 15km/h

12(A) Two objects move according to the equations u x 1 \u003d 5 - 6t (m / s) and x 2 = 1 - 2t + 3t 2 (m). Find the module of their speed relative to each other 3 s after the start of movement.

1) 3 m/s 2) 29 m/s 3) 20 m/s 4) 6 m/s

13(A) When accelerating from a state of rest, the car acquired a speed of 12 m / s, having traveled 36 m. If the acceleration of the car is constant, then 5 seconds after the start, its speed will be equal to

1) 6 m/s 2) 8 m/s 3) 10 m/s 4) 15 m/s

14(A) Two skiers start with an interval ∆t. The speed of the first skier is 1.4 m/s, the speed of the second skier is 2.2 m/s. If the second skier overtakes the first one in 1 minute, then the interval ∆t is equal to

1) 0.15 min 3) 0.8 min

2) 0.6 min 4) 2.4 min

15(A) A ball is thrown with an initial speed of 30 m/s. The time of the entire flight of the ball at the throwing angle α=45º is equal to

1) 1.2 s 2) 2.1 s 3) 3.0 s 4) 4.3 s

16(A) A stone is thrown from a tower with an initial velocity of 8 m/s in a horizontal direction. Its speed will become modulo equal to 10 m/s after

1) 0.6 s 2) 0.7 s 3) 0.8 s 4) 0.9 s

17(B) A material point moves at a constant speed along a circle of radius R. How will the physical quantities listed in the first column change if the point's rotation frequency decreases?

acceleration 3) will not change

B) Period of circulation

around the circumference

BUT	B	AT

18(B) Two material points move along circles with radii R 1 and R 2 and R 2 = 4 R 1 . If the linear velocities of the points are equal, the ratio of their centripetal accelerations a 1 / a 2 equals ……

19(B) Determine the average speed for the entire time of movement from the graph of the body's speed versus time. Specify the accuracy of the result to tenths.

υ, m/s

20(С) The inclined plane intersects the horizontal plane along the straight line AB. Angle between planes α=30º. A small puck starts moving up the inclined plane from point A with an initial speed u 0 = 2 m/s at an angle β=60º to the straight line AB. Find the maximum distance that the puck will move away from the straight line AB during the ascent on the inclined plane. Ignore the friction between the washer and the inclined plane.

Answers to training tasks.

1A	2A	3A	4A	5A	6A	7A	8A	9A	10A

11A	12A	13A	14A	15A	16A	17V	18V	19V	20C
								21.7 m/s	30 cm

Control tasks.

1 (A) The material point is:

1) a body of negligible mass;

2) the body is very small;

3) a point showing the position of the body in space;

4) a body whose dimensions can be neglected under the conditions of a given problem.

2(A) What is the change in position of one body relative to another called?

1) trajectory;

2) movement;

4) mechanical movement.

3(A) What is the displacement of a point moving along a circle of radius R when it is rotated by 180º?

1) R/2 2) R 3) 2R 4) R

4(A) The line that the body describes when moving in space is called:

1) trajectory;

2) movement;

4) mechanical movement.

1) 32 m/s 2) 23 m/s 3) 20 m/s 4) 21.5 m/s

10(A) A body that began to move uniformly accelerated from a state of rest with an acceleration of 2 m / s 2, then in the third second it will pass the path

1) 7 m 2) 5 m 3) 3 m 4) 2 m

11(A) The coordinates of bodies A and B moving along the same straight line change with time, as shown in the graph. What is the speed of body A relative to body B?

1) 40 m/s x.m

12(A) The escalator staircase rises with a speed u, with what speed relative to the walls, should a person go down it in order to rest relative to the people standing on the stairs going down?

