The red light of the traffic light changed to yellow, then green. With a tense roar, they break away from the car, then the sound of the engines subsides for a moment - this is the drivers released the fuel pedal and shift gears, acceleration again, again a moment of calm and again acceleration. Only 100 meters after the intersection, the flow of cars seems to calm down and roll smoothly until the next traffic light. Only one old car"Moskvich" passed the intersection smoothly and silently. The figure shows how he overtook all the cars and pulled far ahead. This car drove up to the intersection just at the moment when the green traffic light turned on, the driver did not have to slow down and stop the car, did not have to accelerate again after that. How is it that one car (and even a low-power "Moskvich" of the old production) easily, without tension, moves at a speed of about 50 km / h, while others with obvious tension gradually pick up speed and reach a speed of 50 km / h far after intersection when Moskvich is already approaching the next traffic light? It is obvious that for uniform motion significantly less effort and power consumption is required than during acceleration or, as they say, during accelerated movement.
Rice. Relatively weak car can overtake more powerful ones if it approaches the intersection at the moment the green light is turned on and does not expend effort on starting off and accelerating.
But before studying the acceleration of the car, you need to remember some concepts.
car acceleration
If the car passes the same number of meters every second, the movement is called uniform or steady. If the distance traveled by the car in every second (speed) changes, the movement is called:
- with increasing speed - accelerated
- when reducing speed - slow
The increase in speed per unit time is called acceleration, decrease in speed per unit time - negative acceleration, or slowdown.
Acceleration is measured by the increase or decrease in speed (in meters per second) in 1 second. If the speed increases by 3 m/s per second, the acceleration is 3 m/s per second or 3 m/s/s or 3 m/s2.
Acceleration is denoted by the letter j.
An acceleration of 9.81 m/s2 (or rounded 10 m/s2) corresponds to the acceleration known from experience for a freely falling body (ignoring air resistance) and is called the acceleration of gravity. It is denoted by the letter g.
Car acceleration
Car acceleration is usually depicted graphically. The path is plotted on the horizontal axis of the graph, and the speed is plotted on the vertical axis, and points are plotted corresponding to each segment of the path traveled. Instead of speed on a vertical scale, you can put off the acceleration time, as shown in the acceleration graph of domestic cars.
Rice. Acceleration path chart.
The acceleration graph is a curve with a gradually decreasing slope. Curve ledges correspond to shift points where acceleration drops off at some point, but they are often not shown.
Inertia
The car cannot develop from a place immediately great speed, because he has to overcome not only the forces of resistance to movement, but also inertia.
Inertia is the property of a body to maintain a state of rest or a state of uniform motion. It is known from mechanics that a stationary body can be set in motion (or the speed of a moving body is changed) only under the action of an external force. Overcoming the action of inertia, the external force changes the speed of the body, in other words, gives it acceleration. The amount of acceleration is proportional to the amount of force. The greater the mass of the body, the greater must be the force to give this body the desired acceleration. Weight is a quantity proportional to the amount of matter in the body; the mass m is equal to the weight of the body G divided by the acceleration of gravity g (9.81 m/s2):
m = G / 9.81, kg/(m/s2)
The mass of the car resists acceleration with the force Pj, this force is called the inertia force. In order for acceleration to occur, an additional traction force must be created on the drive wheels, equal to strength inertia. This means that the force necessary to overcome the inertia of the body and to give the body a certain acceleration j is proportional to the body's mass and acceleration. This force is:
Pj = mj = Gj / 9.81, kg
For the accelerated movement of the car, additional power is required:
Nj \u003d Pj * Va / 75 \u003d Gj * Va / 270 * 9.81 \u003d Gj * Va / 2650, hp
For the accuracy of calculations, the factor b (“delta”) should be included in equations (31) and (32) - the coefficient of rotating masses, which takes into account the influence of the rotating masses of the car (especially the engine flywheel and wheels) on acceleration. Then:
Nj = Gj*Va*b / 2650, h.p.
Rice. Graphs of the acceleration time of domestic cars.
The influence of the rotating masses lies in the fact that, in addition to overcoming the inertia of the mass of the car, it is necessary to "spin" the flywheel, wheels and other rotating parts of the car, spending part of the engine power on this. The value of the coefficient b can be considered approximately equal to:
b = 1.03 + 0.05*ik^2
where ik is the gear ratio in the gearbox.
Now, taking for example a car with a total weight of 2000 kg, it is not difficult to compare the forces required to maintain the movement of this car on asphalt at a speed of 50 km / h (so far without taking into account air resistance) and to start it off with an acceleration of about 2.5 m / sec2, common for modern cars.
