« Physics - 10th grade"
How does uniform motion differ from uniformly accelerated motion?
How is the route schedule different? uniformly accelerated motion from the path schedule at uniform motion?
What is the projection of a vector onto any axis?
In the case of uniform rectilinear motion, you can determine the speed from a graph of the coordinates versus time.
The velocity projection is numerically equal to the tangent of the angle of inclination of the straight line x(t) to the abscissa axis. Moreover, the higher the speed, the larger angle tilt
Rectilinear uniformly accelerated motion.
Figure 1.33 shows graphs of the projection of acceleration versus time for three different meanings acceleration during rectilinear uniformly accelerated motion of a point. They are straight lines parallel to the abscissa axis: a x = const. Graphs 1 and 2 correspond to movement when the acceleration vector is directed along the OX axis, graph 3 - when the acceleration vector is directed in the opposite direction to the OX axis.
With uniformly accelerated motion, the velocity projection depends linearly on time: υ x = υ 0x + a x t. Figure 1.34 shows graphs of this dependence for the indicated three cases. In this case, the initial speed of the point is the same. Let's analyze this graph.
Projection of acceleration From the graph it is clear that, than more acceleration points, the greater the angle of inclination of the straight line to the t axis and, accordingly, the greater the tangent of the angle of inclination, which determines the value of acceleration.
Over the same period of time, with different accelerations, the speed changes to different values.
At positive value projection of acceleration over the same period of time, the projection of velocity in case 2 increases 2 times faster than in case 1. When negative value projection of acceleration onto the OX axis, the velocity projection modulo changes to the same value as in case 1, but the speed decreases.
For cases 1 and 3, the graphs of the velocity modulus versus time will be the same (Fig. 1.35).
Using the graph of speed versus time (Figure 1.36), we find the change in the coordinates of the point. This change is numerically equal to the area of the shaded trapezoid, in in this case coordinate change in 4 s Δx = 16 m.
We found a change in coordinates. If you need to find the coordinate of a point, then you need to add it to the found number initial value. Let in starting moment time x 0 = 2 m, then the value of the coordinate of the point in at the moment time equal to 4 s is equal to 18 m. In this case, the displacement module equal to the path traversed by the point, or a change in its coordinates, i.e. 16 m.
If the movement is uniformly slow, then the point during the selected time interval can stop and begin to move in the direction opposite to the initial one. Figure 1.37 shows the dependence of the velocity projection on time for such a movement. We see that at a time equal to 2 s, the direction of the velocity changes. The change in coordinates will be numerically equal to algebraic sum areas of shaded triangles.
Calculating these areas, we see that the change in coordinate is -6 m, which means that in the direction opposite to the OX axis, the point passed longer distance than in the direction of this axis.
Square over we take the t axis with a plus sign, and the area under the t axis, where the velocity projection is negative, with a minus sign.
If at the initial moment of time the speed of a certain point was equal to 2 m/s, then its coordinate at the moment of time equal to 6 s is equal to -4 m. The modulus of movement of the point in this case is also equal to 6 m - the modulus of change in coordinates. However, the path traveled by this point is equal to 10 m - the sum of the areas of the shaded triangles shown in Figure 1.38.
Let's plot the dependence of the x coordinate of a point on time. According to one of the formulas (1.14), the curve of coordinate versus time - x(t) - is a parabola.
If the point moves at a speed, the graph of which versus time is shown in Figure 1.36, then the branches of the parabola are directed upward, since a x > 0 (Figure 1.39). From this graph we can determine the coordinate of the point, as well as the speed at any time. So, at a time equal to 4 s, the coordinate of the point is 18 m.
For the initial moment of time, drawing a tangent to the curve at point A, we determine the tangent of the angle of inclination α 1, which is numerically equal to initial speed, i.e. 2 m/s.
To determine the speed at point B, draw a tangent to the parabola at this point and determine the tangent of the angle α 2. It is equal to 6, therefore the speed is 6 m/s.