1) u 2) 2u 3) 3u 4) 4u

1) 12.3 m 3) 28.4 m

2) 16.5 m 4) 33.4 m

14(A) A truck and a motorcyclist travel in the same direction along a ring road 5 km long with speeds u 1, respectively. = 40 km/h and u 2 = 100 km/h. If at the initial moment of time they were in the same place, then the motorcyclist will catch up with the car, passing

1) 3.3 km 3) 8.3 km

2) 6.2 km 4) 12.5 km

15(A) A body is thrown from the Earth's surface at an angle α to the horizon with an initial velocity u 0 = 10m/s, if the range of the body is L = 10 m, then the angle α is equal to

1) 15º 2) 22.5º 3) 30º 4) 45º

16(A) The boy threw the ball horizontally from a window at a height of 20 m. The ball fell at a distance of 8 m from the wall of the house. With what initial speed was the ball thrown?

1) 0.4 m/s 2) 2.5 m/s 3) 3 m/s 4) 4 m/s

17(B) A material point moves at a constant speed along a circle of radius R. How will the physical quantities listed in the first column change if the speed of the point increases?

Physical quantities. Their change.

A) Angular velocity 1) will increase

B) Centripetal 2) decrease

acceleration 3) will not change

B) Period of circulation

around the circumference

BUT	B	AT

18(B) Determine the path traveled in 5 s from the graph of body velocity versus time.

υ, m/s

19(B) The centripetal acceleration of a material point moving along a circle, with an increase in linear velocity by 2 times and an angular velocity by 2 times with a constant radius, increased by .... once.

20(С) The inclined plane intersects the horizontal plane along the straight line AB.

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1. Mechanical motion is a change in the position of a body in space relative to other bodies over time. Exist different kinds mechanical movement. If all points of the body move in the same way and any straight line drawn in the body remains parallel to itself during its movement, then such a movement is called progressive(Fig. 1).

The points of a rotating wheel describe circles about the axis of this wheel. The wheel as a whole and all its points make rotational movement (Fig. 2).

If a body, such as a ball suspended from a thread, deviates from vertical position now in one direction, then in the other direction, then its movement is oscillatory(Fig. 3).

2. The definition of the concept of mechanical motion includes the words "relative to other bodies." They mean that a given body can be at rest relative to some bodies and move relative to other bodies. Thus, a passenger sitting in a bus moving relative to buildings also moves relative to them, but is at rest relative to the bus. A raft floating along the river is stationary relative to the water, but moves relative to the shore (Fig. 4). Thus, speaking of the mechanical motion of a body, it is necessary to indicate the body relative to which the given body is moving or at rest. Such a body is called a reference body. In the above example with a moving bus, a house, or a tree, or a pole near the bus stop can be chosen as the reference body.

To determine the position of the body in space, enter coordinate system, which is associated with the reference body. When considering the motion of a body along a straight line, a one-dimensional coordinate system is used, i.e. one coordinate axis is associated with the reference body, for example, the OX axis (Fig. 5).

If the body moves along a curved path, then the coordinate system will already be two-dimensional, since the position of the body is characterized by two coordinates X and Y (Fig. 6). Such a movement is, for example, the movement of the ball from a kick by a football player or an arrow fired from a bow.

If the motion of a body in space is considered, for example, the motion of a flying aircraft, then the coordinate system associated with the reference body will consist of three mutually perpendicular coordinate axes (OX, OY and OZ) (Fig. 7).

Since when the body moves, its position in space, i.e. its coordinates change over time, then a device (clock) is needed that allows you to measure time and determine which point in time corresponds to one or another coordinate.

Thus, to determine the position of the body in space and change this position over time, it is necessary reference body,associated coordinate system and method of measuring time,t.e.watch, which together represent reference system(Fig. 7).

3. To study the movement of a body means to determine how its position changes, i.e. coordinate, over time.

If you know how the coordinate changes over time, you can determine the position (coordinate) of the body at any time.

The main task of mechanics is to determine the position (coordinate) of the body at any time.

To indicate how the position of the body changes over time, it is necessary to establish a relationship between the quantities characterizing this movement, i.e. find mathematical description motion or, in other words, write down the equation of motion of the body.

The branch of mechanics that studies how to describe the motion of bodies is called kinematics.

4. Any moving body has certain dimensions, and its various parts occupy different positions in space. The question arises how to determine the position of the body in space in this case. In a number of cases, there is no need to indicate the position of each point of the body and for each point to write down the equation of motion.

So, since in translational motion all points of the body move in the same way, there is no need to describe the movement of each point of the body.