According to the equation:
Pf \u003d 2000 * 0.015 \u003d 30, kg
To overcome the resistance of inertia on top gear(ik = 1) force is required:
Pj \u003d 2000 * 2.5 * 1.1 / 9.81 \u003d 560, kg
The car cannot develop such a force in the highest gear, you need to turn on the first gear (with gear ratio ik = 3).
Then we get:
Pj \u003d 2000 * 2.5 * 1.5 / 9.81 \u003d 760, kg
which is quite possible for modern cars.
So, the force required to start off is 25 times the force required to maintain motion with constant speed 50 km/h.
To ensure fast acceleration of the car, it is required to install an engine high power. When driving at a constant speed (except the maximum), the engine does not work at full power.
From the foregoing, it is clear why, when starting off, you need to turn on a lower gear. In passing, we note that on trucks, acceleration should usually be started in second gear. The fact is that in first gear (ik is approximately equal to 7.), the influence of rotating masses is very large and the traction force is not enough to tell the car a lot of acceleration; acceleration will be very slow.
On a dry road, with a friction coefficient φ of about 0.7, starting off in low gear does not cause any difficulties, since the adhesion force still exceeds the traction force. But on slippery road it can often turn out that the traction force in low gear is greater than the traction force (especially when the car is not loaded), and the wheels begin to slip. There are two ways out of this situation:
- reduce the traction force by starting off with a low fuel supply or in second gear (for trucks- on third);
- increase the coefficient of adhesion, i.e., pour sand under the drive wheels, put branches, boards, rags, put chains on the wheels, etc.
During acceleration, the unloading of the front wheels and the additional load on the rear wheels are especially affected. You can observe how, at the moment of starting off, the car noticeably, and sometimes very sharply, "crouches" on rear wheels. This redistribution of the load occurs even with a uniform movement of the car. It is due to counteracting torque. Pinion gear teeth main gear put pressure on the teeth of the driven (crown) and, as it were, press rear axle to the ground; in this case, a reaction occurs that pushes the drive gear up; there is a slight rotation rear axle in the direction opposite to the direction of rotation of the wheels. The springs fixed on the crankcase of the axle with their ends raise the front part of the frame or body and lower the rear. By the way, we note that it is precisely due to the unloading of the front wheels that it is easier to turn them while the car is moving with the gear engaged than during coasting, and even more so than in the parking lot. Every driver knows this. However, back to the additionally loaded rear wheels.
Additional, surplus load on the rear wheels Zd from transmitted moment the more than more moment Mk, brought to the wheel and the shorter wheelbase vehicle L (in m):
Naturally, this load is especially high when driving in lower gears, since the moment supplied to the wheels is increased. So, on a GAZ-51 car, the additional load in first gear is:
Zd \u003d 316 / 3.3 \u003d 96, kg
During starting and acceleration, the car is affected by the inertial force Pj applied at the center of gravity of the car and directed backward, i.e., in the opposite direction to acceleration. Since the force Pj is applied at a height hg from the plane of the road, it will tend to overturn the car around rear wheels. In this case, the load on the rear wheels will increase, and on the front wheels it will decrease by:
Rice. When power is transferred from the engine, the load on the rear wheels increases, and the load on the front wheels decreases.
Thus, when starting off, the rear wheels and tires are loaded by the weight of the vehicle, by the increased torque transmitted and by the force of inertia. This load acts on the rear axle bearings and mainly on the tires of the rear wheels. To save them, you need to start off as smoothly as possible. It should be recalled that on the rise, the rear wheels are even more loaded. On steep climb when starting off, and even with a high center of gravity of the car, such an unloading of the front wheels and overloading of the rear wheels can occur, which will lead to damage to the tires and even to the car tipping back.
Rice. In addition to the load from tractive effort, during acceleration, an additional force acts on the rear wheels from the inertia of the mass of the car.
The car moves with acceleration, and its speed increases as long as the traction force is greater than the resistance force. With increasing speed, the resistance to movement increases; when the equality of traction force and resistance is established, the car acquires a uniform movement, the speed of which depends on the amount of pressure on the fuel pedal. If the driver fully presses the fuel pedal, this uniform speed is at the same time the highest speed of the car.
