The lower the acceleration of the body, the graph of displacement. Determination of kinematic characteristics of movement using graphs

Graphic representation of uniformly accelerated linear motion.

Moving during uniformly accelerated motion.

Ilevel.

Many physical quantities that describe the movements of bodies change over time. Therefore, for greater clarity of description, movement is often depicted graphically.

Let us show how the time dependences of kinematic quantities describing rectilinear uniformly accelerated motion are graphically depicted.

Uniformly accelerated linear motion- this is a movement in which the speed of a body changes equally over any equal periods of time, i.e. it is a movement with acceleration constant in magnitude and direction.

a=const - acceleration equation. That is, a has a numerical value that does not change over time.

By definition of acceleration

From here we have already found equations for the dependence of speed on time: v = v0 + at.

Let's see how this equation can be used to graphically represent uniformly accelerated motion.

Let us graphically depict the dependences of kinematic quantities on time for three bodies

.

1, the body moves along the 0X axis, while increasing its speed (acceleration vector a is codirectional with the velocity vector v). vx >0, akh > 0

2, the body moves along the 0X axis, while reducing its speed (the acceleration vector a is not codirectional with the velocity vector v). vx >0, ah< 0

2, the body moves against the 0X axis, while reducing its speed (the acceleration vector is not codirectional with the velocity vector v). vx< 0, ах > 0

Acceleration graph

Acceleration, by definition, is a constant value. Then, for the presented situation, the graph of acceleration versus time a(t) will look like:

From the acceleration graph, you can determine how the speed changed - increased or decreased and by what numerical value the speed changed and which body the speed changed more.

Speed ​​graph

If we compare the dependence of the coordinate on time during uniform motion and the dependence of the velocity projection on time during uniformly accelerated motion, we can see that these dependencies are the same:

x= x0 + vx t vx = v 0 x + a X t

This means that the dependency graphs have the same appearance.

To construct this graph, the time of movement is plotted on the abscissa axis, and the speed (projection of speed) of the body is plotted on the ordinate axis. In uniformly accelerated motion, the speed of a body changes over time.

Moving during uniformly accelerated motion.

With uniformly accelerated rectilinear motion, the speed of a body is determined by the formula

vx = v 0 x + a X t

In this formula, υ0 is the speed of the body at t = 0 (starting speed ), a= const – acceleration. On the speed graph υ ( t) this dependence looks like a straight line (Fig.).

Acceleration can be determined from the slope of the velocity graph a bodies. The corresponding constructions are shown in Fig. for graph I. Acceleration is numerically equal to the ratio of the sides of the triangle ABC: MsoNormalTable">

The greater the angle β that the velocity graph forms with the time axis, i.e., the greater the slope of the graph ( steepness), the greater the acceleration of the body.

For graph I: υ0 = –2 m/s, a= 1/2 m/s2.

For graph II: υ0 = 3 m/s, a= –1/3 m/s2.

The velocity graph also allows you to determine the projection of movement s bodies for some time t. Let us select on the time axis a certain small period of time Δ t. If this period of time is small enough, then the change in speed over this period is small, i.e. the movement during this period of time can be considered uniform with a certain average speed, which is equal to the instantaneous speed υ of the body in the middle of the interval Δ t. Therefore, displacement Δ s in time Δ t will be equal to Δ s = υΔ t. This movement is equal to the area of ​​the shaded strip (Fig.). Breaking down the time period from 0 to some point t for small intervals Δ t, we find that the movement s for a given time t with uniformly accelerated rectilinear motion is equal to the area of ​​the trapezoid ODEF. The corresponding constructions were made for graph II in Fig. 1.4.2. Time t taken equal to 5.5 s.

Since υ – υ0 = at s t will be written in the form:

To find the coordinates y body at any time t need to the starting coordinate y 0 add movement in time t: DIV_ADBLOCK189">

Since υ – υ0 = at, the final formula for moving s body with uniformly accelerated motion over a time interval from 0 to t will be written in the form: https://pandia.ru/text/78/516/images/image009_57.gif" width="146 height=55" height="55">

When analyzing uniformly accelerated motion, sometimes the problem arises of determining the movement of a body based on the given values ​​of the initial υ0 and final υ velocities and acceleration a. This problem can be solved using the equations written above by eliminating time from them t. The result is written in the form

If the initial speed υ0 is zero, these formulas take the form MsoNormalTable">

It should be noted once again that the quantities υ0, υ, included in the formulas for uniformly accelerated rectilinear motion s, a, y 0 are algebraic quantities. Depending on the specific type of movement, each of these quantities can take on both positive and negative values.