The movement of each point of the body does not need to be described when solving such problems, when the dimensions of the body can be neglected. For example, if we are interested in the speed with which a swimmer swims his distance, then there is no need to consider the movement of each point of the swimmer. If it is necessary to determine the buoyancy force acting on the ball, then it is no longer possible to neglect the size of the swimmer. If we want to calculate the travel time of a spacecraft from Earth to a space station, then the ship can be considered as a single entity and represented as a point. If, however, the docking mode of the ship with the station is calculated, then by representing the ship as a point, it is impossible to solve this problem.

Thus, to solve a number of problems related to the movement of bodies, the concept is introduced material point .

A material point is a body whose dimensions can be neglected under the conditions of a given problem.

In the above examples, a swimmer can be considered a material point when calculating the speed of his movement, spaceship when determining the time of its movement.

A material point is a model of real objects, real bodies. Considering the body as a material point, we abstract from signs that are not essential for solving a specific problem, in particular, from the dimensions of the body.

5. When moving, the body sequentially passes through points in space, connecting which, you can get a line. This line along which the body moves is called the trajectory.. The path can be visible or invisible. The visible trajectory is described by a tram when moving along rails, a skier gliding along a ski track, chalk, which is used to write on a blackboard. The trajectory of a flying aircraft is invisible in most cases; the trajectory of a crawling insect is invisible.

The trajectory of a body is relative: its shape depends on the choice of reference system. Thus, the trajectory of the rim points of a bicycle wheel moving along a straight road is a circle relative to the wheel axis, and a helix relative to the Earth (Fig. 8 a, b).

6. One of the characteristics of mechanical motion is the path traveled by the body. A path is a physical quantity equal to the distance traveled by the body along the trajectory.

If the trajectory of the body, its initial position and the path it has traveled in the time \(t \) are known, then the position of the body at the time \(t \) can be found. (Fig. 9)

The path is denoted by the letter \(l \) (sometimes \(s \) ), the basic unit of the path is 1 m: \([\,\mathrm(l)\,] \) = 1 m. The multiple unit of the path is kilometer (1 km = 1000 m); submultiples - decimeter (1 dm = 0.1 m), centimeter (1 cm = 0.01 m) and millimeter (1 mm = 0.001 m).

The path is a relative value, the value of the path depends on the choice of reference system. Thus, the path of a passenger passing from the end of a moving bus to its front door is equal to the length of the bus in the reference frame associated with the bus. In the reference frame associated with the Earth, it is equal to the sum of the length of the bus and the path that the bus traveled relative to the Earth.

7. If the trajectory of the body is unknown, then the value of the path will not allow you to set its position at any time, since the direction of the body is not defined. In this case, another characteristic of mechanical movement is used - moving.

Displacement - a vector connecting the initial position of the body with its final position(Fig. 10)

Moving - vector physical quantity, has direction and numerical value, denoted by \(\overrightarrow(s) \) . Unit of displacement \([\,\mathrm(s)\,] \) = 1 m.

Knowing the initial position of the body, its movement (direction and module) for a certain period of time, it is possible to determine the position of the body at the end of this period of time.

It should be borne in mind that the displacement in the general case does not coincide with the trajectory, and the displacement module does not coincide with the path traveled. This coincidence takes place only when the body moves along a rectilinear trajectory in one direction. For example, if a swimmer swam a 100-meter distance in a pool with a 50-m lane, then his path is 100 m, and the displacement modulus is zero.

Displacement, as well as the path, is a relative value, depending on the choice of reference system.

When solving problems, projections of the displacement vector are used. Figure 10 shows the coordinate system and the displacement vector in this coordinate system.

Move start coordinates - \(x_0, y_0 \) ; move end coordinates - \(x_1, y_1 \) . The projection of the displacement vector onto the OX axis is: \(s_x=x_1-x_0 \) . The projection of the displacement vector onto the OY axis is: \(s_y=y_1-y_0 \) .

The modulus of the displacement vector is: \(s=\sqrt(s^2_x-s^2_y) \) .