The graph of the path versus time is the same parabola, but drawn from the origin (Fig. 1.40). We see that the path continuously increases over time, the movement occurs in one direction.
If the point moves at a speed, the graph of the projection of which versus time is shown in Figure 1.37, then the branches of the parabola are directed downward, since a x< 0 (рис. 1.41). При этом моменту времени, равному 2 с, соответствует вершина параболы. Касательная в точке В параллельна оси t, угол наклона касательной к этой оси равен нулю, и скорость также равна нулю. До этого момента времени тангенс угла наклона касательной уменьшался, но был положителен, движение точки происходило в направлении оси ОХ.
Starting from the moment of time t = 2 s, the tangent of the angle of inclination becomes negative, and its module increases, this means that the point moves in the direction opposite to the initial one, while the module of the movement speed increases.
Motion module equal to modulus the difference between the coordinates of a point at the final and initial moments of time and is equal to 6 m.
The graph of the distance traveled by a point versus time, shown in Figure 1.42, differs from the graph of displacement versus time (see Figure 1.41).
Regardless of the direction of the speed, the path traveled by the point continuously increases.
Let us derive the dependence of the point coordinates on the velocity projection. Speed υx = υ 0x + a x t, hence
In the case of x 0 = 0 and x > 0 and υ x > υ 0x, the graph of the coordinate versus speed is a parabola (Fig. 1.43).
In this case, the greater the acceleration, the less steep the branch of the parabola will be. This is easy to explain, since the greater the acceleration, the less the distance that the point must travel for the speed to increase by the same amount as when moving with less acceleration.
In case a x< 0 и υ 0x >0 the velocity projection will decrease. Let us rewrite equation (1.17) in the form where a = |a x |. The graph of this relationship is a parabola with branches directed downward (Fig. 1.44).
Accelerated movement.
Using graphs of the velocity projection versus time, you can determine the coordinate and acceleration projection of a point at any time for any type of movement.
Let the projection of the point's velocity depend on time as shown in Figure 1.45. It is obvious that in the time interval from 0 to t 3 the movement of the point along the X axis occurred with variable acceleration. Starting from the moment of time equal to t 3, the movement is uniform with constant speedυ Dx. According to the graph, we see that the acceleration with which the point moved continuously decreased (compare the angle of inclination of the tangent at points B and C).
The change in the x coordinate of a point during time t 1 is numerically equal to the area curved trapezoid OABt 1, for time t 2 - area OACt 2, etc. As we can see from the graph of the velocity projection versus time, you can determine the change in the coordinates of the body over any period of time.
From a graph of coordinates versus time, you can determine the value of speed at any point in time by calculating the tangent of the tangent to the curve at the point corresponding to a given point in time. From Figure 1.46 it follows that at time t 1 the velocity projection is positive. In the time interval from t 2 to t 3, the speed is zero, the body is motionless. At time t 4 the speed is also zero (the tangent to the curve at point D is parallel to the x-axis). Then the velocity projection becomes negative, the direction of motion of the point changes to the opposite.
If the graph of the velocity projection versus time is known, you can determine the acceleration of the point, and also, knowing the initial position, determine the coordinate of the body at any time, i.e., solve the main problem of kinematics. From the graph of coordinates versus time, one can determine one of the most important kinematic characteristics movement - speed. In addition, from the indicated graphs you can determine the type of movement along the selected axis: uniform, with constant acceleration or movement with variable acceleration.
Questions.
1. Write down a formula that can be used to calculate the projection of a vector instantaneous speed rectilinear uniformly accelerated motion, if the following are known: a) the projection of the initial velocity vector and the projection of the acceleration vector; b) projection of the acceleration vector given that the initial speed is zero.
2. What is the projection graph of the velocity vector of uniformly accelerated motion at the initial speed: a) equal to zero; b) not equal to zero?