An example of solving a problem:

Petya slides down the mountainside from a state of rest with an acceleration of 0.5 m/s2 in 20 s and then moves along a horizontal section. Having traveled 40 m, he crashes into the gaping Vasya and falls into a snowdrift, reducing his speed to 0 m/s. With what acceleration did Petya move along the horizontal surface to the snowdrift? What is the length of the mountain slope from which Petya so unsuccessfully slid down?

Stage 1.

The equation for Petit's speed at the end of the descent from the mountain is:

v 1 = v 01 + a 1t 1.

In projections onto the axis X we get:

v 1x = a 1xt.

Let us write an equation connecting the projections of Petya’s velocity, acceleration and displacement at the first stage of movement:

or because Petya was driving from the very top of the hill with an initial speed of V01=0

(If I were Petya, I would be careful about driving down such high hills)

Considering that Petya’s initial speed at this 2nd stage of movement is equal to his final speed at the first stage:

v 02 x = v 1 x, v 2x = 0, where v1 is the speed with which Petya reached the foot of the hill and began to move towards Vasya. V2x - Petya's speed in a snowdrift.

2. Using this acceleration graph, tell us how the speed of the body changes. Write down the equations for the dependence of speed on time if at the moment of the start of movement (t=0) the speed of the body is v0х =0. Please note that with each subsequent section of movement, the body begins to pass at a certain speed (which was achieved during the previous time!).

3. A metro train, leaving the station, can reach a speed of 72 km/h in 20 s. Determine with what acceleration a bag, forgotten in a subway car, is moving away from you. How far will she travel?

4. A cyclist moving at a speed of 3 m/s begins to descend down a mountain with an acceleration of 0.8 m/s2. Find the length of the mountain if the descent took 6 s.

5. Having started braking with an acceleration of 0.5 m/s2, the train traveled 225 m to the stop. What was its speed before braking began?

6. Having started to move, the soccer ball reached a speed of 50 m/s, covered a distance of 50 m and crashed into the window. Determine the time it took the ball to travel this path and the acceleration with which it moved.

7. Reaction time of Uncle Oleg’s neighbor = 1.5 minutes, during which time he will figure out what happened to his window and will have time to run out into the yard. Determine what speed young football players should develop so that the joyful owners of the window do not catch up with them, if they need to run 350 m to their entrance.

8. Two cyclists are riding towards each other. The first one, having a speed of 36 km/h, began to climb up the mountain with an acceleration of 0.2 m/s2, and the second, having a speed of 9 km/h, began to descend the mountain with an acceleration of 0.2 m/s2. After how long and in what place will they collide due to their absent-mindedness, if the length of the mountain is 100 m?

Instructions

Consider the function f(x) = |x|. To begin with, this is an unsigned modulus, that is, the graph of the function g(x) = x. This graph is a straight line passing through the origin and the angle between this straight line and the positive direction of the x-axis is 45 degrees.

Since the modulus is a non-negative quantity, the part that is below the abscissa axis must be mirrored relative to it. For the function g(x) = x, we find that the graph after such a mapping will look like V. This new graph will be a graphical interpretation of the function f(x) = |x|.

Video on the topic

note

The graph of the modulus of a function will never be in the 3rd and 4th quarters, since the modulus cannot take negative values.

Helpful advice

If a function contains several modules, then they need to be expanded sequentially and then stacked on top of each other. The result will be the desired graph.

Sources:

• how to graph a function with modules

Kinematics problems in which you need to calculate speed, time or the path of uniformly and rectilinearly moving bodies, found in the school course of algebra and physics. To solve them, find in the condition quantities that can be equalized. If the condition requires defining time at a known speed, use the following instructions.

You will need

• - pen;
• - paper for notes.

Instructions

The simplest case is the movement of one body with a given uniform speed Yu. The distance that the body has traveled is known. Find on the way: t = S/v, hour, where S is the distance, v is the average speed bodies.

The second is for oncoming movement of bodies. A car moves from point A to point B speed 50 km/h. A moped with a speed 30 km/h. The distance between points A and B is 100 km. Need to find time through which they will meet.

Label the meeting point K. Let the distance AK of the car be x km. Then the motorcyclist’s path will be 100 km. From the problem conditions it follows that time On the road, a car and a moped have the same experience. Make up the equation: x/v = (S-x)/v’, where v, v’ – and the moped. Substituting the data, solve the equation: x = 62.5 km. Now time: t = 62.5/50 = 1.25 hours or 1 hour 15 minutes.

