In this lesson, the topic of which is: “Equation of motion with constant acceleration. Forward movement,” we will remember what movement is, what it happens. Let’s also remember what acceleration is, consider the equation of motion with constant acceleration and how to use it to determine the coordinates of a moving body. Let's consider an example of a task for consolidating material.
the main task kinematics - determine the position of the body at any time. The body can be at rest, then its position will not change (see Fig. 1).
Rice. 1. Body at rest
A body can move in a straight line with constant speed. Then its movement will change uniformly, that is, equally over equal periods of time (see Fig. 2).
Rice. 2. Movement of a body when moving at a constant speed
Movement, speed multiplied by time, we have been able to do this for a long time. A body can move with constant acceleration; consider such a case (see Fig. 3).
Rice. 3. Body motion with constant acceleration
Acceleration
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Acceleration is the change in speed per unit time(see Fig. 4) :
Rice. 4. Acceleration Speed - vector quantity, therefore the change in speed, i.e. the difference between the vectors of the final and initial speed, is a vector. Acceleration is also a vector, directed in the same direction as the vector of the speed difference (see Fig. 5).
We are considering linear motion, so we can select a coordinate axis along the straight line along which the motion occurs, and consider the projections of the velocity and acceleration vectors onto this axis:
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Then its speed changes uniformly: (if its initial speed was zero). How to find the displacement now? It is impossible to multiply speed by time: the speed was constantly changing; which one to take? How to determine where during such a movement the body will be at any moment in time - today we will solve this problem.
Let’s immediately define the model: we are considering the rectilinear translational motion of a body. In this case, we can use the model material point. Acceleration is directed along the same straight line along which the material point moves (see Fig. 6).
Forward movement
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Translational motion is a motion in which all points of the body move equally: with same speed, making the same movement (see Fig. 7).
Rice. 7. Forward movement How else could it be? Wave your hand and observe: it is clear that the palm and shoulder moved differently. Look at the Ferris wheel: the points near the axis hardly move, but the cabins move at different speeds and along different trajectories (see Fig. 8).
Rice. 8. Movement of selected points on the Ferris wheel Look at a moving car: if you do not take into account the rotation of the wheels and the movement of engine parts, all points of the car move equally, we consider the movement of the car to be translational (see Fig. 9).
Rice. 9. Car movement Then there is no point in describing the movement of each point; you can describe the movement of one. We consider a car to be a material point. Please note that during translational movement, the line connecting any two points of the body during movement remains parallel to itself (see Fig. 10).
Rice. 10. Position of the line connecting two points |
The car drove straight for an hour. At the beginning of the hour his speed was 10 km/h, and at the end - 100 km/h (see Fig. 11).
Rice. 11. Drawing for the problem
The speed changed uniformly. How many kilometers did the car travel?
Let us analyze the condition of the problem.
The speed of the car changed uniformly, that is, its acceleration was constant throughout the journey. Acceleration by definition is equal to:
The car was driving straight, so we can consider its movement in projection onto one coordinate axis:
Let's find the displacement.
Increasing speed example
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Nuts are placed on the table, one nut per minute. It’s clear: no matter how many minutes pass, so many nuts will appear on the table. Now let’s imagine that the rate of placing nuts increases uniformly from zero: the first minute no nuts are placed, the second minute they put one nut, then two, three, and so on. How many nuts will be on the table after some time? It is clear that it is less than if maximum speed always supported. Moreover, it is clearly visible that it is 2 times less (see Fig. 12).
Rice. 12. Number of nuts at different laying speeds It’s the same with uniformly accelerated motion: let’s say that at first the speed was zero, but at the end it became equal (see Fig. 13).
Rice. 13. Change speed If the body were constantly moving at such a speed, its displacement would be equal to , but since the speed increased uniformly, it would be 2 times less. |
We know how to find displacement during UNIFORM movement: . How to work around this problem? If the speed does not change much, then the movement can be approximately considered uniform. The change in speed will be small over a short period of time (see Fig. 14).
Rice. 14. Change speed
Therefore, we divide the travel time T into N small sections duration (see Fig. 15).
Rice. 15. Splitting a period of time
Let's calculate the displacement at each time interval. The speed increases at each interval by:
On each segment we will consider the movement uniform and the speed approximately equal to the initial speed on this segment time. Let's see if our approximation will lead to an error if we assume the motion to be uniform over a short interval. The maximum error will be:
and the total error for the entire journey -> . For large N we assume the error is close to zero. We will see this on the graph (see Fig. 16): there will be an error at each interval, but the total error for sufficiently large quantities intervals will be negligible.
