Find a symmetric vector about a straight line. The simplest problems with a straight line on a plane
1439. The motorcycle can increase its speed from 0 to 72 km / h within 5 seconds. Determine the acceleration of the motorcycle.
1440. Determine the acceleration of an elevator in a high-rise building if it increases its speed by 3.2 m / s for 2 s.
1441. A car moving at a speed of 72 km / h brakes evenly and stops after 10 seconds. What is the acceleration of a car?
1442. What to call the movement, at which the acceleration is constant? equal to zero?
Equally accelerated, uniform.
1443. The sleds, rolling down the mountain, move uniformly and at the end of the third second from the start of movement they have a speed of 10.8 km / h. Determine with what acceleration the sled is moving.
1444. The speed of the car for 1.5 minutes of movement increased from 0 to 60 km / h. Find the acceleration of the car in m / s2, in cm / s2.
1445. Motorcycle "Honda", moving at a speed of 90 km / h, began to brake evenly and after 5 seconds dropped the speed to 18 km / h. What is the acceleration of a motorcycle?
1446. The object from the state of rest begins to move with a constant acceleration equal to 6 10-3 m / s2. Determine the speed 5 minutes after starting the movement. What way did the object go during this time?
1447. The yacht is lowered into the water along the inclined slipways. She covered the first 80 cm in 10 seconds. How long did it take for the yacht to cover the remaining 30 m if its movement remained uniformly accelerated?
1448. The truck starts off with an acceleration of 0.6 m / s2. How long will it take to cover the path of 30 m?
1449. The train departs from the station, moving uniformly for 1 minute 20 seconds. What is the acceleration of the electric train if during this time its speed has become 57.6 km / h? What way did she go in the specified time?
1450. The aircraft for take-off is accelerated uniformly within 6 seconds to a speed of 172.8 km / h. Find the acceleration of the plane. How far did the plane go during acceleration?
1451. A freight train, moving away, moved with an acceleration of 0.5 m / s2 and accelerated to a speed of 36 km / h. What way did he go?
1452. A high-speed train set off from the station with uniform acceleration and, having covered 500 m, reached a speed of 72 km / h. What is the acceleration of the train? Determine its acceleration time.
1453. When leaving the cannon barrel, the projectile has a speed of 1100 m / s. The length of the gun barrel is 2.5 m. Inside the barrel, the projectile moved uniformly. What is its acceleration? How long did it take for the projectile to travel the entire length of the barrel?
1454. The train traveling at a speed of 72 km / h began to brake with a constant acceleration equal to 2 m / s2 in modulus. How long will it take to stop? How far will it travel to a complete stop?
1455. The city bus moved uniformly at a speed of 6 m / s, and then began to brake with an acceleration of 0.6 m / s2. How long before the stop and at what distance from it should you start braking?
1456. Sleds slide along the ice path with an initial speed of 8 m / s, and for each second their speed decreases by 0.25 m / s. How long will it take for the sled to stop?
1457. A scooter moving at a speed of 46.8 km / h stops with uniform braking for 2 s. What is the acceleration of a scooter? What is its stopping distance?
1458. The motor ship, sailing at a speed of 32.4 km / h, began to slow down evenly and, approaching the pier in 36 s, completely stopped. What is the ship's acceleration? What way did he go during the braking time?
1459. The freight train, passing by the barrier, began to brake. After 3 minutes, he stopped at the junction. What is the initial speed of the freight train and the modulus of its acceleration if the barrier is located 1.8 km from the junction?
1460. The braking distance of the train is 150 m, the braking time is 30 s. Find the initial train speed and acceleration.
1461. The electric train, moving at a speed of 64.8 km / h, after the start of braking to a complete stop, covered 180 m. Determine its acceleration and deceleration time.
1462. The airplane flew uniformly at a speed of 360 km / h, then for 10 s it moved uniformly: its speed increased by 9 m / s per second. Determine what speed the airplane has acquired. What distance did he fly with uniformly accelerated motion?