Create an equation similar to the previous one. But in this case time a moped's travel time will be 20 minutes faster than that of a car. To equalize the parts, subtract one third of an hour from the right side of the expression: x/v = (S-x)/v’-1/3. Find x – 56.25. Calculate time: t = 56.25/50 = 1.125 hours or 1 hour 7 minutes 30 seconds.

The fourth example is a problem involving the movement of bodies in one direction. A car and a moped are moving from point A at the same speeds. It is known that the car left half an hour later. After what time will he catch up with the moped?

In this case, the distance traveled by the vehicles will be the same. Let time the car will travel x hours, then time the moped's journey will be x+0.5 hours. You have the equation: vx = v’(x+0.5). Solve the equation by substituting , and find x – 0.75 hours or 45 minutes.

Fifth example – a car and a moped are moving at the same speeds in the same direction, but the moped left point B, located 10 km from point A, half an hour earlier. Calculate through what time After the start, the car will catch up with the moped.

The distance traveled by the car is 10 km more. Add this difference to the motorcyclist’s path and equalize the parts of the expression: vx = v’(x+0.5)-10. Substituting the speed values ​​and solving it, you get: t = 1.25 hours or 1 hour 15 minutes.

Sources:

• what is the speed of the time machine

Instructions

Calculate the average of a body moving uniformly along a section of path. Such speed is the easiest to calculate, since it does not change over the entire segment movement and equals the average. This can be expressed in the form: Vрд = Vср, where Vрд – speed uniform movement, and Vav – average speed.

Calculate the average speed uniformly slow (uniformly accelerated) movement in this area, for which it is necessary to add the initial and final speed. Divide the result by two, which is the average speed Yu. This can be written more clearly as a formula: Vср = (Vн + Vк)/2, where Vн represents

§ 14. GRAPHICS OF PATH AND SPEED

Determining the path using the speed graph

In physics and mathematics, three ways of presenting information about the relationship between various quantities are used: a) in the form of a formula, for example, s =v ∙ t; b) in the form of a table; c) in the form of a graph (drawing).

Dependence of speed on time v(t) - the speed graph is depicted using two mutually perpendicular axes. We will plot time along the horizontal axis, and speed along the vertical axis (Fig. 14.1). It is necessary to think about the scale in advance so that the drawing is not too large or too small. A letter is indicated at the end of the axis, which is a designation that is numerically equal to the area of ​​the shaded rectangle abcd of the value that is plotted on it. The unit of measurement of this quantity is indicated next to the letter. For example, near the time axis indicate t, s, and near the speed axis v(t), months. Select a scale and apply divisions on each axis.

Rice. 14.1. Graph of the speed of a body moving uniformly at a speed of 3 m/sec. The path traveled by the body from the 2nd to the 6th second is

Representation of uniform motion by table and graphs

Let us consider the uniform movement of a body with a speed of 3 m/s, that is, the numerical value of the speed will be constant throughout the entire time of movement. This is abbreviated as follows: v = const (constant, that is, a constant value). In our example, it is equal to three: v = 3. You already know that information about the dependence of one quantity on another can be presented in the form of a table (array, as they say in computer science):

The table shows that at all specified times the speed is 3 m/sec. Let the scale of the time axis be 2 cells. = 1 s, and the speed axis is 2 cells. = 1 m/sec. A graph of speed versus time (abbreviated as speed graph) is shown in Figure 14.1.

Using a velocity graph, you can find the path a body travels over a certain time interval. To do this, you need to compare two facts: on the one hand, the path can be found by multiplying the speed by time, and on the other hand, the product of speed by time, as can be seen from the figure, is the area of ​​a rectangle with sides t and v.

For example, from the second to the sixth second the body moved for four seconds and traveled 3 m/s ∙ 4 s = 12 m. This is the area of ​​the rectangle abcd, the length of which is 4 s (segment ad along the time axis) and height 3 m/s ( segment ab along the vertical). The area, however, is somewhat unusual, since it is measured not in m 2, but in g. Therefore, the area under the speed graph is numerically equal to the distance traveled.

Path graph

The graph of the path s(t) can be depicted using the formula s = v ∙ t, that is, in our case, when the speed is 3 m/s: s = 3 ∙ t. Let's build a table:

Time (t, s) is again plotted along the horizontal axis, and the path is plotted along the vertical axis. Near the axis of the path we write: s, m (Fig. 14.2).

