What is the formula for acceleration? Centripetal acceleration - formula derivation and practical application

And why is it needed? We already know what a reference system, relativity of motion and a material point are. Well, it's time to move on! Here we will look at the basic concepts of kinematics, put together the most useful formulas for the basics of kinematics, and give a practical example of solving the problem.

Let's solve this problem: a point moves in a circle with a radius of 4 meters. The law of its motion is expressed by the equation S=A+Bt^2. A=8m, B=-2m/s^2. At what point in time is the normal acceleration of a point equal to 9 m/s^2? Find the speed, tangential and total acceleration of the point for this moment in time.

Solution: we know that in order to find the speed we need to take the first time derivative of the law of motion, and the normal acceleration is equal to the quotient of the square of the speed and the radius of the circle along which the point is moving. Armed with this knowledge, we will find the required quantities.

Need help solving problems? Professional student service is ready to provide it.

Acceleration characterizes the rate of change in the speed of a moving body. If the speed of a body remains constant, then it does not accelerate. Acceleration occurs only when the speed of a body changes. If the speed of a body increases or decreases by a certain constant amount, then such a body moves with constant acceleration. Acceleration is measured in meters per second per second (m/s2) and is calculated from the values ​​of two speeds and time or from the value of the force applied to the body.

Steps

Calculation of average acceleration over two speeds

Formula for calculating average acceleration. The average acceleration of a body is calculated from its initial and final speeds (speed is the speed of movement in a certain direction) and the time it takes the body to reach its final speed. Formula for calculating acceleration: a = Δv / Δt, where a is acceleration, Δv is the change in speed, Δt is the time required to reach the final speed.

Definition of variables. You can calculate Δv And Δt in the following way: Δv = v k - v n And Δt = t k - t n, Where v to– final speed, v n- starting speed, t to– final time, t n– initial time.

• Since acceleration has a direction, always subtract the initial velocity from the final velocity; otherwise the direction of the calculated acceleration will be incorrect.
• If the initial time is not given in the problem, then it is assumed that tn = 0.

1. Find the acceleration using the formula. First, write the formula and the variables given to you. Formula: . Subtract the initial speed from the final speed, and then divide the result by the time interval (time change). You will get the average acceleration over a given period of time.

   • If the final speed is less than the initial speed, then the acceleration has a negative value, that is, the body slows down.
   • Example 1: A car accelerates from 18.5 m/s to 46.1 m/s in 2.47 s. Find the average acceleration.
     • Write the formula: a = Δv / Δt = (v k - v n)/(t k - t n)
     • Write the variables: v to= 46.1 m/s, v n= 18.5 m/s, t to= 2.47 s, t n= 0 s.
     • Calculation: a= (46.1 - 18.5)/2.47 = 11.17 m/s 2 .
   • Example 2: A motorcycle starts braking at a speed of 22.4 m/s and stops after 2.55 s. Find the average acceleration.
     • Write the formula: a = Δv / Δt = (v k - v n)/(t k - t n)
     • Write the variables: v to= 0 m/s, v n= 22.4 m/s, t to= 2.55 s, t n= 0 s.
     • Calculation: A= (0 - 22.4)/2.55 = -8.78 m/s 2 .

   Calculation of acceleration by force

   1. Newton's second law. According to Newton's second law, a body will accelerate if the forces acting on it do not balance each other. This acceleration depends on the net force acting on the body. Using Newton's second law, you can find the acceleration of a body if you know its mass and the force acting on that body.

      • Newton's second law is described by the formula: F res = m x a, Where F cut– resultant force acting on the body, m- body mass, a– acceleration of the body.
      • When working with this formula, use metric units, which measure mass in kilograms (kg), force in newtons (N), and acceleration in meters per second per second (m/s2).

   2. Find the mass of the body. To do this, place the body on the scale and find its mass in grams. If you are considering a very large body, look up its mass in reference books or on the Internet. The mass of large bodies is measured in kilograms.

      • To calculate acceleration using the above formula, you need to convert grams to kilograms. Divide the mass in grams by 1000 to get the mass in kilograms.

   3. Find the net force acting on the body. The resulting force is not balanced by other forces. If two differently directed forces act on a body, and one of them is greater than the other, then the direction of the resulting force coincides with the direction of the larger force. Acceleration occurs when a force acts on a body that is not balanced by other forces and which leads to a change in the speed of the body in the direction of action of this force.

      Rearrange the formula F = ma to calculate the acceleration. To do this, divide both sides of this formula by m (mass) and get: a = F/m. Thus, to find acceleration, divide the force by the mass of the accelerating body.