1463. The motorcycle, moving at a speed of 27 km / h, began to accelerate uniformly and in 10 seconds reached a speed of 63 km / h. Determine the average speed of the motorcycle at a uniform acceleration. What path did he travel during the time of uniformly accelerated movement?
1464. The device counts time intervals equal to 0.75 s. The ball rolls off the inclined chute for three such times. After rolling off the inclined chute, it continues to move along the horizontal chute and travels for the first period of time 45 cm. Determine the instantaneous velocity of the ball at the end of the inclined chute and the acceleration of the ball when moving along this chute.
1465. Departing from the station, the train moves uniformly with an acceleration of 5 cm / s2. How long does it take for the train to reach a speed of 36 km / h?
1466. When the train departs from the station, its speed during the first 4 s increased to 0.2 m / s, during the next 6 s by another 30 cm / s and over the next 10 s by 1.8 km / h. How did the train move during these 20 seconds?
1467. Sleds, rolling down the mountain, move uniformly. On some part of the route, the speed of the sled increased from 0.8 m / s to 14.4 km / h for 4 s. Determine the acceleration of the sled.
1468. The cyclist begins to move with an acceleration of 20 cm / s2. How long will it take for a cyclist to reach 7.2 km / h?
1469. Figure 184 shows a graph of the speed of some uniformly accelerated movement. Using the scale given in the figure, determine the distance traveled in this movement for 3.5 seconds.
1470. Figure 185 shows a graph of the speed of some variable movement. Trace the drawing into a notebook and mark with shading an area numerically equal to the path traveled for 3 s. What is this path roughly equal to?
1471. During the first period of time from the beginning of uniformly accelerated motion, the ball passes along the groove of 8 cm. What distance will the ball cover during three similar intervals that have passed from the beginning of the motion?
1472. During 10 equal intervals of time from the beginning of movement, the body, moving uniformly accelerated, has passed 75 cm. How many centimeters has passed this body during the first two same intervals of time?
1473. The train, departing from the station, moves at uniform acceleration and during the first two seconds passes 12 cm. What distance will the train cover during 1 minute, counting from the beginning of movement?
1474. The train, leaving the station, moves uniformly with an acceleration of 5 cm / s2. How long will it take to develop a speed of 28.8 km / h and how far will the train travel during this time?
1475. A locomotive along a horizontal path approaches a slope at a speed of 8 m / s, then moves down the slope with an acceleration of 0.2 m / s. Determine the length of the slope if the locomotive passes it in 30 seconds.
1476. The initial speed of the cart moving down the inclined board is 10 cm / s. The trolley traveled the entire length of the board, equal to 2 m, within 5 seconds. Determine the acceleration of the cart.
1477. The bullet takes off from the barrel of the gun at a speed of 800 m / s. Barrel length 64 cm. Assuming uniformly accelerated movement of the bullet inside the barrel, determine the acceleration and movement time.
1478. The bus, moving at a speed of 4 m / s, begins to accelerate uniformly by 1 m / s per second. How far will the bus travel in six seconds?
1479. The truck, having a certain initial speed, began to move with uniform acceleration: in the first 5 s it covered 40 m, and in the first 10 s - 130 m. Find the initial speed of the truck and its acceleration.
1480. The boat, departing from the pier, began to move uniformly. After covering some distance, he reached a speed of 20 m / s. What was the speed of the boat at the moment when it sailed half this distance?
1481. A skier is sliding down a mountain with zero initial speed. In the middle of the mountain, its speed was 5 m / s, after 2 s the speed became 6 m / s. Assuming that it increases evenly, determine the speed of the skier 8 seconds after starting the movement.
1482. The car started off and moves at uniform acceleration. In what second from the start of the movement the path covered by the car is twice the distance covered by it in the previous second?