Rice. 16. Interval error
So, each subsequent speed value is the same amount greater than the previous one. From algebra we know that this is an arithmetic progression with a progression difference:
The path in sections (with uniform straight motion(see Fig. 17) is equal to:
Rice. 17. Consideration of areas of body movement
On the second section:
On n-th section the path is:
Arithmetic progression
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Arithmetic progression it's called this number sequence, in which each next number differs from the previous one by the same amount. An arithmetic progression is specified by two parameters: the initial term of the progression and the difference of the progression. Then the sequence is written like this: Sum of first terms arithmetic progression calculated by the formula:
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Let's sum up all the paths. This will be the sum of the first N terms of the arithmetic progression:
Since we have divided the movement into many intervals, we can assume that then:
We had many formulas, and in order not to get confused, we did not write the x indices each time, but considered everything in projection onto the coordinate axis.
So, we got the main formula equal to accelerated movement: displacement during uniformly accelerated motion in time T, which we, along with the definition of acceleration (change in speed per unit time), will use to solve problems:
We were working on solving a problem about a car. Let's substitute numbers into the solution and get the answer: the car traveled 55.4 km.
Mathematical part of solving the problem
We figured out the movement. How to determine the coordinate of a body at any moment in time?
By definition, the movement of a body over time is a vector, the beginning of which is at the initial point of movement, and the end at end point, in which the body will be after time. We need to find the coordinate of the body, so we write an expression for the projection of displacement onto the coordinate axis (see Fig. 18):
Rice. 18. Motion projection
Let's express the coordinate:
That is, the coordinate of the body at the moment of time is equal to initial coordinate plus the projection of the movement that the body has made in time. We have already found the projection of displacement during uniformly accelerated motion, all that remains is to substitute and write:
This is the equation of motion with constant acceleration. It allows you to find out the coordinates of a moving material point at any time. It is clear that we choose the moment of time within the interval when the model works: the acceleration is constant, the movement is rectilinear.
Why the equation of motion cannot be used to find a path
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In what cases can we consider movement modulo equal to path? When a body moves along a straight line and does not change direction. For example, with uniform rectilinear motion, we do not always clearly define whether we are finding a path or a displacement; they still coincide. With uniformly accelerated motion, the speed changes. If speed and acceleration are directed in opposite sides(see Fig. 19), then the velocity modulus decreases, and at some point it will become equal to zero and the speed will change direction, that is, the body will begin to move in the opposite direction.
Rice. 19. Velocity modulus decreases And then, if in this moment time the body is at a distance of 3 m from the beginning of observation, then its displacement is 3 m, but if the body first traveled 5 m, then turned around and traveled another 2 m, then the path will be 7 m. And how can you find it if you don’t know these numbers? You just need to find the moment when the speed is zero, that is, when the body turns around, and find the path to and from this point (see Fig. 20).
Rice. 20. The moment when the speed is 0 |
Bibliography
- Sokolovich Yu.A., Bogdanova G.S. Physics: A reference book with examples of problem solving. - 2nd edition repartition. - X.: Vesta: Ranok Publishing House, 2005. - 464 p.
- Landsberg G.S. Elementary textbook physicists; v.1. Mechanics. Heat. Molecular physics- M.: Publishing house "Science", 1985.
- Internet portal “kaf-fiz-1586.narod.ru” ()
- Internet portal “Study - Easy” ()
- Internet portal "Knowledge Hypermarket" ()
Homework
- What is an arithmetic progression?
- What kind of movement is called translational?
- What is a vector quantity characterized by?
- Write down the formula for acceleration through a change in speed.
- What is the form of the equation of motion with constant acceleration?
- The acceleration vector is directed towards the movement of the body. How will the body change its speed?
Studying classical mechanical movement Physics deals with kinematics. Unlike dynamics, science studies why bodies move. She answers the question of how they do it. In this article we will look at what acceleration and motion with constant acceleration are.
The concept of acceleration
When a body moves in space, over a period of time it covers a certain path, which is the length of the trajectory. To calculate this path, we use the concepts of speed and acceleration.
Speed as a physical quantity characterizes the rapidity in time of changes in the distance traveled. The speed is directed tangentially to the trajectory in the direction of the body movement.