      • Force is directly proportional to acceleration, that is, the greater the force acting on a body, the faster it accelerates.
      • Mass is inversely proportional to acceleration, that is, the greater the mass of a body, the slower it accelerates.

   4. Calculate the acceleration using the resulting formula. Acceleration is equal to the quotient of the resulting force acting on the body divided by its mass. Substitute the values ​​given to you into this formula to calculate the acceleration of the body.

      • For example: a force equal to 10 N acts on a body weighing 2 kg. Find the acceleration of the body.
      • a = F/m = 10/2 = 5 m/s 2

   Testing your knowledge

   1. Direction of acceleration. The scientific concept of acceleration does not always coincide with the use of this quantity in everyday life. Remember that acceleration has a direction; acceleration is positive if it is directed upward or to the right; acceleration is negative if it is directed downward or to the left. Check your solution based on the following table:

   2. Example: a toy boat with a mass of 10 kg is moving north with an acceleration of 2 m/s 2 . A wind blowing in a westerly direction exerts a force of 100 N on the boat. Find the acceleration of the boat in a northerly direction.
   3. Solution: Since the force is perpendicular to the direction of movement, it does not affect the movement in that direction. Therefore, the acceleration of the boat in the north direction will not change and will be equal to 2 m/s 2.

2. Resultant force. If several forces act on a body at once, find the resulting force, and then proceed to calculate the acceleration. Consider the following problem (in two-dimensional space):

   • Vladimir pulls (on the right) a container with a mass of 400 kg with a force of 150 N. Dmitry pushes (on the left) a container with a force of 200 N. The wind blows from right to left and acts on the container with a force of 10 N. Find the acceleration of the container.
   • Solution: The conditions of this problem are designed to confuse you. In fact, everything is very simple. Draw a diagram of the direction of forces, so you will see that a force of 150 N is directed to the right, a force of 200 N is also directed to the right, but a force of 10 N is directed to the left. Thus, the resulting force is: 150 + 200 - 10 = 340 N. The acceleration is: a = F/m = 340/400 = 0.85 m/s 2.

Content:

Acceleration characterizes the rate of change in the speed of a moving body. If the speed of a body remains constant, then it does not accelerate. Acceleration occurs only when the speed of a body changes. If the speed of a body increases or decreases by a certain constant amount, then such a body moves with constant acceleration. Acceleration is measured in meters per second per second (m/s2) and is calculated from the values ​​of two speeds and time or from the value of the force applied to the body.

Steps

1 Calculation of average acceleration at two speeds

1. 1 Formula for calculating average acceleration. The average acceleration of a body is calculated from its initial and final speeds (speed is the speed of movement in a certain direction) and the time it takes the body to reach its final speed. Formula for calculating acceleration: a = Δv / Δt, where a is acceleration, Δv is the change in speed, Δt is the time required to reach the final speed.
   • The units of acceleration are meters per second per second, that is, m/s 2 .
   • Acceleration is a vector quantity, that is, it is given by both value and direction. Value is a numerical characteristic of acceleration, and direction is the direction of movement of the body. If the body slows down, then the acceleration will be negative.
2. 2 Definition of variables. You can calculate Δv And Δt in the following way: Δv = v k - v n And Δt = t k - t n, Where v to– final speed, v n- starting speed, t to– final time, t n– initial time.
   • Since acceleration has a direction, always subtract the initial velocity from the final velocity; otherwise the direction of the calculated acceleration will be incorrect.
   • If the initial time is not given in the problem, then it is assumed that tn = 0.
3. 3 Find the acceleration using the formula. First, write the formula and the variables given to you. Formula: . Subtract the initial speed from the final speed, and then divide the result by the time interval (time change). You will get the average acceleration over a given period of time.
   • If the final speed is less than the initial speed, then the acceleration has a negative value, that is, the body slows down.
   • Example 1: A car accelerates from 18.5 m/s to 46.1 m/s in 2.47 s. Find the average acceleration.
     • Write the formula: a = Δv / Δt = (v k - v n)/(t k - t n)
     • Write the variables: v to= 46.1 m/s, v n= 18.5 m/s, t to= 2.47 s, t n= 0 s.
     • Calculation: a= (46.1 - 18.5)/2.47 = 11.17 m/s 2 .
   • Example 2: A motorcycle starts braking at a speed of 22.4 m/s and stops after 2.55 s. Find the average acceleration.
     • Write the formula: a = Δv / Δt = (v k - v n)/(t k - t n)
     • Write the variables: v to= 0 m/s, v n= 22.4 m/s, t to= 2.55 s, t n= 0 s.
     • Calculation: A= (0 - 22.4)/2.55 = -8.78 m/s 2 .