Part 1

1. The reference system includes

1) only the reference body
2) only reference body and coordinate system
3) only the reference body and clock
4) reference body, coordinate system, clock

2. The relative value is: A. Way; B. Moving. Correct answer

1) only A
2) only B
3) both A and B
4) neither A nor B

3. A subway passenger stands on an upward moving escalator. He is relatively immobile.

1) passengers standing on another escalator moving down
2) other passengers standing on the same escalator
3) passengers walking up the same escalator
4) Escalator balustrade fixtures

4. Relative to what body is a car moving along a freeway at rest?

1) relative to another car moving at the same speed in the opposite direction
2) relative to another car moving at the same speed in the same direction
3) regarding the traffic light
4) relative to a pedestrian walking along the road

5. Two cars are moving at the same speed of 20 m/s relative to the Earth in the same direction. What is the speed of one car in the reference frame associated with the other car?

1) 0
2) 20 m/s
3) 40 m/s
4) -20 m/s

6. Two cars are moving at the same speed of 15 m/s relative to the Earth towards each other. What is the speed of one car in the reference frame associated with the other car?

1) 0
2) 15 m/s
3) 30 m/s
4) -15 m/s

7. What is the trajectory of the point of the propeller blade of a flying helicopter relative to the Earth?

8. The ball falls from a height of 2 m and after hitting the floor rises to a height of 1.3 m.

1) \ (l \) \u003d 3.3 m, \ (s \) \u003d 3.3 m
2) \(l \) \u003d 3.3 m,\ (s\) \u003d 0.7 m
3) \(l \) \u003d 0.7 m,\ (s\) \u003d 0.7 m
4) \(l \) \u003d 0.7 m,\ (s\) \u003d 3.3 m

9. Solve two problems. 1. Calculate the speed of the train between two stations. 2. Determine the friction force acting on the train. When solving what problem can a train be considered a material point?

1) only the first
2) only the second
3) both first and second
4) neither first nor second

10. The wheel rim point describes half a circle of radius \(R\) when the bicycle is moving. What are the path \(l \) and the displacement modulus \(s \) of the rim point?

1)\(l=2R \) , \(s=2R \)
2)\(l=\pi R \) ,\(s=2R \)
3)\(l=2R \) ,\(s=\pi R \)
4) \(l=\pi R \) , \(s=\pi R \) .

11. Match the knowledge items in the left column with the concepts in the right column. In the table under the number of the knowledge element in the left column, write down the corresponding number of the concept you have chosen in the right column.

KNOWLEDGE ELEMENT
A) physical quantity
B) unit of magnitude
B) measuring instrument

CONCEPT
1) trajectory
2) path
3) stopwatch
4) kilometer
5) reference system

12. Establish a correspondence between the values in the left column and the nature of the value in the right column. In the table under the number of the knowledge element in the left column, write down the corresponding number of the concept you have chosen in the right column.

VALUE
A) way
B) moving
B) displacement projection

CHARACTER OF VALUE
1) scalar
2) vector

Part 2

13. The car turned onto the road making an angle of 30° with the main road and moved along it, the modulus of which is 20 m. Determine the projection of the car's movement onto the main road and onto the road perpendicular to the main road.

Answers

On the this lesson, the topic of which is: “Determining the coordinates of a moving body”, we will talk about how you can determine the location of the body, its coordinate. Let's talk about reference systems, consider the problem as an example, and also remember what displacement is

Imagine: you threw the ball with all your might. How to determine where it will be in two seconds? You can wait two seconds and just see where he is. But even without looking, you can approximately predict where the ball will be: the throw was stronger than usual, directed at a large angle to the horizon, which means it will fly high, but not far ... Using the laws of physics, it will be possible to accurately determine the position of our ball.

To determine the position of a moving body at any moment in time is the main task of kinematics.

Let's start with the fact that we have a body: how to determine its position, how to explain to someone where it is? We will say about a car: it is on the road 150 meters before a traffic light or 100 meters behind an intersection (see Fig. 1).

Rice. 1. Locating the machine

Or on the highway 30 km south of Moscow. Let's say about the phone on the table: it is 30 centimeters to the right of the keyboard or near the far corner of the table (see fig. 2).