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3. How are the movements, the graphs of which are presented in Figures 11 and 12, similar and different from each other?
In both cases, the movement occurs with acceleration, but in the first case the acceleration is positive, and in the second case it is negative.
Exercises.
1. A hockey player lightly hit the puck with his stick, giving it a speed of 2 m/s. What will be the speed of the puck 4 s after impact if, as a result of friction with ice, it moves with an acceleration of 0.25 m/s 2?
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2. A skier slides down a mountain from a state of rest with an acceleration equal to 0.2 m/s 2 . After what period of time will its speed increase to 2 m/s?
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3. In the same coordinate axes construct graphs of the projection of the velocity vector (on the X-axis, co-directed with the initial velocity vector) for rectilinear uniformly accelerated motion for the cases: a) v ox = 1 m/s, a x = 0.5 m/s 2 ; b) v ox = 1 m/s, a x = 1 m/s 2; c) v ox = 2 m/s, a x = 1 m/s 2.
The scale is the same in all cases: 1 cm - 1 m/s; 1cm - 1s.
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4. In the same coordinate axes, construct graphs of the projection of the velocity vector (on the X axis, codirectional with the initial velocity vector) for rectilinear uniformly accelerated motion for the cases: a) v ox = 4.5 m/s, a x = -1.5 m/s 2 ; b) v ox = 3 m/s, a x = -1 m/s 2
Choose the scale yourself.
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5. Figure 13 shows graphs of the dependence of the magnitude of the velocity vector on time during the rectilinear motion of two bodies. With what absolute acceleration does body I move? body II?
Uniform rectilinear movement - This special case uneven movement.
Uneven movement- this is a movement in which a body (material point) makes unequal movements over equal periods of time. For example, a city bus moves unevenly, since its movement consists mainly of acceleration and deceleration.
Equally alternating motion is a movement in which the speed of the body ( material point) changes equally over any equal periods of time.
Body acceleration at uniformly alternating motion remains constant in magnitude and direction (a = const).
Uniform motion can be uniformly accelerated or uniformly decelerated.
Uniformly accelerated motion- this is the movement of a body (material point) with positive acceleration, that is, with such movement the body accelerates with constant acceleration. In the case of uniformly accelerated motion, the velocity module of the body increases over time, the direction of acceleration coincides with the direction of the speed of movement.
Equal slow motion is the movement of a body (material point) with negative acceleration, that is, with such movement the body slows down uniformly. In uniformly slow motion, the velocity and acceleration vectors are opposite, and the velocity modulus decreases over time.
In mechanics, any rectilinear motion is accelerated, therefore slow motion differs from accelerated motion only in the sign of the projection of the acceleration vector onto the selected axis of the coordinate system.
Average speed variable motion is determined by dividing the movement of the body by the time during which this movement was made. The unit of average speed is m/s.
V cp = s/t
– this is the speed of the body (material point) at a given moment in time or at a given point of the trajectory, that is, the limit to which it tends average speed with an infinite decrease in the time period Δt:
Instantaneous velocity vector uniformly alternating motion can be found as the first derivative of the displacement vector with respect to time:
Velocity vector projection on the OX axis:
V x = x’
this is the derivative of the coordinate with respect to time (the projections of the velocity vector onto other coordinate axes are similarly obtained).
is a quantity that determines the rate of change in the speed of a body, that is, the limit to which the change in speed tends with an infinite decrease in the time period Δt:
Acceleration vector of uniformly alternating motion can be found as the first derivative of the velocity vector with respect to time or as the second derivative of the displacement vector with respect to time:
If a body moves rectilinearly along the OX axis rectilinear Cartesian system coordinates coinciding in direction with the trajectory of the body, then the projection of the velocity vector onto this axis is determined by the formula:
V x = v 0x ± a x t
The “-” (minus) sign in front of the projection of the acceleration vector refers to uniformly slow motion. The equations for projections of the velocity vector onto other coordinate axes are written similarly.