2 Calculation of acceleration by force

1. 1 Newton's second law. According to Newton's second law, a body will accelerate if the forces acting on it do not balance each other. This acceleration depends on the net force acting on the body. Using Newton's second law, you can find the acceleration of a body if you know its mass and the force acting on that body.
   • Newton's second law is described by the formula: F res = m x a, Where F cut– resultant force acting on the body, m- body mass, a– acceleration of the body.
   • When working with this formula, use metric units, which measure mass in kilograms (kg), force in newtons (N), and acceleration in meters per second per second (m/s2).
2. 2 Find the mass of the body. To do this, place the body on the scale and find its mass in grams. If you are considering a very large body, look up its mass in reference books or on the Internet. The mass of large bodies is measured in kilograms.
   • To calculate acceleration using the above formula, you need to convert grams to kilograms. Divide the mass in grams by 1000 to get the mass in kilograms.
3. 3 Find the net force acting on the body. The resulting force is not balanced by other forces. If two differently directed forces act on a body, and one of them is greater than the other, then the direction of the resulting force coincides with the direction of the larger force. Acceleration occurs when a force acts on a body that is not balanced by other forces and which leads to a change in the speed of the body in the direction of action of this force.
   • For example, you and your brother are in a tug of war. You are pulling the rope with a force of 5 N, and your brother is pulling the rope (in the opposite direction) with a force of 7 N. The resulting force is 2 N and is directed towards your brother.
   • Remember that 1 N = 1 kg∙m/s 2.
4. 4 Rearrange the formula F = ma to calculate the acceleration. To do this, divide both sides of this formula by m (mass) and get: a = F/m. Thus, to find acceleration, divide the force by the mass of the accelerating body.
   • Force is directly proportional to acceleration, that is, the greater the force acting on a body, the faster it accelerates.
   • Mass is inversely proportional to acceleration, that is, the greater the mass of a body, the slower it accelerates.
5. 5 Calculate the acceleration using the resulting formula. Acceleration is equal to the quotient of the resulting force acting on the body divided by its mass. Substitute the values ​​given to you into this formula to calculate the acceleration of the body.
   • For example: a force equal to 10 N acts on a body weighing 2 kg. Find the acceleration of the body.
   • a = F/m = 10/2 = 5 m/s 2

3 Testing your knowledge

1. 1 Direction of acceleration. The scientific concept of acceleration does not always coincide with the use of this quantity in everyday life. Remember that acceleration has a direction; acceleration is positive if it is directed upward or to the right; acceleration is negative if it is directed downward or to the left. Check your solution based on the following table:
2. 2 Direction of force. Remember that acceleration is always co-directional with the force acting on the body. Some problems provide data that is intended to mislead you.
   • Example: a toy boat with a mass of 10 kg is moving north with an acceleration of 2 m/s 2 . A wind blowing in a westerly direction exerts a force of 100 N on the boat. Find the acceleration of the boat in a northerly direction.
   • Solution: Since the force is perpendicular to the direction of movement, it does not affect the movement in that direction. Therefore, the acceleration of the boat in the north direction will not change and will be equal to 2 m/s 2.
3. 3 Resultant force. If several forces act on a body at once, find the resulting force, and then proceed to calculate the acceleration. Consider the following problem (in two-dimensional space):
   • Vladimir pulls (on the right) a container with a mass of 400 kg with a force of 150 N. Dmitry pushes (on the left) a container with a force of 200 N. The wind blows from right to left and acts on the container with a force of 10 N. Find the acceleration of the container.
   • Solution: The conditions of this problem are designed to confuse you. In fact, everything is very simple. Draw a diagram of the direction of forces, so you will see that a force of 150 N is directed to the right, a force of 200 N is also directed to the right, but a force of 10 N is directed to the left. Thus, the resulting force is: 150 + 200 - 10 = 340 N. The acceleration is: a = F/m = 340/400 = 0.85 m/s 2.

Page 6 of 12

§ 5. Acceleration.
Uniformly accelerated linear motion

1. With uneven movement, the speed of a body changes over time. Let's consider the simplest case of uneven motion.

Movement in which the speed of a body changes by the same value over any equal intervals of time is called uniformly accelerated.

For example, if for every 2 s the speed of a body changed by 4 m/s, then the movement of the body is uniformly accelerated. The velocity module during such movement can either increase or decrease.

2. Let at the initial moment of time t 0 = 0 the speed of the body is v 0 . At some point in time t she became equal v. Then the change in speed over a period of time t – t 0 = t equals v– v 0, and per unit time - . This relationship is called acceleration. Acceleration characterizes the rate of change in speed.

