This is a vector physical quantity, numerically equal to the limit to which the average speed tends over an infinitely small period of time:
In other words, the instantaneous velocity is the radius vector in time.
The instantaneous velocity vector is always directed tangentially to the trajectory of the body in the direction of the movement of the body.
Instantaneous speed provides accurate information about movement at a specific point in time. For example, when driving in a car at some point in time, the driver looks at the speedometer and sees that the device shows 100 km / h. After a while, the speedometer needle points to 90 km / h, and a few minutes later - to 110 km / h. All of the listed speedometer readings are the values of the instantaneous vehicle speed at certain points in time. The speed at each moment of time and at each point of the trajectory must be known when docking space stations, when landing aircraft, etc.
Does the concept of "instantaneous speed" have a physical meaning? Velocity is a characteristic of change in space. However, in order to determine how the displacement has changed, it is necessary to observe the movement for some time. Even the most advanced speed measuring devices, such as radar systems, measure speed over a period of time - albeit small enough, but it is still a finite time interval, not a moment in time. The expression “the speed of a body at a given moment of time” is not correct from the point of view of physics. However, the concept of instantaneous speed is very convenient in mathematical calculations, and it is constantly used.
Examples of solving problems on the topic "Instantaneous speed"
EXAMPLE 1
EXAMPLE 2
| Exercise | The law of motion of a point along a straight line is given by the equation. Find the instantaneous speed of the point 10 seconds after the start of the movement. |
| Solution | The instantaneous speed of a point is the radius vector in time. Therefore, for the instantaneous speed, you can write: 10 seconds after the start of movement, the instantaneous speed will have a value: |
| Answer | In 10 seconds after the start of movement, the instantaneous speed of the point is m / s. |
EXAMPLE 3
| Exercise | The body moves in a straight line so that its coordinate (in meters) changes according to the law. How many seconds after the start of movement will the body stop? |
| Solution | Let's find the instantaneous speed of the body: |
And why is it needed. We already know what a frame of reference, 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 on the basics of kinematics, and give a practical example of solving the problem.
Let's solve the following problem: the 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 a 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 values.
Need help solving problems? Professional student service is ready to provide it.
1.2. Straight motion
1.2.4. average speed
A material point (body) retains its speed unchanged only with uniform rectilinear motion. If the movement is uneven (including equally variable), then the speed of the body changes. This movement is characterized by an average speed. Distinguish between average travel speed and average ground speed.
Average travel speed is a vector physical quantity, which is determined by the formula
v → r = Δ r → Δ t,
where Δ r → is the displacement vector; ∆t is the time interval during which this movement took place.
Average ground speed is a scalar physical quantity and is calculated by the formula
v s = S total t total,
where S total = S 1 + S 1 + ... + S n; t total = t 1 + t 2 + ... + t N.
Here S 1 = v 1 t 1 - the first section of the path; v 1 - the speed of passage of the first section of the path (Fig. 1.18); t 1 - the time of movement on the first section of the path, etc.
Rice. 1.18
Example 7. One quarter of the way the bus moves at a speed of 36 km / h, the second quarter of the way - 54 km / h, the rest of the way - at a speed of 72 km / h. Calculate the average road speed of the bus.
Solution. The total path traveled by the bus is denoted by S:
S total = S.
S 1 = S / 4 - the path traveled by the bus on the first section,
S 2 = S / 4 - the path traveled by the bus on the second section,
S 3 = S / 2 - the path traveled by the bus in the third section.
The bus travel time is determined by the formulas:
- in the first section (S 1 = S / 4) -
t 1 = S 1 v 1 = S 4 v 1;
- in the second section (S 2 = S / 4) -
t 2 = S 2 v 2 = S 4 v 2;
- in the third section (S 3 = S / 2) -
t 3 = S 3 v 3 = S 2 v 3.
The total travel time of the bus is:
t total = t 1 + t 2 + t 3 = S 4 v 1 + S 4 v 2 + S 2 v 3 = S (1 4 v 1 + 1 4 v 2 + 1 2 v 3).
v s = S total t total = S S (1 4 v 1 + 1 4 v 2 + 1 2 v 3) =
1 (1 4 v 1 + 1 4 v 2 + 1 2 v 3) = 4 v 1 v 2 v 3 v 2 v 3 + v 1 v 3 + 2 v 1 v 2.
v s = 4 ⋅ 36 ⋅ 54 ⋅ 72 54 ⋅ 72 + 36 ⋅ 72 + 2 ⋅ 36 ⋅ 54 = 54 km / h.