Determining speed from the path graph

Let us now depict in one figure two graphs that will correspond to movements with speeds of 3 m/s (line 2) and 6 m/s (line 1) (Fig. 14.3). It can be seen that the greater the speed of the body, the steeper the line of points on the graph.

There is also an inverse problem: having a motion graph, you need to determine the speed and write down the equation of the path (Fig. 14.3). Let's consider straight line 2. From the beginning of the movement until the moment of time t = 2 s, the body has traveled a distance of s = 6 m. Therefore, its speed: v = = 3. Choosing a different time interval will not change anything, for example, at the moment t = 4 s, the path traveled by the body from the beginning of the movement is s = 12 m. The ratio is again 3 m/sec. But this is how it should be, since the body moves at a constant speed. Therefore, the easiest way would be to choose a time interval of 1 s, because the path traveled by the body in one second is numerically equal to the speed. The path traveled by the first body (graph 1) in 1 s is 6 m, that is, the speed of the first body is 6 m/sec. The corresponding dependences of the path on time in these two bodies will be:

s 1 = 6 ∙ t and s 2 =3 ∙ t.

Rice. 14.2. Path schedule. The remaining points, except for the six indicated in the table, were set in the task, so that the movement of the rain was uniform throughout the entire time

Rice. 14.3. Path graph for different speeds

Let's sum it up

In physics, three methods of presenting information are used: graphical, analytical (using formulas) and tables (arrays). The third method is more suitable for solving on a computer.

The path is numerically equal to the area under the speed graph.

The steeper the s(t) graph, the greater the speed.

Creative tasks

14.1. Draw graphs of speed and distance when the speed of a body increases or decreases uniformly.

Exercise 14

1. How is the path determined on the speed graph?

2. Is it possible to write down a formula for the dependence of the path on time, given the graph s(t)?

3. Or will the slope of the path graph change if the scale on the axes is halved?

4. Why is the graph of the path of uniform motion depicted as a straight line?

5. Which of the bodies (Fig. 14.4) has the highest speed?

6. Name three ways of representing information about body movement, and (in your opinion) their advantages and disadvantages.

7. How can you determine the path from the speed graph?

8. a) How do the path graphs differ for bodies moving at different speeds? b) What do they have in common?

9. Using the graph (Fig. 14.1), find the path traveled by the body from the beginning of the first to the end of the third second.

10. What distance did the body travel (Fig. 14.2) in: a) two seconds; b) four seconds? c) Indicate where the third second of movement begins and where it ends.

11. Draw on the velocity and path graphs motion at a speed of a) 4 m/s; b) 2 m/sec.

12. Write down the formula for the dependence of the path on time for the movements shown in Fig. 14.3.

13. a) Find the velocities of the bodies using the graphs (Fig. 14.4); b) write down the corresponding equations for path and speed. c) Draw graphs of the speed of these bodies.

14. Construct graphs of path and velocity for bodies whose movements are given by the equations: s 1 = 5 ∙ t and s 2 = 6 ∙ t. What are the speeds of the bodies?

15. Using the graphs (Fig. 14.5), determine: a) the speed of the body; b) the paths they covered in the first 5 seconds. c) Write down the equation of path and plot the corresponding graphs for all three movements.

16. Draw a graph of the path for the movement of the first body relative to the second (Fig. 14.3).

Let's show how you can find the path traveled by a body using a graph of speed versus time.

Let's start with the simplest case - uniform motion. Figure 6.1 shows a graph of v(t) – speed versus time. It represents 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

the path is numerically equal to the area of ​​the figure enclosed under the graph of speed versus time.

Let us now show that uneven motion also has this remarkable property.

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 of 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 case of 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 set off, the car covered a distance of 20 m in the first 4 s. Consider the car’s motion to be rectilinear 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 when solving simple problems. 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, the 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 started, 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 beginning of the 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 the case of braking of a 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 journey 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 carriage to the stop less than the distance covered by the train in the same time?

14. Having left the station, the train traveled at a uniform acceleration for some time, then for 1 minute at a uniform speed of 60 km/h, and then again at a uniform acceleration 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) What distance 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 so, at what point in time, and what will be the distance traveled?

16. After the push, the ball rolls up an inclined plane, after which it returns to the starting point. The ball was at a distance b from the initial point twice at time intervals t 1 and t 2 after the push. The ball moved up and down along the inclined plane with the same acceleration.
a) Direct the x-axis upward along the inclined plane, select the origin at the initial position of the 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 the 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 the numerical values ​​of v 0, a and l for 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 graph of sx(t), determine the moment when the ball’s modulus of displacement was maximum.