The acceleration of a body during uniformly accelerated motion is a vector physical quantity equal to the ratio of the change in the speed of the body to the period of time during which this change occurred.

The SI unit of acceleration is meters per second squared (1 ):

[a] === 1 .

The unit of acceleration is taken to be the acceleration of such uniformly accelerated motion, at which the speed of the body is 1 s changes to 1 m/s.

3. Since acceleration is a vector quantity, it is necessary to find out how it is directed.

Let the car move in a straight line with an initial speed v 0 (speed at time t= 0) and speed v at some point in time t. The vehicle's speed modulus increases. In Figure 22, A depicts a vector of car speed. From the definition of acceleration, it follows that the acceleration vector is directed in the same direction as the vector difference v–v 0 . Therefore, in this case, the direction of the acceleration vector coincides with the direction of motion of the body (with the direction of the velocity vector).

Let now the speed modulus of the car decrease (Fig. 22 b). In this case, the direction of the acceleration vector is opposite to the direction of motion of the body (the direction of the velocity vector).

4. By transforming the acceleration formula for uniformly accelerated rectilinear motion, you can obtain a formula for finding the speed of a body at any time:

If the initial speed of the body is zero, i.e. at the initial moment of time it was at rest, then this formula takes the form:

v = at.

5. When calculating speed or acceleration, formulas are used that include not vectors, but projections of these quantities onto the coordinate axis. Since the projection of the sum of vectors is equal to the sum of their projections, the formula for the projection of velocity onto the axis X has the form:

v x = v 0x + a x t,

Where v x- projection of speed at a moment in time t, v 0x- projection of initial speed, a x- acceleration projection.

When solving problems, it is necessary to take into account the signs of projections. So, in the case shown in Figure 22, A, projections of velocities and acceleration onto the axis X positive; The speed modulus increases over time. In the case shown in Figure 22, b, projections onto the axis X velocities are positive, and the projection of acceleration is negative; the speed modulus decreases over time.

6. Example of problem solution

The vehicle speed during braking decreased from 23 to 15 m/s. What is the acceleration of the body if the braking lasts 5 s?

Let's write down the formula for finding the speed for uniformly accelerated rectilinear motion:

v = v 0 + at.

In projections onto the axis X we get

v x = v 0x + a x t.

Considering that the projection of the body’s acceleration onto the axis X is negative, and the projections of velocities on this axis are positive, we write: v = v 0 – at.

Where:

a = ;

a== 1.6 m/s 2 .

Answer: a= 1.6 m/s 2.

Self-test questions

1. What kind of motion is called uniformly accelerated?

2. What is the acceleration of uniformly accelerated motion called?

3. What formula is used to calculate acceleration during uniformly accelerated motion?

4. What is the SI unit of acceleration?

5. What formula is used to calculate the speed of a body in uniformly accelerated linear motion?

6. What is the sign of the acceleration projection onto the axis X in relation to the projection of the body’s speed onto the same axis, if the module of its speed increases; is it decreasing?

Task 5

1. What is the acceleration of the car if, 2 minutes after it started moving from rest, it acquired a speed of 72 km/h?

2. A train whose initial speed is 36 km/h accelerates with an acceleration of 0.5 m/s 2 . What speed will the train acquire in 20 s?

3. A car moving at a speed of 54 km/h stops at a traffic light for 15 s. What is the acceleration of the car?

4. What speed will the cyclist acquire 5 s after the start of braking, if his initial speed is 10 m/s and the acceleration during braking is 1.2 m/s 2?

Allows us to exist on this planet. How can we understand what centripetal acceleration is? The definition of this physical quantity is presented below.

Observations

The simplest example of the acceleration of a body moving in a circle can be observed by rotating a stone on a rope. You pull the rope, and the rope pulls the stone towards the center. At each moment of time, the rope imparts a certain amount of movement to the stone, and each time in a new direction. You can imagine the movement of the rope as a series of weak jerks. A jerk - and the rope changes its direction, another jerk - another change, and so on in a circle. If you suddenly release the rope, the jerking will stop, and with it the change in direction of speed will stop. The stone will move in the direction tangent to the circle. The question arises: “With what acceleration will the body move at this instant?”

Formula for centripetal acceleration

First of all, it is worth noting that the movement of a body in a circle is complex. The stone participates in two types of motion simultaneously: under the influence of force it moves towards the center of rotation, and at the same time along a tangent to the circle, moving away from this center. According to Newton's Second Law, the force holding a stone on a rope is directed toward the center of rotation along the rope. The acceleration vector will also be directed there.