1483. Find the path traversed by the body in the eighth second of motion, if it starts to move uniformly accelerated without initial speed and in the fifth second passes the path of 27 m.
1484. Those seeing off stand at the beginning of the head car of the train. The train starts off and moves at uniform acceleration. For 3 seconds, the entire lead carriage passes by the mourners. How long does it take for the entire train, consisting of 9 carriages, to pass by those seeing off?
1485. The material point moves according to the law x \u003d 0,5t². What kind of movement is this? What is the point acceleration? Plot a graph against time:
a) point coordinates;
b) point speed;
c) acceleration.
1486. \u200b\u200bThe train stopped 20 s after the start of braking, having covered 120 m during this time. Determine the initial train speed and train acceleration.
1488. Plot the graphs of the speed of uniformly slowed motion for the cases:
1) V0 \u003d 10 m / s, a \u003d - 1.5 m / s2;
2) V0 \u003d 10 m / s; a \u003d - 2 m / s2.
The scale is the same in both cases: 0.5 cm - 1 m / s; about, 5 cm - 1 sec.
1489. Draw the distance traveled in time t on the graph of the speed of an equal slow motion. Take V0 \u003d 10 m / s, a \u003d 2 m / s2.
1490. Describe the movements, the speed graphs of which are given in Figure 186, a and b.
a) the movement will be equally slow;
b) first, the body will move uniformly, then evenly. On the 3rd section, the movement will be equally slow.
What is uniformly accelerated motion?
Equally accelerated motion in physics is such a motion, the acceleration vector of which does not change in magnitude and direction. In simple terms, uniformly accelerated motion is an uneven motion (that is, moving at different speeds), the acceleration of which is constant over a certain period of time. Imagine who starts to move, the first 2 seconds his speed is 10 m / s, the next 2 seconds he is already moving at a speed of 20 m / s, and after another 2 seconds already at a speed of 30 m / s. That is, every 2 seconds it accelerates by 10 m / s, such a movement is uniformly accelerated.
From this it is possible to derive an extremely simple definition of uniformly accelerated motion: it is such a motion of any physical body in which its speed changes in the same way over equal intervals of time.
Examples of uniformly accelerated motion
A good example of uniformly accelerated movement in everyday life can be a bicycle riding down a hill (but not a bicycle driven by a cyclist), or a stone thrown at a certain angle to the horizon.
By the way, the example with a stone can be considered in more detail. At any point of the flight path, the stone is acted upon by the acceleration of gravity g. The acceleration g does not change, that is, it remains constant and is always directed in one direction (in fact, this is the main condition for uniformly accelerated motion).
The flight of a thrown stone is conveniently represented as a sum of movements relative to the vertical and horizontal axes of the coordinate system.
If along the X axis the movement of the stone is uniform and rectilinear, then along the Y axis it will be uniformly accelerated and rectilinear.
Formula of uniformly accelerated motion
The speed formula for uniformly accelerated motion will look like this:
Where V 0 is the initial velocity of the body, a is the acceleration (as we remember, this value is a constant), t is the total flight time of the stone.
With uniformly accelerated motion, the V (t) dependence will have the form of a straight line.
Acceleration can be determined from the slope of the velocity graph. In this figure, it is equal to the ratio of the sides of the triangle ABC.
The larger the angle β, the greater the slope and, as a consequence, the slope of the graph with respect to the time axis, and the greater the acceleration of the body.
- Sivukhin D.V. General course of physics. - Moscow: Fizmatlit, 2005. - T. I. Mechanics. - P. 37 .-- 560 p. - ISBN 5-9221-0225-7.
- Targ S. M. A short course in theoretical mechanics. - 11th ed. - M .: "High school", 1995. - S. 214. - 416 p. - ISBN 5-06-003117-9.
Equal acceleration, video
One of the most common types of movement of objects in space, which a person encounters on a daily basis, is uniformly accelerated rectilinear motion. In the 9th grade of secondary schools, in the physics course, this type of movement is studied in detail. Let's consider it in the article.