Acceleration is somewhat more complex quantity. In short, it describes the change in speed at a given point in time. The math looks like this:
To understand this formula more clearly, let’s give a simple example: suppose that in 1 second of movement the speed of the body increased by 1 m/s. These numbers, substituted into the expression above, lead to the result: the acceleration of the body during this second was equal to 1 m/s 2 .
The direction of acceleration is completely independent of the direction of velocity. Its vector coincides with the vector of the resulting force that causes this acceleration.
It should be noted important point in the given definition of acceleration. This value characterizes not only the change in speed in magnitude, but also in direction. Last fact should be taken into account in case curvilinear movement. Further in the article only rectilinear motion will be considered.
Speed when moving with constant acceleration
Acceleration is constant if it maintains its magnitude and direction during movement. Such motion is called uniformly accelerated or uniformly decelerated - it all depends on whether acceleration leads to an increase in speed or to a decrease in speed.
In the case of a body moving with constant acceleration, the speed can be determined by one of following formulas:
The first two equations characterize uniformly accelerated movement. The difference between them is that the second expression is applicable for the case of non-zero initial velocity.
The third equation is an expression for the speed of uniformly slow motion with constant acceleration. Acceleration is directed against speed.
The graphs of all three functions v(t) are straight lines. In the first two cases, the straight lines have a positive slope relative to the x-axis; in the third case, this slope is negative.
Formulas for the distance traveled
For a path in the case of motion with constant acceleration (acceleration a = const), it is not difficult to obtain formulas if you calculate the integral of the speed over time. Having done this mathematical operation for the three equations written above, we obtain the following expressions for the path L:
L = v 0 *t + a*t 2 /2;
L = v 0 *t - a*t 2 /2.
The graphs of all three path functions versus time are parabolas. In the first two cases, the right branch of the parabola increases, and for the third function it gradually reaches a certain constant, which corresponds to the distance traveled until the body stops completely.
The solution of the problem
Moving at a speed of 30 km/h, the car began to accelerate. In 30 seconds he covered a distance of 600 meters. What was the acceleration of the car?
First of all, let's translate initial speed from km/h to m/s:
v 0 = 30 km/h = 30000/3600 = 8.333 m/s.
Now let's write the equation of motion:
L = v 0 *t + a*t 2 /2.
From this equality we express the acceleration, we get:
a = 2*(L - v 0 *t)/t 2 .
All physical quantities in this equation are known from the problem conditions. We substitute them into the formula and get the answer: a ≈ 0.78 m/s 2 . Thus, moving with constant acceleration, the car increased its speed by 0.78 m/s every second.
Let's also calculate (for fun) what speed he acquired after 30 seconds of accelerated movement, we get:
v = v 0 + a*t = 8.333 + 0.78*30 = 31.733 m/s.
The resulting speed is 114.2 km/h.
The position of bodies relative to the selected coordinate system is usually characterized by a radius vector depending on time. Then the position of the body in space at any time can be found using the formula:
.
(Recall that this is the main task of mechanics.)
Among the many various types the simplest movement is uniform– movement at a constant speed (zero acceleration), and the velocity vector () must remain unchanged. Obviously, such a movement can only be rectilinear. Precisely when uniform motion the movement is calculated by the formula:
Sometimes the body moves curvilinear trajectory so that the velocity module remains constant () (such movement cannot be called uniform and the formula cannot be applied to it). In this case distance traveled can be calculated using a simple formula:
An example of such a movement is movement in a circle with a constant absolute speed.
More difficult is uniformly accelerated motion– movement with constant acceleration (). For such a movement, two kinematic formulas are valid:
of which you can get two additional formulas, which can often be useful in solving problems:
;
Uniformly accelerated motion does not have to be rectilinear. It is only necessary that vector acceleration remained constant. An example of uniformly accelerated, but not always rectilinear motion is motion with acceleration free fall (g= 9.81 m/s 2), directed vertically downwards.
From school course physics is familiar and more complex movement – harmonic vibrations a pendulum for which the formulas are not valid.
At movement of a body in a circle with a constant absolute speed it moves with the so-called normal (centripetal) acceleration
directed towards the center of the circle and perpendicular to the speed of movement.
In more general case motion along a curved path with varying speed, the acceleration of a body can be decomposed into two mutually perpendicular components and represented as the sum of tangential (tangent) and normal (perpendicular, centripetal) acceleration:
,
where is the unit vector of the velocity vector and the unit unit normal to the trajectory; R– radius of curvature of the trajectory.