Let us assume that after some time t our stone, moving uniformly with speed V, gets from point A to point B. Let us assume that at the moment of time when the body crossed point B, the centripetal force ceased to act on it. Then, in a period of time, it would get to point K. It lies on the tangent. If at the same moment of time only centripetal forces acted on the body, then during time t, moving with the same acceleration, it would end up at point O, which is located on a straight line representing the diameter of a circle. Both segments are vectors and obey the rule of vector addition. As a result of summing these two movements over a period of time t, we obtain the resulting movement along the arc AB.

If the time interval t is taken to be negligibly small, then the arc AB will differ little from the chord AB. Thus, it is possible to replace movement along an arc with movement along a chord. In this case, the movement of the stone along the chord will obey the laws of rectilinear motion, that is, the distance AB traveled will be equal to the product of the speed of the stone and the time of its movement. AB = V x t.

Let us denote the desired centripetal acceleration by the letter a. Then the path traveled only under the influence of centripetal acceleration can be calculated using the formula for uniformly accelerated motion:

Distance AB is equal to the product of speed and time, that is, AB = V x t,

AO - calculated earlier using the formula of uniformly accelerated motion for moving in a straight line: AO = at 2 / 2.

Substituting this data into the formula and transforming it, we get a simple and elegant formula for centripetal acceleration:

In words, this can be expressed as follows: the centripetal acceleration of a body moving in a circle is equal to the quotient of linear velocity squared by the radius of the circle along which the body rotates. The centripetal force in this case will look like the picture below.

Angular velocity

Angular velocity is equal to the linear velocity divided by the radius of the circle. The converse statement is also true: V = ωR, where ω is the angular velocity

If we substitute this value into the formula, we can obtain an expression for the centrifugal acceleration for the angular velocity. It will look like this:

Acceleration without changing speed

And yet, why does a body with acceleration directed towards the center not move faster and move closer to the center of rotation? The answer lies in the very formulation of acceleration. The facts show that circular motion is real, but to maintain it requires acceleration directed towards the center. Under the influence of the force caused by this acceleration, a change in the amount of motion occurs, as a result of which the trajectory of motion is constantly curved, all the time changing the direction of the velocity vector, but without changing its absolute value. Moving in a circle, our long-suffering stone rushes inward, otherwise it would continue to move tangentially. Every moment of time, going tangentially, the stone is attracted to the center, but does not fall into it. Another example of centripetal acceleration would be a water skier making small circles on the water. The athlete's figure is tilted; he seems to fall, continuing to move and leaning forward.

Thus, we can conclude that acceleration does not increase the speed of the body, since the velocity and acceleration vectors are perpendicular to each other. Added to the velocity vector, acceleration only changes the direction of movement and keeps the body in orbit.

Exceeding the safety factor

In the previous experiment we were dealing with a perfect rope that did not break. But let’s say our rope is the most ordinary, and you can even calculate the force after which it will simply break. In order to calculate this force, it is enough to compare the strength of the rope with the load it experiences during the rotation of the stone. By rotating the stone at a higher speed, you impart to it a greater amount of motion, and therefore greater acceleration.

With a jute rope diameter of about 20 mm, its tensile strength is about 26 kN. It is noteworthy that the length of the rope does not appear anywhere. By rotating a 1 kg load on a rope with a radius of 1 m, we can calculate that the linear speed required to break it is 26 x 10 3 = 1 kg x V 2 / 1 m. Thus, the speed that is dangerous to exceed will be equal to √ 26 x 10 3 = 161 m/s.

Gravity

When considering the experiment, we neglected the effect of gravity, since at such high speeds its influence is negligible. But you can notice that when unwinding a long rope, the body describes a more complex trajectory and gradually approaches the ground.

Celestial bodies

If we transfer the laws of circular motion into space and apply them to the movement of celestial bodies, we can rediscover several long-familiar formulas. For example, the force with which a body is attracted to the Earth is known by the formula:

In our case, the factor g is the same centripetal acceleration that was derived from the previous formula. Only in this case, the role of the stone will be played by a celestial body attracted to the Earth, and the role of the rope will be played by the force of gravity. The g factor will be expressed in terms of the radius of our planet and its rotation speed.

Results

The essence of centripetal acceleration is the hard and thankless work of keeping a moving body in orbit. A paradoxical case is observed when, with constant acceleration, a body does not change the value of its speed. To the untrained mind, such a statement is quite paradoxical. Nevertheless, both when calculating the motion of an electron around the nucleus, and when calculating the speed of rotation of a star around a black hole, centripetal acceleration plays an important role.