Kinematic characteristics of movement
Before presenting formulas describing uniformly accelerated rectilinear motion in physics, let us consider the quantities that characterize it.
First of all, this is the path traveled. We will denote it by the letter S. According to the definition, the path is the distance that the body has traveled along the trajectory of movement. In the case of rectilinear motion, the trajectory is a straight line. Accordingly, the path S is the length of the straight segment on this line. It is measured in the SI system of physical units in meters (m).
Velocity, or as it is often called linear velocity, is the rate at which the position of a body in space changes along its trajectory. Let us denote the speed by the letter v. It is measured in meters per second (m / s).
Acceleration is the third important quantity for describing rectilinear uniformly accelerated motion. It shows how quickly the body's speed changes over time. Acceleration is denoted by the symbol a and is expressed in meters per square second (m / s 2).
The path S and the speed v are variable characteristics during uniformly accelerated rectilinear motion. Acceleration is a constant value.
The relationship between speed and acceleration
Let's imagine that some car is moving along a straight road without changing its speed v 0. This movement is called uniform. At some point in time, the driver began to press the gas pedal, and the car began to increase its speed, acquiring acceleration a. If we start counting the time from the moment when the car has acquired non-zero acceleration, then the equation of the dependence of speed on time will take the form:
Here the second term describes the increase in speed for each period of time. Since v 0 and a are constant values, and v and t are variable parameters, the graph of the function v will be a straight line intersecting the ordinate at the point (0; v 0), and having a certain angle of inclination to the abscissa axis (the tangent of this angle is value of acceleration a).
The figure shows two graphs. The only difference between them is that the upper graph corresponds to the speed in the presence of some initial value v 0, and the lower one describes the speed of uniformly accelerated rectilinear motion when the body starts accelerating from a state of rest (for example, a starting car).
Note that if in the example above the driver pressed the brake pedal instead of the accelerator pedal, then the braking movement would be described by the following formula:
This type of movement is called rectilinear equidistant.
Distance traveled formulas
In practice, it is often important to know not only the acceleration, but also the value of the path that the body travels over a given period of time. In the case of rectilinear uniformly accelerated motion, this formula has the following general form:
S \u003d v 0 * t + a * t 2/2.
The first term corresponds to uniform motion without acceleration. The second term is the contribution to the traversed path of pure accelerated motion.
In the case of braking a moving object, the expression for the path will take the form:
S \u003d v 0 * t - a * t 2/2.
In contrast to the previous case, here the acceleration is directed against the speed of movement, which leads to the vanishing of the latter some time after the start of braking.
It is not difficult to guess that the graphs of the functions S (t) will be the branches of the parabola. The figure below shows these graphs in schematic form.
Parabolas 1 and 3 correspond to the accelerated movement of the body, parabola 2 describes the braking process. It can be seen that the distance traveled for 1 and 3 is constantly increasing, while for 2 it goes to some constant value. The latter means that the body has stopped moving.
The task of determining the time of movement
The car must take the passenger from point A to point B. The distance between them is 30 km. It is known that a car moves with an acceleration of 1 m / s 2 for 20 seconds. Then its speed does not change. How long does it take for the car to take the passenger to point B?
The distance that the car will cover in 20 seconds will be:
In this case, the speed that he picks up in 20 seconds is:
Then the required travel time t can be calculated using the following formula:
t \u003d (S - S 1) / v + t 1 \u003d (S - a * t 1 2/2) / (a \u200b\u200b* t 1) + t 1.
Here S is the distance between A and B.
Let's translate all known data into SI system and substitute it into the written expression. We get the answer: t \u003d 1510 seconds, or approximately 25 minutes.
The task of calculating the braking distance
Now we will solve the problem of equally slow motion. Suppose the truck was traveling at 70 km / h. Ahead, the driver saw a red traffic light and began to stop. What is the braking distance of a car if it stopped in 15 seconds.