The work of overcoming the forces of rolling resistance and air does not create a reserve of energy - energy is spent on combating these forces. The work of overcoming the forces of inertia during the acceleration of the car goes into the energy of motion. This energy is called kinetic energy. The energy reserve created in this case can be used if, after some acceleration, the drive wheels are disconnected from the engine, the gearshift lever is set to the neutral position, i.e., allow the car to move by inertia, coasting. Coasting occurs until the energy reserve is used up to overcome the forces of resistance to movement. It is appropriate to recall that on the same segment of the path, the energy consumption for acceleration is much greater than the consumption for overcoming the forces of resistance to movement. Therefore, due to the accumulated energy, the overrun path can be several times longer than the acceleration path. So, the roll-on path from a speed of 50 km / h is about 450 m for the Pobeda car, about 720 m for the GAZ-51 car, while the acceleration path to this speed is 150-200 m and 250-300 m, respectively. If the driver does not seek to drive a car with a very high speed, it can coast the vehicle for a significant part of the way and thus save energy and, thus, fuel.
The speed of a car accelerating from the starting point to straight line segment path length km with constant acceleration km/h 2 is calculated by the formula . Determine the minimum acceleration with which the car must move in order to acquire a speed of at least km / h after driving a kilometer. Express your answer in km / h 2.
The solution of the problem
This lesson demonstrates an example of calculating the least acceleration of a car under given conditions. This decision can be used for the purpose successful preparation to the Unified State Examination in mathematics, in particular, when solving problems of type B12.
The condition is the formula for determining the speed of the car: with a known path length and constant acceleration . To solve the problem, all known quantities are substituted into the above formula for determining the speed. As a result, an irrational inequality with one unknown is obtained. Since both parts of this inequality are greater than zero, they are squared according to the main property of the inequality. Expressing the value from the resulting linear inequality, the acceleration range is determined. According to the condition of the problem, the lower limit of this range is the desired least acceleration vehicle under given conditions.
One of the most important indicators of the dynamic qualities of a car is the intensity of acceleration - acceleration.
When changing the speed of movement, inertia forces arise that the car must overcome in order to provide a given acceleration. These forces are caused both by the progressively moving masses of the car m, and the moments of inertia of the rotating parts of the engine, transmission and wheels.
For the convenience of carrying out calculations, a complex indicator is used - reduced inertial forces:
Where δ vr- coefficient of accounting for rotating masses.
Acceleration amount j = dv/dt, which a car can develop when driving along a horizontal section of the road in a given gear and at a given speed, is found as a result of transforming the formula for determining the power reserve that is spent on acceleration:
,
or according to the dynamic characteristic:
D=f+
.
From here: j =
.
To determine the acceleration on an ascent or descent, use the formula:
The vehicle's ability to fast acceleration especially important in city driving. Increased acceleration for a car can be obtained by increasing the gear ratio u 0 main gear and the corresponding selection of the engine torque change characteristic.
The maximum acceleration during acceleration is within:
For cars in first gear 2.0 ... 3.5 m/s 2 ;
For passenger cars in direct gear 0.8 ... 2.0 m/s 2 ;
For trucks in second gear 1.8 ... 2.8 m/s 2 ;
For trucks in direct gear 0.4 ... 0.8 m/s 2 .
Time and path of car acceleration
The magnitude of acceleration in some cases is not a sufficiently clear indicator of the ability of the car to accelerate. For this purpose, it is convenient to use indicators such as acceleration time and path before set speed and graphs showing the dependence of speed on time and acceleration path.
Because j =
, That dt =
.
From here, by integrating the resulting equation, we find the acceleration time t in a given range of speed changes from v 1 before v 2 :
.
Determination of the acceleration path S in a given range of speed changes is carried out as follows. Since the speed is the first derivative of the path with respect to time, the path differential dS=v dt, or the acceleration path in the range of speed changes from v 1 before v 2 equals:
.
In the conditions of actual operation of the car, the time spent on gear shifting operations and clutch slipping increase the acceleration time compared to its theoretical (calculated) value. The time it takes to shift gears depends on the design of the gearbox. When using an automatic transmission, this time is practically zero.
In addition, overclocking does not always occur at full fuel supply, as it is supposed in the stated method. This also increases real time overclocking.
When using a manual transmission, an important point is the correct choice of the most favorable gear shift speeds. v 1-2 , v 2-3 etc. (see section "Traction calculation of the vehicle").
To assess the ability of a car to accelerate, the acceleration time after starting off on the way to 100 and 500 is also used as an indicator. m.
Plotting Accelerations
In practical calculations, it is assumed that acceleration occurs on a horizontal paved road. Clutch engaged and not slipping. The engine control is in the full fuel position. This ensures the grip of the wheels with the road without slipping. It is also assumed that the engine parameters change according to the external speed characteristic.