Since in uniform motion the acceleration is constant (a = const), the acceleration graph is a straight line parallel to the 0t axis (time axis, Fig. 1.15).
Rice. 1.15. Dependence of body acceleration on time.
Dependence of speed on time- This linear function, the graph of which is a straight line (Fig. 1.16).
Rice. 1.16. Dependence of body speed on time.
Speed versus time graph(Fig. 1.16) shows that
In this case, the displacement is numerically equal to the area of the figure 0abc (Fig. 1.16).
The area of a trapezoid is equal to the product of half the sum of the lengths of its bases and its height. The bases of the trapezoid 0abc are numerically equal:
0a = v 0 bc = v
The height of the trapezoid is t. Thus, the area of the trapezoid, and therefore the projection of displacement onto the OX axis is equal to:
In the case of uniformly slow motion, the acceleration projection is negative and in the formula for the displacement projection a “–” (minus) sign is placed before the acceleration.
A graph of the velocity of a body versus time at various accelerations is shown in Fig. 1.17. The graph of displacement versus time for v0 = 0 is shown in Fig. 1.18.
Rice. 1.17. Dependence of body speed on time for different meanings acceleration.
Rice. 1.18. Dependence of body movement on time.
The speed of the body at a given time t 1 is equal to the tangent of the angle of inclination between the tangent to the graph and the time axis v = tg α, and the displacement is determined by the formula:
If the time of movement of the body is unknown, you can use another displacement formula by solving a system of two equations:
It will help us derive the formula for displacement projection:
Since the coordinate of the body at any time is determined by the sum of the initial coordinate and the displacement projection, it will look like this:
The graph of the coordinate x(t) is also a parabola (like the displacement graph), but the vertex of the parabola is at general case does not coincide with the origin. When a x< 0 и х 0 = 0 ветви параболы направлены вниз (рис. 1.18).
Let's show how you can find the path traveled by a body using a graph of speed versus time.
Let's start from the very beginning simple case– uniform movement. Figure 6.1 shows a graph of v(t) – speed versus time. It is a segment of a straight line parallel to the base of time, since with uniform motion the speed is constant.
The figure enclosed under this graph is a rectangle (it is shaded in the figure). Its area is numerically equal to the product of speed v and time of movement t. On the other hand, the product vt is equal to the path l traversed by the body. So, with uniform motion
way numerically equal to area the figure enclosed under the graph of speed versus time.
Let us now show that this remarkable property It also has uneven movement.
Let, for example, the graph of speed versus time look like the curve shown in Figure 6.2. 
Let us mentally divide the entire time of movement into such small intervals that during each of them the movement of the body can be considered almost uniform (this division is shown by dashed lines in Figure 6.2).
Then the path traveled during each such interval is numerically equal to the area of the figure under the corresponding lump of the graph. Therefore, the entire path is equal to the area of the figures contained under the entire graph. (The technique we used is the basis integral calculus, the basics of which you will study in the course “Beginnings of Mathematical Analysis”.)
2. Path and displacement during rectilinear uniformly accelerated motion
Let us now apply the method described above for finding the path to rectilinear uniformly accelerated motion.
The initial speed of the body is zero
Let's direct the x axis in the direction of body acceleration. Then a x = a, v x = v. Hence,
Figure 6.3 shows a graph of v(t). 
1. Using Figure 6.3, prove that in case of rectilinear uniformly accelerated motion without initial speed, the path l is expressed in terms of the acceleration module a and the time of movement t by the formula
l = at 2 /2. (2)
Main conclusion:
in rectilinear uniformly accelerated motion without initial speed, the distance traveled by the body is proportional to the square of the time of movement.
In this way, uniformly accelerated motion differs significantly from uniform motion.
Figure 6.4 shows graphs of the path versus time for two bodies, one of which moves uniformly, and the other uniformly accelerates without an initial speed. 