S \u003d v 0 * t - a * t 2/2.
We know the braking time t and the initial speed v 0. Acceleration a can be found from the expression for the velocity, given that its final value is zero. We have:
Substituting the resulting expression into the equation, we arrive at the final formula for the path S:
S \u003d v 0 * t - v 0 * t / 2 \u003d v 0 * t / 2.
We substitute the values \u200b\u200bfrom the condition and write down the answer: S \u003d 145.8 meters.
The problem of determining the speed in free fall
Perhaps the most common rectilinear uniformly accelerated motion in nature is the free fall of bodies in the gravitational field of planets. Let's solve the following problem: the body was released from a height of 30 meters. What speed will it have when it falls to the surface of the earth?
Where g \u003d 9.81 m / s 2.
The body fall time is determined from the corresponding expression for the path S:
S \u003d g * t 2/2;
t \u003d √ (2 * S / g).
We substitute the time t into the formula for v, we get:
v \u003d g * √ (2 * S / g) \u003d √ (2 * S * g).
The value of the path S traveled by the body is known from the condition, we substitute it into the equality, we get: v \u003d 24.26 m / s or about 87 km / h.
Definition 1
A movement in which a body travels an unequal distance in equal intervals of time is called uneven (or variable).
With variable motion, the speed of the body changes over time, for this reason, the definitions of the average and instantaneous speeds are used to characterize such a movement.
The average speed of variable motion $ v_ (cp) $ is called a vector quantity equal to the ratio of the movement of the body $ s $ to the time interval $ t $, during which it moved:
$ v_ (cp) \u003d lim \\ left (\\ frac (Ds) (Dt) \\ right) $.
Variable movement introduces into the process only the time interval for which this speed is set. Instantaneous speed is the speed that the body possesses in a certain period of time (and therefore, at a specific point on the trajectory). The instantaneous velocity $ v $ is the limit to which the average velocity of the point $ v_ (cp) $ tends, while the time interval of the point's movement tends to 0:
$ v \u003d lim \\ left (\\ frac (Ds) (Dt) \\ right) $.
From the course of mathematics it is known that the limit of the ratio of the increment of a function to the increment of the argument, when the latter tends to 0 (if this threshold exists), acts as the main derivative of this function with respect to this argument.
Let's study how the ball rolls off an inclined plane. The ball moves unevenly: the paths traversed by it in successive equal intervals of the period increase. Thus, the rate of movement of the ball increases. The movement of an object rolling off an oblique plane is considered a classic example of uniformly accelerated rectilinear movement.
Consider the definition of uniformly accelerated motion.
Definition 2
Rectilinear uniformly accelerated motion is called rectilinear motion, in which the speed of the body changes by the same amount for any equal time intervals.
Directly uniformly accelerated is able to move, for example, vehicles during the acceleration period. But it may seem unusual in this case that during braking the car is also able to move in a straight line with uniform acceleration! Since in the definition of uniformly accelerated movement, we are not talking about an increase in impetus, but only about a change in speed.
The bottom line is that the concept of acceleration in physics is broader than in the ordinary sense. In everyday speech, acceleration usually means only an increase in speed. In physics we will begin to say that a body moves with acceleration constantly, if the speed of the body changes in any way (increases or decreases according to the modulus, changes according to direction, etc.).
The question may arise: for what reason do we pay attention to directly rectilinear uniformly accelerated motion? Running a little ahead, we will say that we will often deal with this displacement when considering the laws of mechanics.
Recall that under the influence of a stable force, the body moves directly and uniformly. (If the initial velocity of the body is equal to zero or is oriented along the line of action of the force.) And in numerous problems from the sphere of mechanics, a situation is directly considered in which the equations of rectilinear uniformly accelerated motion, the formulas of the final speed and the formulas of the path without time are applied.