The motion of bodies is always described relative to some reference system (FR). When solving problems, it is necessary to choose the most convenient SO. For progressively moving COs, the formula is
allows you to easily move from one CO to another. In the formula – the speed of the body relative to one CO; – body speed relative to the second reference point; – speed of the second CO relative to the first.
Self-test questions and tasks
1) Model of a material point: what is its essence and meaning?
2) Formulate the definition of uniform, uniformly accelerated motion.
3) Formulate definitions of the main kinematic quantities(radius vector, displacement, velocity, acceleration, tangential and normal acceleration).
4) Write the formulas for the kinematics of uniformly accelerated motion and derive them.
5) Formulate Galileo’s principle of relativity.
2.1.1. Straight-line movement
Problem 22.(1) A car moves along a straight section of road at a constant speed of 90. Find the movement of the car in 3.3 minutes and its position at the same time, if in starting moment time the car was at a point whose coordinate is 12.23 km, and the axis Ox directed 1) along the movement of the car; 2) against the movement of the car.
Problem 23.(1) A cyclist moves along a country road to the north at a speed of 12 for 8.5 minutes, then he turns right at the intersection and travels another 4.5 km. Find the displacement of the cyclist during his movement.
Problem 24.(1) A skater moves in a straight line with an acceleration of 2.6, and in 5.3 s his speed increases to 18. Find initial value speed skater speed. How far will the athlete run in this time?
Problem 25.(1) The car moves in a straight line, slowing down in front of a speed limit sign of 40 with an acceleration of 2.3 How long did this movement last if before braking the car’s speed was 70? At what distance from the sign did the driver start to brake?
Problem 26.(1) With what acceleration is the train moving if its speed increases from 10 to 20 along a journey of 1200 m? How long did the train take on this journey?
Problem 27.(1) A body thrown vertically upward returns to the ground after 3 s. What was the initial speed of the body? What is the maximum height it has been to?
Problem 28.(2) A body on a rope is lifted from the surface of the earth with an acceleration of 2.7 m/s 2 vertically upward from a state of rest. After 5.8 s the rope broke. How long did it take the body to reach the ground after the rope broke? Neglect air resistance.
Problem 29.(2) The body begins to move without an initial speed with an acceleration of 2.4. Determine the path traveled by the body in the first 16 s from the beginning of the movement, and the path traveled over the next 16 s. At what average speed did the body move during these 32 s?
2.1.2. Uniformly accelerated motion in a plane
Problem 30.(1) A basketball player throws a ball into a hoop at a speed of 8.5 at an angle of 63° to the horizontal. At what speed did the ball hit the hoop if it reached it in 0.93 s?
Problem 31.(1) A basketball player throws the ball into the hoop. At the moment of the throw, the ball is at a height of 2.05 m, and after 0.88 s it falls into the ring located at a height of 3.05 m. From what distance from the ring (horizontally) was the throw made if the ball was thrown at an angle of 56 o to the horizon?
Problem 32.(2) The ball is thrown horizontally with a speed of 13, after some time its speed turns out to be equal to 18. Find the movement of the ball during this time. Neglect air resistance.
Problem 33.(2) A body is thrown at a certain angle to the horizon with an initial speed of 17 m/s. Find the value of this angle if the body’s flight range is 4.3 times greater than the maximum lift height.
Problem 34.(2) A bomber diving at a speed of 360 km/h drops a bomb from a height of 430 m, being horizontally at a distance of 250 m from the target. At what angle should a bomber dive? At what height will the bomb be 2 seconds after the start of its fall? What speed will it have at this point?
Problem 35.(2) An airplane flying at an altitude of 2940 m at a speed of 410 km/h dropped a bomb. How long before passing over the target and at what distance from it must the plane release the bomb in order to hit the target? Find the magnitude and direction of the bomb’s velocity after 8.5 s from the beginning of its fall. Neglect air resistance.
Problem 36.(2) A projectile fired at an angle of 36.6 degrees to the horizontal was at the same height twice: 13 and 66 seconds after departure. Determine the initial speed, maximum height lift and range of the projectile. Neglect air resistance.
2.1.3. Circular movement
Problem 37.(2) A sinker moving on a line in a circle with constant tangential acceleration, by the end of the eighth revolution had a speed of 6.4 m/s, and after 30 s of movement it normal acceleration became 92 m/s 2 . Find the radius of this circle.
Problem 38.(2) A boy riding on a carousel moves when the carousel stops along a circle with a radius of 9.5 m and covers a path of 8.8 m, having a speed of 3.6 m/s at the beginning of this arc and 1.4 m/s at the end. With. Determine the total acceleration of the boy at the beginning and end of the arc, as well as the time of his movement along this arc.