Example 8. One fifth of the time a city bus spends on stops, the rest of the time it moves at a speed of 36 km / h. Determine the average road speed of the bus.
Solution. The total travel time of the bus on the route is denoted by t:
t total = t.
t 1 = t / 5 - time spent on stops,
t 2 = 4t / 5 - bus travel time.
The route traveled by the bus:
- for time t 1 = t / 5 -
S 1 = v 1 t 1 = 0,
since the speed of the bus v 1 in this time interval is equal to zero (v 1 = 0);
- in time t 2 = 4t / 5 -
S 2 = v 2 t 2 = v 2 4 t 5 = 4 5 v 2 t,
where v 2 is the speed of the bus at a given time interval (v 2 = = 36 km / h).
The total bus route is:
S total = S 1 + S 2 = 0 + 4 5 v 2 t = 4 5 v 2 t.
We calculate the average ground speed of the bus using the formula
v s = S total t total = 4 5 v 2 t t = 4 5 v 2.
The calculation gives the value of the average ground speed:
v s = 4 5 ⋅ 36 = 30 km / h.
Example 9. The equation of motion of a material point has the form x (t) = (9.0 - 6.0t + 2.0t 2) m, where the coordinate is given in meters, time - in seconds. Determine the average ground speed and the value of the average speed of movement of a material point in the first three seconds of movement.
Solution. For determining average travel speed it is necessary to calculate the movement of the material point. The modulus of movement of a material point in the time interval from t 1 = 0 s to t 2 = 3.0 s is calculated as the difference of coordinates:
| Δ r → | = | x (t 2) - x (t 1) | ,
Substituting values into the formula for calculating the displacement modulus gives:
| Δ r → | = | x (t 2) - x (t 1) | = 9.0 - 9.0 = 0 m.
Thus, the displacement of the material point is zero. Therefore, the modulus of the average movement speed is also zero:
| v → r | = | Δ r → | t 2 - t 1 = 0 3.0 - 0 = 0 m / s.
For determining average ground speed it is necessary to calculate the path traversed by the material point during the time interval from t 1 = 0 s to t 2 = 3.0 s. The point's movement is uniformly slow, so you need to find out if the stopping point falls within the specified interval.
To do this, we write down the law of change in the speed of a material point over time in the form:
v x = v 0 x + a x t = - 6.0 + 4.0 t,
where v 0 x = −6.0 m / s is the projection of the initial velocity onto the Ox axis; a x = = 4.0 m / s 2 - projection of acceleration onto the specified axis.
Find a stopping point from the condition
v (τ rest) = 0,
those.
τ rest = v 0 a = 6.0 4.0 = 1.5 s.
The stopping point falls within the time interval from t 1 = 0 s to t 2 = 3.0 s. Thus, the traversed path is calculated by the formula
S = S 1 + S 2,
where S 1 = | x (τ rest) - x (t 1) | - the path traversed by a material point to a stop, i.e. for the time from t 1 = 0 s to τ rest = 1.5 s; S 2 = | x (t 2) - x (τ rest) | - the path traversed by the material point after stopping, i.e. for the time from τ rest = 1.5 s to t 1 = 3.0 s.
Let's calculate the values of the coordinates at the specified times:
x (t 1) = 9.0 - 6.0 t 1 + 2.0 t 1 2 = 9.0 - 6.0 ⋅ 0 + 2.0 ⋅ 0 2 = 9.0 m;
x (τ rest) = 9.0 - 6.0 τ rest + 2.0 τ rest 2 = 9.0 - 6.0 ⋅ 1.5 + 2.0 ⋅ (1.5) 2 = 4.5 m ;
x (t 2) = 9.0 - 6.0 t 2 + 2.0 t 2 2 = 9.0 - 6.0 ⋅ 3.0 + 2.0 ⋅ (3.0) 2 = 9.0 m ...
The coordinate values allow you to calculate the paths S 1 and S 2:
S 1 = | x (τ rest) - x (t 1) | = | 4.5 - 9.0 | = 4.5 m;
S 2 = | x (t 2) - x (τ rest) | = | 9.0 - 4.5 | = 4.5 m,
as well as the total distance traveled:
S = S 1 + S 2 = 4.5 + 4.5 = 9.0 m.
Therefore, the desired value of the average ground speed of a material point is
v s = S t 2 - t 1 = 9.0 3.0 - 0 = 3.0 m / s.
Example 10. The graph of the dependence of the projection of the speed of a material point on time is a straight line and passes through the points (0; 8.0) and (12; 0), where the speed is set in meters per second, time - in seconds. How many times does the average ground speed for 16 seconds of movement exceed the value of the average speed of movement for the same time?