2. Look at Figure 6.4 and answer the questions.
a) What color is the graph for a body moving with uniform acceleration?
b) What is the acceleration of this body?
c) What are the speeds of the bodies at the moment when they have covered the same path?
d) At what point in time are the velocities of the bodies equal?
3. Having started, the car covered a distance of 20 m in the first 4 s. Consider the car’s motion to be linear and uniformly accelerated. Without calculating the acceleration of the car, determine how far the car will travel:
a) in 8 s? b) in 16 s? c) in 2 s?
Let us now find the dependence of the projection of displacement s x on time. In this case, the projection of acceleration onto the x axis is positive, so s x = l, a x = a. Thus, from formula (2) it follows:
s x = a x t 2 /2. (3)
Formulas (2) and (3) are very similar, which sometimes leads to errors in solving simple tasks. The fact is that the displacement projection value can be negative. This will happen if the x axis is directed opposite to the displacement: then s x< 0. А путь отрицательным быть не может!
4. Figure 6.5 shows graphs of travel time and displacement projection for a certain body. What color is the displacement projection graph?
The initial speed of the body is not zero
Let us recall that in this case the dependence of the velocity projection on time is expressed by the formula
v x = v 0x + a x t, (4)
where v 0x is the projection of the initial velocity onto the x axis.
We will further consider the case when v 0x > 0, a x > 0. In this case, we can again take advantage of the fact that the path is numerically equal to the area of the figure under the graph of speed versus time. (Consider other combinations of signs for the projection of initial velocity and acceleration yourself: the result will be the same general formula (5).
Figure 6.6 shows a graph of v x (t) for v 0x > 0, a x > 0. 
5. Using Figure 6.6, prove that in case of rectilinear uniformly accelerated motion with an initial speed, the projection of displacement
s x = v 0x + a x t 2 /2. (5)
This formula allows you to find the dependence of the x coordinate of the body on time. Let us recall (see formula (6), § 2) that the coordinate x of a body is related to the projection of its displacement s x by the relation
s x = x – x 0 ,
where x 0 is the initial coordinate of the body. Hence,
x = x 0 + s x , (6)
From formulas (5), (6) we obtain:
x = x 0 + v 0x t + a x t 2 /2. (7)
6. The dependence of the coordinate on time for a certain body moving along the x axis is expressed in SI units by the formula x = 6 – 5t + t 2.
a) What is the initial coordinate of the body?
b) What is the projection of the initial velocity onto the x-axis?
c) What is the projection of acceleration on the x-axis?
d) Draw a graph of the x coordinate versus time.
e) Draw a graph of the projected velocity versus time.
f) At what moment is the speed of the body equal to zero?
g) Will the body return to the starting point? If so, at what point(s) in time?
h) Will the body pass through the origin? If so, at what point(s) in time?
i) Draw a graph of the displacement projection versus time.
j) Draw a graph of the distance versus time.
3. Relationship between path and speed
When solving problems, relationships between path, acceleration and speed (initial v 0, final v or both) are often used. Let us derive these relations. Let's start with movement without an initial speed. From formula (1) we obtain for the time of movement:
Let's substitute this expression into formula (2) for the path:
l = at 2 /2 = a/2(v/a) 2 = v 2 /2a. (9)
Main conclusion:
in rectilinear uniformly accelerated motion without initial speed, the distance traveled by the body is proportional to the square of the final speed.
7. Having set off, the car picked up a speed of 10 m/s over a distance of 40 m. Consider the car’s motion to be linear and uniformly accelerated. Without calculating the acceleration of the car, determine how far from the start of movement the car traveled when its speed was equal to: a) 20 m/s? b) 40 m/s? c) 5 m/s?
Relationship (9) can also be obtained by remembering that the path is numerically equal to the area of the figure enclosed under the graph of speed versus time (Fig. 6.7). 
This consideration will help you easily cope with the next task.