Problem 39.(2) A fly sitting on the edge of a fan blade, when it is turned on, moves in a circle of radius 32 cm with a constant tangential acceleration of 4.6 cm/s 2 . How long after the start of motion will the normal acceleration be twice as large as the tangential acceleration and what will it be equal to? linear speed flies at this point in time? How many revolutions will the fly make during this time?
Problem 40.(2) When the door is opened, the handle moves from rest in a circle of radius 68 cm with a constant tangential acceleration equal to 0.32 m/s 2 . Find the dependence of the total acceleration of the handle on time.
Problem 41.(3) To save space, the entrance to one of the highest bridges in Japan is arranged in the form of a helical line wrapping around a cylinder with a radius of 65 m. The roadbed makes an angle of 4.8 degrees with the horizontal plane. Find the acceleration of a car moving along this road at a constant absolute speed of 85 km/h?
2.1.4. Relativity of motion
Problem 42.(2) Two ships are moving relative to the shores at a speed of 9.00 and 12.0 knots (1 knot = 0.514 m/s), directed at an angle of 30 and 60 o to the meridian, respectively. At what speed is the second ship moving relative to the first?
Problem 43.(3) A boy who can swim at a speed 2.5 times slower than the speed of the river current wants to swim across this river so that he is carried downstream as little as possible. At what angle to the shore should the boy swim? How far will it be carried if the width of the river is 190 m?
Problem 44.(3) Two bodies simultaneously begin to move from one point in the gravity field with the same speed equal to 2.6 m/s. The speed of one body is directed at an angle π/4, and the other – at an angle –π/4 to the horizon. Define relative speed of these bodies 2.9 s after the start of their movement.
Lesson objectives:
Educational:
Educational:
Vos nutritious
Lesson type : Combined lesson.
View document contents
“Lesson topic: “Acceleration. Rectilinear motion with constant acceleration."
Prepared by Marina Nikolaevna Pogrebnyak, physics teacher at MBOU “Secondary School No. 4”
Class -11
Lesson 5/4 Lesson topic: “Acceleration. Rectilinear motion with constant acceleration».
Lesson objectives:
Educational: Introduce students to characteristic features rectilinear uniformly accelerated motion. Give the concept of acceleration as a basic physical quantity, characterizing uneven movement. Enter the formula to determine instantaneous speed body at any time, calculate the instantaneous speed of the body at any time,
improve students' ability to solve problems analytically and graphically.
Educational: development of schoolchildren's theoretical, creative thinking, formation of operational thinking aimed at choosing optimal solutions
Vosnutritious : bring up conscious attitude to study and interest in studying physics.
Lesson type : Combined lesson.
Demos:
1. Uniformly accelerated motion of the ball along inclined plane.
2. Multimedia application “Fundamentals of Kinematics”: fragment “Uniformly accelerated motion”.
Progress.
1.Organizational moment.
2. Test of knowledge: Independent work(“Displacement.” “Graphs of rectilinear uniform motion") - 12 min.
3. Studying new material.
Plan for presenting new material:
1. Instantaneous speed.
2. Acceleration.
3. Speed during rectilinear uniformly accelerated motion.
1. Instantaneous speed. If the speed of a body changes with time, to describe the movement you need to know what the speed of the body is at a given moment in time (or at a given point in the trajectory). This speed is called instantaneous speed.
We can also say that instantaneous speed is the average speed over a very short time interval. When driving at a variable speed, the average speed measured over different time intervals will be different.
However, if when measuring average speed take smaller and smaller time intervals, the value of the average speed will tend to a certain a certain value. This is the instantaneous speed at a given moment in time. In the future, when speaking about the speed of a body, we will mean its instantaneous speed.
2. Acceleration. With uneven movement, the instantaneous speed of a body is a variable quantity; it is different in modulus and (or) in direction different moments time and in different points trajectories. All speedometers of cars and motorcycles show us only the instantaneous speed module.
If the instantaneous speed of uneven motion changes unequally over equal periods of time, then it is very difficult to calculate it.
Such complex uneven movements are not studied at school. Therefore, we will consider only the simplest non-uniform motion - uniformly accelerated rectilinear motion.
Rectilinear motion, in which the instantaneous speed changes equally over any equal time intervals, is called uniformly accelerated rectilinear motion.