It is believed that acceleration for passenger cars begins with a minimum sustained speed in the lowest gear of the order v 0 = 1,5…2,0m/s up to values v T = 27,8m/s(100km/h). For trucks accept: v T = 16,7m/s(60km/h).
Sequentially starting with speed v 0 = 1,5…2,0m/s on the first gear and subsequent gears, on the dynamic characteristic (Fig. 1) for the abscissas selected along the abscissa v calculated points (at least five) determine the reserve of the dynamic factor during acceleration as the difference in ordinates ( D-f) on various transmissions. Rotating mass factor ( δ vr) for each transmission is calculated by the formula:
δ vr= 1.04 + 0.05 i kp 2 .
Vehicle accelerations are determined by the formula:
j =
.
Based on the data obtained, acceleration graphs are built j=f(v)(Fig. 2).
Fig.2. Characteristics of the acceleration of the car.
With correct calculation and construction, the acceleration curve in top gear will cross the abscissa at the point of maximum speed. Achieving the maximum speed occurs with the full use of the dynamic factor reserve: D–f=0.
Plotting the acceleration timet = f(v)
This graph is built using the acceleration graph of the car j=f(v)(Fig. 2). The speed scale of the acceleration graph is divided into equal sections, for example, every 1 m/s, and from the beginning of each section, perpendiculars are drawn to the intersection with the acceleration curves (Fig. 3).
The area of each of the obtained elementary trapezoids on the accepted scale is equal to the acceleration time for a given section of speed, if we assume that in each section of the speed acceleration occurs with a constant (average) acceleration:
j Wed = (j 1 + j 2 )/2 ,
Where j 1 , j 2 - accelerations, respectively, at the beginning and at the end of the considered section of speeds, m/s 2 .
This calculation does not take into account the time for gear shifting and other factors that lead to an overestimation of the acceleration time. Therefore, instead of the average acceleration, take the acceleration j i at the beginning of an arbitrarily taken section (determined by a scale).
Given the assumption made acceleration time on each section of the speed increment Δv defined as:
t i = Δv/j i ,With.
Rice. 3. Plotting the acceleration time
Based on the data obtained, a graph of the acceleration time is built. t = f(v). Full time acceleration from v 0 up to values v T is defined as the sum of the acceleration time (with a cumulative total) for all sections:
t 1 =Δv/j 1 , t 2 =t 1 +(Δv/j 2 ) ,t 3 = t 2 +(Δv/j 3 ) and so on until t T final acceleration time:
.
When plotting the acceleration time graph, it is convenient to use the table and take Δv= 1m/s.
|
Plots of speed v i , m/s |
||||||||
|
No. of plots | ||||||||
|
j i , m/s 2 | ||||||||
|
t i , With | ||||||||
|
Ramp up time | ||||||||
Recall that the constructed (theoretical) acceleration graph (Fig. 4) differs from the actual one in that the real time for gear shifting is not taken into account. In Fig. 4, time (1.0 With) to the gear change is displayed conditionally to illustrate the moment of the shift.
When using a mechanical (speed) transmission on a car, the actual acceleration time graph is characterized by a loss of speed at the moments of gear shifting. It also increases the acceleration time. A car with a gearbox with synchronizers has a higher acceleration rate. The greatest intensity in a car with an automatic continuously variable transmission.
Acceleration time of domestic passenger cars of a small class from a standstill to a speed of 100 km/h(28m/s) is about 13…20 With. For medium and big class it does not exceed 8…10 With.
Rice. 4. Characteristics of the car's acceleration over time.
Acceleration time for trucks up to speed 60 km/h(17m/s) is 35…45 With and higher, which indicates their insufficient dynamism.
km/h is 500…800 m.
Comparative data on the acceleration time of cars of domestic and foreign production are given in table. 3.4.
Table 3.4.
Acceleration time of passenger cars up to a speed of 100 km/h (28 m/s)
|
Automobile |
Time, With |
Automobile |
Time, With |
|
VAZ-2106 1.6 (74) |
Alfa Romeo-156 2.0 (155) | ||
|
VAZ-2121 1.6 (74) |
Audi A6 Tdi 2,5 (150) | ||
|
Moskvich 2.0 (113) |
BMW-320i 2.0 (150) | ||
|
Cadillac Sevilie 4.6 (395) | |||
|
GAZelle-3302 D 2.1 (95) |
Mercedes S 220 CD (125) | ||
|
ZAZ-1102 1.1 (51) |
Peugeot-406 3.0 (191) | ||
|
VAZ-2110 1.5 (94) |
Porsche-911 3.4 (300) | ||
|
Ford Focus 2.0 (130) |
VW Polo Sdi 1.7 (60) | ||
|
Fiat Marea 2.0 (147) |
Honda Civic 1.6 (160) |
Note: The working volume is indicated next to the type of vehicle ( l) and power (in brackets) of the engine ( hp).