Solution. The graph of the dependence of the projection of the body's velocity on time is shown in the figure.
To graphically calculate the path traversed by the material point and the modulus of its movement, it is necessary to determine the value of the projection of the velocity at the moment of time, equal to 16 s.
There are two ways to determine the value of v x at a specified moment in time: analytical (through the equation of a straight line) and graphical (through the similarity of triangles). To find v x, we will use the first method and compose the equation of a straight line at two points:
t - t 1 t 2 - t 1 = v x - v x 1 v x 2 - v x 1,
where (t 1; v x 1) - coordinates of the first point; (t 2; v x 2) - coordinates of the second point. By the condition of the problem: t 1 = 0, v x 1 = 8.0, t 2 = 12, v x 2 = 0. Taking into account the specific values of the coordinates, this equation takes the form:
t - 0 12 - 0 = v x - 8.0 0 - 8.0,
v x = 8.0 - 2 3 t.
At t = 16 s, the velocity projection value is
| v x | = 8 3 m / s.
This value can also be obtained from the similarity of triangles.
- Let's calculate the path traversed by the material point as the sum of the values S 1 and S 2:
S = S 1 + S 2,
where S 1 = 1 2 ⋅ 8.0 ⋅ 12 = 48 m - the path traversed by the material point during the time interval from 0 s to 12 s; S 2 = 1 2 ⋅ (16 - 12) ⋅ | v x | = 1 2 ⋅ 4.0 ⋅ 8 3 = = 16 3 m - the path traversed by the material point in the time interval from 12 s to 16 s.
The total distance traveled is
S = S 1 + S 2 = 48 + 16 3 = 160 3 m.
The average ground speed of a material point is
v s = S t 2 - t 1 = 160 3 ⋅ 16 = 10 3 m / s.
- We calculate the value of the displacement of a material point as the modulus of the difference between the values S 1 and S 2:
S = | S 1 - S 2 | = | 48 - 16 3 | = 128 3 m.
The value of the average movement speed is
| v → r | = | Δ r → | t 2 - t 1 = 128 3 ⋅ 16 = 8 3 m / s.
The sought speed ratio is
v s | v → r | = 10 3 ⋅ 3 8 = 10 8 = 1.25.
The average ground speed of a material point is 1.25 times the modulus of the average speed of movement.
If a material point is in motion, then its coordinates are subject to changes. This process can be fast or slow.
Definition 1
The quantity that characterizes the rate of change in the position of the coordinate is called speed.
Definition 2
average speed Is a vector quantity, numerically equal to the displacement per unit of time, and co-directional with the displacement vector υ = ∆ r ∆ t; υ ∆ r.
Picture 1 . Average speed is in the same direction as movement
The modulus of the average speed along the way is υ = S ∆ t.
Instantaneous speed characterizes movement at a certain point in time. The expression "body speed at a given time" is considered not correct, but applicable in mathematical calculations.
Definition 3
The instantaneous speed is called the limit to which the average speed υ tends when the time interval ∆ t tends to 0:
υ = l i m ∆ t ∆ r ∆ t = d r d t = r ˙.
The direction of the vector υ is tangential to the curvilinear trajectory, because the infinitesimal displacement d r coincides with the infinitesimal element of the trajectory d s.
Figure 2. Instantaneous velocity vector υ
The available expression υ = l i m ∆ t ∆ r ∆ t = d r d t = r ˙ in Cartesian coordinates is identical to the equations proposed below:
υ x = d x d t = x ˙ υ y = d y d t = y ˙ υ z = d z d t = z ˙.
The record of the modulus of the vector υ will take the form:
υ = υ = υ x 2 + υ y 2 + υ z 2 = x 2 + y 2 + z 2.
To pass from Cartesian rectangular coordinates to curvilinear ones, the rules for differentiating complex functions are applied. If the radius vector r is a function of curvilinear coordinates r = r q 1, q 2, q 3, then the velocity value will be written as:
υ = d r d t = ∑ i = 1 3 ∂ r ∂ q i ∂ q i ∂ r = ∑ i = 1 3 ∂ r ∂ q i q ˙ i.
Figure 3. Displacement and instantaneous velocity in curvilinear coordinate systems
For spherical coordinates, assume that q 1 = r; q 2 = φ; q 3 = θ, then we get υ presented in the following form:
υ = υ r e r + υ φ e φ + υ θ φ θ, where υ r = r ˙; υ φ = r φ ˙ sin θ; υ θ = r θ ˙; r ˙ = d r d t; φ ˙ = d φ d t; θ ˙ = d θ d t; υ = r 1 + φ 2 sin 2 θ + θ 2.