Rice. 2. The position of the phone on the table

Note that we will not be able to determine the position of the car without mentioning other objects, without being attached to them: a traffic light, a city, a keyboard. We define position, or coordinates, always relative to something.

Coordinates are a set of data that determines the position of an object, its address.

Examples of ordered and unordered names

The body coordinate is its address where we can find it. He is orderly. For example, knowing the row and place, we determine exactly where our place is in the cinema hall (see Fig. 3).

Rice. 3. Cinema hall

A letter and a number, such as e2, precisely specifies the position of the piece on the chessboard (see Fig. 4).

Rice. 4. The position of the piece on the board

Knowing the address of the house, for example Solnechnaya street 14, we will look for it on this street, on the even side, between houses 12 and 16 (see Fig. 5).

Rice. 5. Home search

Street names are not ordered, we will not search alphabetically for Solnechnaya Street between Rozovaya and Turgenev streets. Phone numbers and license plates of cars are also not ordered (see Fig. 6).

Rice. 6. Unordered names

These consecutive numbers are just a coincidence, not a neighborhood.

We can set the position of the body in different coordinate systems, as we like. For the same car, you can set the exact geographical coordinates(latitude and longitude) (see Fig. 7).

Rice. 7. Longitude and latitude of the area

Rice. 8. Location relative to the point

Moreover, if we choose different such points, we will get different coordinates, although they will set the position of the same car.

So, the position of the body relative to different bodies in different coordinate systems will be different. What is movement? Movement is the change in body position over time. Therefore, we will describe the motion in different frames of reference in different ways, and it makes no sense to consider the motion of a body without a frame of reference.

For example, how does a glass of tea move on a table on a train if the train itself is moving? Looking at what. Relative to the table or the passenger sitting next to the seat, the glass is at rest (see Fig. 9).

Rice. 9. The movement of the glass relative to the passenger

About a tree railway the glass moves along with the train (see Fig. 10).

Rice. 10. The movement of the glass along with the train relative to the tree

Relative to the earth's axis, the glass and the train, together with all points on the earth's surface, will also move in a circle (see Fig. 11).

Rice. 11. The movement of the glass with the rotation of the Earth relative to the earth's axis

Therefore, it makes no sense to talk about motion in general, motion is considered in relation to the frame of reference.

Everything we know about the motion of a body can be divided into observable and calculated. Consider the example of the ball we threw. The observable is its position in the chosen coordinate system when we just drop it (see Fig. 12).

Rice. 12. Surveillance

This is the point in time that we abandoned it; the time that has elapsed since the throw. Let there be no speedometer on the ball that would show the speed of the ball, but its module, like the direction, can also be found using, for example, slow motion.

With the help of observed data, we can predict, for example, that the ball will fall 20 m from the place of the throw in 5 seconds or hit the top of the tree in 3 seconds. The position of the ball at any given time is, in our case, the computed data.

What determines each new position of a moving body? It is determined by displacement, because displacement is a vector that characterizes a change in position. If the beginning of the vector is aligned with the initial position of the body, then the end of the vector will indicate the new position of the moved body (see Fig. 13).

Rice. 13. Displacement vector

Let's consider a few examples of determining the coordinates of a moving body by its displacement.

Let the body move in a straight line from point 1 to point 2. Let's construct a displacement vector and denote it (see Fig. 14).

Rice. 14. Body movement

The body moved along one straight line, which means that we will need only one coordinate axis directed along the movement of the body. Let's say we are watching the movement from the side, let's match the reference point with the observer.

Displacement is a vector, it is more convenient to work with projections of vectors on the coordinate axes (we have one). - vector projection (see Fig. 15).

Rice. 15. Vector projection

How to determine the coordinate of the starting point, point 1? We lower the perpendicular from point 1 to the coordinate axis. This perpendicular will intersect the axis and mark the coordinate of point 1 on the axis. We also determine the coordinate of point 2 (see Fig. 16).

Rice. 16. Lowering perpendiculars to the OX axis

The displacement projection is equal to:

With this direction of the axis and displacement will be modulo equal to the displacement itself.