If the speed of a body changes during movement, the question arises: what is the “rate of change of speed”? This quantity, called acceleration, plays vital role in all mechanics: we will soon see that the acceleration of a body is determined by the forces acting on this body.
Acceleration is the ratio of the change in the speed of a body to the time interval during which this change occurred.
The SI unit of acceleration is m/s2.
If a body moves in one direction with an acceleration of 1 m/s 2 , its speed changes by 1 m/s every second.
The term "acceleration" is used in physics when talking about any change in speed, including when the velocity modulus decreases or when the velocity modulus remains unchanged and the speed changes only in direction.
3. Speed during rectilinear uniformly accelerated motion.
From the definition of acceleration it follows that v = v 0 + at.
If we direct the x axis along the straight line along which the body moves, then in projections onto the x axis we obtain v x = v 0 x + a x t.
Thus, with rectilinear uniformly accelerated motion, the projection of velocity depends linearly on time. This means that the graph of v x (t) is a straight line segment.
Movement formula:
Speed graph of an accelerating car:
Speed graph of a braking car
4. Consolidation of new material.
What is the instantaneous speed of a stone thrown vertically upward at the top point of its trajectory?
About what speed - average or instantaneous - we're talking about in the following cases:
a) the train traveled between stations at a speed of 70 km/h;
b) the speed of movement of the hammer upon impact is 5 m/s;
c) the speedometer on the electric locomotive shows 60 km/h;
d) a bullet leaves a rifle at a speed of 600 m/s.
TASKS SOLVED IN THE LESSON
The OX axis is directed along the trajectory of the rectilinear motion of the body. What can you say about the movement in which: a) v x 0, and x 0; b) v x 0, a x v x x 0;
d) v x x v x x = 0?
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?
2. The train, 10 s after the start of movement, acquires a speed of 0.6 m/s. How long after the start of movement will the speed of the train become 3 m/s?
5. HOMEWORK: §5,6, ex. 5 No. 2, ex. 6 No. 2.
For uniformly accelerated motion, the following equations are valid, which we present without derivation:
As you understand, vector formula on the left and two scalar formulas on the right are equal. From the point of view of algebra, scalar formulas mean that with uniformly accelerated motion, projections of displacement depend on time according to a quadratic law. Compare this with the nature of instantaneous velocity projections (see § 12-h).
Knowing that sx = x – xo and sy = y – yo (see § 12th), of the two scalar formulas from the upper right column we obtain equations for the coordinates:
Since the acceleration during uniformly accelerated motion of a body is constant, then coordinate axes can always be positioned so that the acceleration vector is directed parallel to one axis, for example the Y axis. Consequently, the equation of motion along the X axis will be noticeably simplified:
x = xo + υox t + (0) and y = yo + υoy t + ½ ay t²
Please note that the left equation coincides with the equation of uniform rectilinear motion (see § 12-g). This means that uniformly accelerated motion can “compose” from uniform motion along one axis and uniformly accelerated motion along the other. This is confirmed by the experience with the core on a yacht (see § 12-b).
Task. Stretching out her arms, the girl tossed the ball. He rose 80 cm and soon fell at the girl’s feet, flying 180 cm. At what speed was the ball thrown and what speed did the ball have when it hit the ground?
Let us square both sides of the equation for the projection of the instantaneous velocity onto the Y axis: υy = υoy + ay t (see § 12). We get the equality:
υy² = ( υoy + ay t )² = υoy² + 2 υoy ay t + ay² t²
Let’s take out of brackets the factor 2 ay only for the two right-hand terms:
υy² = υoy² + 2 ay ( υoy t + ½ ay t² )
Note that in brackets we get the formula for calculating the displacement projection: sy = υoy t + ½ ay t². Replacing it with sy, we get:
Solution. Let's make a drawing: direct the Y axis upward, and place the origin of coordinates on the ground at the girl's feet. Let us apply the formula we derived for the square of the velocity projection, first at the top point of the ball’s rise:
0 = υoy² + 2·(–g)·(+h) ⇒ υoy = ±√¯2gh = +4 m/s
Then, when starting to move from the top point down:
υy² = 0 + 2·(–g)·(–H) ⇒ υy = ±√¯2gh = –6 m/s
Answer: the ball was thrown upward with a speed of 4 m/s, and at the moment of landing it had a speed of 6 m/s, directed against the Y axis.
Note. We hope you understand that the formula for the squared projection of instantaneous velocity will be correct by analogy for the X axis:
If the movement is one-dimensional, that is, it occurs only along one axis, you can use either of the two formulas in the framework.