Definition 4
Instant speed is the value of the derivative of the time displacement function at a given moment associated with elementary displacement by the relation d r = υ (t) d t
Example 1
A law of rectilinear motion of a point x (t) = 0, 15 t 2 - 2 t + 8 is given. Determine its instantaneous speed 10 seconds after the start of movement.
Solution
It is customary to call the instantaneous velocity the first derivative of the radius vector with respect to time. Then its record will take the form:
υ (t) = x ˙ (t) = 0. 3 t - 2; υ (10) = 0. 3 × 10 - 2 = 1 m / s.
Answer: 1 m / s.
Example 2
The movement of a material point is given by the equation x = 4 t - 0.05 t 2. Calculate the moment of time t o s t, when the point stops moving, and its average ground speed υ.
Solution
Let's calculate the equation of the instantaneous velocity, substitute the numerical expressions:
υ (t) = x ˙ (t) = 4 - 0, 1 t.
4 - 0, 1 t = 0; t about with t = 40 s; υ 0 = υ (0) = 4; υ = ∆ υ ∆ t = 0 - 4 40 - 0 = 0.1 m / s.
Answer: the set point will stop after 40 seconds; the value of the average speed is 0.1 m / s.
If you notice an error in the text, please select it and press Ctrl + Enter
The speed of a point is a vector that determines the speed and direction of movement of the point at any given time.
The speed of uniform movement is determined by the ratio of the path traversed by a point in a certain period of time to the value of this period of time.
Speed; S-path; t- time.
The speed is measured in units of length, divided by a unit of time: m / s; cm / s; km / h, etc.
In the case of rectilinear motion, the velocity vector is directed along the trajectory in the direction of its motion.
If a point travels uneven paths at equal intervals of time, then this movement is called uneven. Speed is a variable and is a function of time.
The average speed of a point over a given period of time is the speed of such a uniform rectilinear movement at which the point during this period of time would receive the same movement as in its considered movement.
Consider a point M that moves along a curved trajectory given by the law
During the time interval Δt, point M will move to position M 1 along the arc MM 1. If the time interval Δt is small, then the arc MM 1 can be replaced by a chord and, in the first approximation, find the average speed of movement of the point
This speed is directed along the chord from point M to point M 1. We find the true speed by passing to the limit at Δt> 0
When? T> 0, the direction of the chord in the limit coincides with the direction of the tangent to the trajectory at point M.
Thus, the value of the point velocity is defined as the limit of the ratio of the path increment to the corresponding time interval when the latter tends to zero. The direction of the velocity coincides with the tangent to the trajectory at this point.
Point acceleration
Note that in the general case, when moving along a curved trajectory, the speed of a point changes both in direction and in magnitude. The change in speed per unit of time is determined by the acceleration. In other words, the acceleration of a point is a value that characterizes the rate at which the speed changes over time. If during the time interval? T the speed changes by an amount, then the average acceleration
The true acceleration of a point at a given time t is the value to which the average acceleration tends at? T> 0, that is
With a time interval tending to zero, the acceleration vector will change both in magnitude and in direction, tending to its limit.
Acceleration dimension
Acceleration can be expressed in m / s 2; cm / s 2, etc.
In the general case, when the motion of a point is given in a natural way, the acceleration vector is usually decomposed into two components directed tangentially and along the normal to the trajectory of the point.
Then the acceleration of a point at time t can be represented as follows
Let us denote the constituent limits by and.
The direction of the vector does not depend on the value of the time interval Δt.
This acceleration always coincides with the direction of the velocity, that is, it is directed tangentially to the trajectory of the point's motion and is therefore called tangential or tangential acceleration.
The second component of the point acceleration is directed perpendicular to the tangent to the trajectory at this point in the direction of the concavity of the curve and affects the change in the direction of the velocity vector. This component of acceleration is called normal acceleration.
Since the numerical value of the vector is equal to the increment in the velocity of the point over the considered time interval Δt, the numerical value of the tangential acceleration
The numerical value of the tangential acceleration of a point is equal to the time derivative of the numerical value of the velocity. The numerical value of the normal acceleration of a point is equal to the square of the speed of the point divided by the radius of curvature of the trajectory at the corresponding point on the curve
The full acceleration at an uneven curvilinear motion of a point is added geometrically from the tangential and normal accelerations.