Knowing the initial coordinate and displacement, finding the final coordinate of the body is a matter of mathematics:

The equation

An equation is an equality containing an unknown term. What is its meaning?

Any task is that we know something, but something is not, and the unknown must be found. For example, a body from a certain point moved 6 m in the direction of the coordinate axis and ended up at a point with coordinate 9 (see Fig. 17).

Rice. 17. Initial point position

How to find from what point the body started moving?

We have a pattern: the displacement projection is the difference between the final and initial coordinates:

The meaning of the equation will be that we know the displacement and the final coordinate () and we can substitute these values, but we don’t know the initial coordinate, it will be unknown in this equation:

And already solving the equation, we get the answer: the initial coordinate.

Consider another case: the displacement is directed in the direction opposite to the direction of the coordinate axis.

The coordinates of the start and end points are determined in the same way as before - the perpendiculars are lowered onto the axis (see Fig. 18).

Rice. 18. The axis is directed in the opposite direction

The projection of displacement (nothing changes) is equal to:

Note that is greater than , and the displacement projection , when directed against the coordinate axis, will be negative.

The final coordinate of the body from the equation for the displacement projection is:

As you can see, nothing changes: in the projection onto the coordinate axis, the final position is equal to initial position plus displacement projection. Depending on the direction in which the body has moved, the displacement projection will be positive or negative in the given coordinate system.

Consider the case when the displacement and the coordinate axis are directed at an angle to each other. Now one coordinate axis is not enough for us, we need a second axis (see Fig. 19).

Rice. 19. The axis is directed in the opposite direction

Now the displacement will have a non-zero projection on each coordinate axis. These displacement projections will be defined as before:

Note that the modulus of each of the projections in this case is less than the displacement modulus. We can easily find the displacement modulus using the Pythagorean theorem. It can be seen that if we build right triangle(see Fig. 20), then its legs will be equal and, and the hypotenuse is equal to the displacement modulus or, as is often written, simply.

Rice. 20. Triangle of Pythagoras

Then, according to the Pythagorean theorem, we write:

The car is located 4 km east of the garage. Use one axis of coordinates pointing east, with the origin in the garage. Specify the coordinate of the car in the given system after 3 minutes, if the car was driving at a speed of 0.5 km/min to the west during this time.

The problem does not say anything about the fact that the car turned or changed speed, so we consider the movement to be uniform and straight.

Let's draw a coordinate system: the origin is at the garage, the x-axis is directed to the east (see Fig. 21).

The car was initially at a point and moved to the west according to the condition of the problem (see Fig. 22).

Rice. 22. Vehicle movement to the west

The displacement projection, as we have repeatedly written, is equal to:

We know that the car traveled 0.5 km every minute, so to find the total movement, you need to multiply the speed by the number of minutes:

This is where physics ended, it remains to mathematically express the desired coordinate. We express it from the first equation:

Let's substitute displacement:

It remains to substitute the numbers and get the answer. Remember that the car was moving west against the direction of the x-axis, which means that the projection of the speed is negative: .

Problem solved.

The main thing that we used today to determine the coordinate is the expression for the displacement projection:

And from it we have already expressed the coordinate:

In this case, the displacement projection itself can be set, can be calculated as , as it was in the problem of uniform rectilinear motion, it can be calculated more difficult, which we still have to study, but in any case, the coordinate of the moving body (where the body ended up) can be determined from the initial coordinate (where the body was) and according to the displacement projection (where it moved).

This concludes our lesson, goodbye!

Bibliography

Sokolovich Yu.A., Bogdanova GS Physics: A Handbook with Examples of Problem Solving. - 2nd edition, redistribution. - X .: Vesta: Publishing house "Ranok", 2005. - 464 p.
Peryshkin A.V., Gutnik E.M. Physics: Grade 9. Tutorial for educational institutions. - 14th ed. - M.: Bustard, 2009.

Class-fizika.narod.ru ().
Av-physics.narod.ru ().
Class-fizika.narod.ru ().

Homework

What is movement, path, trajectory?
How can body coordinates be determined?
Write down the formula for determining the displacement projection.
How will the displacement module be determined if the displacement has projections onto two coordinate axes?