Uniformly accelerated body movement
Definition 3
Uniformly accelerated motion is the movement of a body, at which its speed changes (is able to increase or decrease) in the same way for all possible equal time intervals.
Equally accelerated movement does not have the same speed throughout the entire path. In this case, there is acceleration, which is responsible for the continuous increase in speed. The acceleration of travel remains constant, and the pace increases regularly and equally.
In addition to uniformly accelerated movement, there is also uniformly slowed movement, where the tempo module equally decreases. Thus, uniformly accelerated movement is capable of taking place in some dimensions. It happens:
- one-dimensional;
- multidimensional.
In the case of the first, the movement is carried out along one axis location. In the case of the second, other measurements may be added.
Body acceleration
It is possible to apply displacement formulas for uniformly accelerated motion, as well as acceleration formulas without time, in completely different planes. For example, in order to calculate the fall of rigid bodies in free fall, the place of the fall. In particular, for a variety of precise and geometric calculations.
Based on the opposition to uniform movement, uneven is movement at different speeds according to each trajectory. What is its peculiarity? This is an uneven movement, but it “accelerates equally”.
We associate acceleration with increased speed. Since it accelerates in the same way, an equal increase in speed is obtained. How to understand whether the speed is increasing equally or not? We need to measure the time, to estimate the speed after the same period of time, using the formulas for acceleration at uniformly accelerated motion.
Example 1
For example, the car began to move, in the first 2 seconds it developed a speed of up to 10 m / s, in the next 2 seconds 20 m / s. After another 2 seconds, he is already traveling at a speed of 30 m / s. Every 2 seconds, the pace increases and each time by 10 m / s.
This movement is uniformly accelerated. Acceleration is the quantity that determines how much speed increases each time. In addition, it is necessary to pay attention to the formula for the speed at uniformly accelerated motion.
Moving with decreasing speed - slow movement. However, physicists call each movement with varying speed as accelerated movement. Whether the car starts off the site (the pace increases), or slows down, the speed decreases, in each case it moves with acceleration.
The rate of change in speed is characterized by acceleration. This is the number by which the speed changes every second. If the point acceleration in absolute value is large, then the point is rapidly gaining speed (during acceleration) or rapidly dropping it (during braking). Acceleration $ a $ is a physical vector quantity that is equal to the ratio of the change in speed $ \\ delta V $ to the time interval $ \\ delta t $ during which it occurred
$ \\ vec (a) \u003d \\ frac (\\ delta V) (\\ delta t) $
Uniform movement
Mechanical movement, in which the body passes the same distance at all possible equal intervals of time, is uniform. With uniform movement, the value of the point's speed remains stable (the formula for uniform and uniformly accelerated movement).
$ υ \u003d \\ frac (l) (\\ delta t) $, where:
- $ υ $ - uniform motion speed (m / s)
- $ l $ - path covered by the body (m)
- $ \\ delta t $ - movement time interval (s)
Uniform movement is present if the speed of the object remains equal in each interval of the traveled path, in which case the period of passage of different two identical sections will be the same.
If the movement is not only uniform, but also rectilinear, then the path of the body is the same as the movement module. For this reason, using the analogy with the previous formula of uniformly accelerated motion, the speed of uniform rectilinear motion is determined in physics:
$ \\ vec (v) \u003d \\ frac (\\ vec s) (\\ vec \\ delta t) $, where:
- $ \\ vec (v) $ - speed is equal to rectilinear motion, m / s
- $ \\ vec (s) $ - body movement, m
- $ (\\ vec \\ delta t) $ - movement time interval, s
The speed of uniform rectilinear motion is a vector, since displacement is a vector quantity. This means that it has not only a numerical value, but also a spatial direction.
Remark 1
Equally accelerated movement differs from uniform in that the speed in this movement is regularly and equally increased, up to a specific limit. In a uniform movement, the speed does not change in any way; otherwise, such a movement will in no way be called uniform.