8. Using Figure 6.8, prove that when braking with constant acceleration, the body travels the distance l t = v 0 2 /2a to a complete stop, where v 0 is the initial speed of the body, a is the acceleration modulus.
In case of braking vehicle(car, train) the distance traveled to a complete stop is called the braking distance. Please note: the braking distance at the initial speed v 0 and the distance traveled during acceleration from standstill to speed v 0 with the same acceleration a are the same.
9. During emergency braking on dry asphalt, the acceleration of the car is equal in absolute value to 5 m/s 2 . What is the braking distance of a car at initial speed: a) 60 km/h (maximum permitted speed in the city); b) 120 km/h? Find the braking distance at the indicated speeds during icy conditions, when the acceleration modulus is 2 m/s 2 . Compare the braking distances you found with the length of the classroom.
10. Using Figure 6.9 and the formula expressing the area of a trapezoid through its height and half the sum of the bases, prove that for rectilinear uniformly accelerated motion:
a) l = (v 2 – v 0 2)/2a, if the speed of the body increases;
b) l = (v 0 2 – v 2)/2a, if the speed of the body decreases.
11. Prove that the projections of displacement, initial and final velocity, as well as acceleration are related by the relation
s x = (v x 2 – v 0x 2)/2ax (10)
12. A car on a path of 200 m accelerated from a speed of 10 m/s to 30 m/s.
a) How fast was the car moving?
b) How long did it take the car to travel the indicated distance?
c) What is the average speed of the car?
Additional questions and tasks
13. The last car is uncoupled from a moving train, after which the train moves uniformly, and the car moves with constant acceleration until it comes to a complete stop.
a) Draw on one drawing graphs of speed versus time for a train and a carriage.
b) How many times is the distance covered by the car to the stop? less way traveled by the train in the same time?
14. Having left the station, the train drove uniformly accelerated for some time, then for 1 minute – uniformly at a speed of 60 km/h, after which again uniformly accelerated until it stopped at the next station. The acceleration modules during acceleration and braking were different. The train covered the distance between stations in 2 minutes.
a) Draw a schematic graph of the projection of the speed of the train as a function of time.
b) Using this graph, find the distance between the stations.
c) How far would the train travel if it accelerated on the first section of the route and slowed down on the second? What would be its maximum speed?
15. A body moves uniformly accelerated along the x axis. At the initial moment it was at the origin of coordinates, and the projection of its speed was equal to 8 m/s. After 2 s, the coordinate of the body became 12 m.
a) What is the projection of the acceleration of the body?
b) Plot a graph of v x (t).
c) Write a formula expressing the dependence x(t) in SI units.
d) Will the speed of the body be zero? If yes, at what point in time?
e) Will the body visit the point with coordinate 12 m a second time? If yes, at what point in time?
f) Will the body return to the starting point? If yes, then at what point in time, and what will be the distance traveled?
16. After the push, the ball rolls up inclined plane, after which it returns to the starting point. At a distance b from starting point the ball visited twice at intervals t 1 and t 2 after the push. The ball moved up and down along the inclined plane with the same magnitude of acceleration.
a) Direct the x-axis up along the inclined plane, select the origin at the point initial position ball and write a formula expressing the dependence x(t), which includes the modulus of the initial velocity of the ball v0 and the modulus of acceleration of the ball a.
b) Using this formula and the fact that the ball was at a distance b from the starting point at times t 1 and t 2, create a system of two equations with two unknowns v 0 and a.
c) Having solved this system of equations, express v 0 and a in terms of b, t 1 and t 2.
d) Express the entire path l traveled by the ball in terms of b, t 1 and t 2.
e) Find numeric values v 0 , a and l at b = 30 cm, t 1 = 1 s, t 2 = 2 s.
f) Plot graphs of v x (t), s x (t), l(t).
g) Using the sx(t) graph, determine the moment when the ball’s modulus of displacement was maximum.