Building a graph of the path of acceleration of the carS = f(v)
Similarly, the graphical integration of the previously constructed dependence is carried out t = f(V) to get the acceleration path dependency S on vehicle speed. IN this case the curve of the car acceleration time graph (Fig. 5) is divided into time intervals, for each of which the corresponding values are found V c R k .
Fig.5. Diagram explaining the use of the vehicle acceleration time graph t = f ( V ) to build a graph of the acceleration pathS = f( V ) .
The area of an elementary rectangle, for example, in the interval Δ t 5 there is a path that the car passes from the mark t 4 up to the mark t 5 , moving at a constant speed V c R 5 .
The area of an elementary rectangle is determined as follows:
Δ S k = V c R k (t k - t k -1 ) = V c R k · Δ t k .
Where k=l... m- sequence number of the interval, m is chosen arbitrarily, but is considered convenient for calculation when m = n.
For example (Fig. 5), if V cf5 =12,5 m/s; t 4 =10 With; t 5 =14 With, That Δ S 5 = 12,5(14 - 10) = 5 m.
Acceleration path from speed V 0 up to speed V 1 : S 1 = Δ S 1 ;
up to speed V 2 : S 2 = Δ S 1 + Δ S 2 ;
up to speed V n
: S n
= Δ
S 1
+ Δ
S 2
+ ... + Δ
S n
=
.
The calculation results are entered into a table and presented in the form of a graph (Fig. 6).
Acceleration path for cars up to speed 100 km/h is 300…600 m. For trucks, the path of acceleration to speed 50 km/h equal to 150…300 m.
Fig.6. Graphic artsacceleration pathscar.
- studying various movements, we can distinguish one relatively simple and common type of movement - movement with constant acceleration. Let us give a definition and a precise description of this movement. Galileo was the first to discover motion with constant acceleration.
A simple case of non-uniform motion is motion with constant acceleration, in which the modulus and direction of acceleration do not change with time. It can be straight and curvilinear. A bus or train moves with approximately constant acceleration when setting off or when braking, a puck sliding on ice, etc. All bodies under the influence of attraction to the Earth fall near its surface with constant acceleration, if air resistance can be neglected. This will be discussed further. We will study mainly motion with constant acceleration.
When moving with constant acceleration, the velocity vector changes in the same way for any equal time intervals. If the time interval is halved, then the modulus of the velocity change vector will also be halved. Indeed, for the first half of the interval, the speed changes in exactly the same way as for the second. In this case, the direction of the velocity change vector remains unchanged. The ratio of speed change to time interval will be the same for any time interval. Therefore, the expression for acceleration can be written as:
Let's explain this with a picture. Let the trajectory be curvilinear, the acceleration is constant and directed downwards. Then the velocity change vectors for equal time intervals, for example, for every second, will be directed downwards. Let us find the changes in speed for successive time intervals equal to 1 s. To do this, we set aside from one point A the speeds 0, 1, 2, 3, etc., which the body acquires after 1 s, and subtract the initial speed from the final one. Since = const, then all the velocity increment vectors for each second lie on the same vertical and have the same modules (Fig. 1.48), i.e., the module of the velocity change vector A increases uniformly.
Rice. 1.48
If acceleration is constant, then it can be understood as a change in speed per unit time. This allows you to set units for the acceleration module and its projections. Let's write an expression for the acceleration module:
Hence it follows that
Therefore, the unit of acceleration is the constant acceleration of the movement of the body (point), at which the module of speed changes per unit of speed per unit of time:
These units of acceleration are read as one meter per second squared and one centimeter per second squared.
The unit of acceleration 1 m/s 2 is such a constant acceleration at which the modulus of change in speed for each second is 1 m/s.
If the acceleration of a point is not constant and at some instant becomes equal to 1 m/s 2, then this does not mean that the modulus of the speed increment is 1 m/s per second. In this case, the value of 1 m / s 2 should be understood as follows: if starting from this moment the acceleration became constant, then for every second the modulus of speed change would be equal to 1 m / s.
The Zhiguli car, when accelerating from a standstill, acquires an acceleration of 1.5 m / s 2, and the train - about 0.7 m / s 2. A stone falling to the ground moves with an acceleration of 9.8 m/s 2 .
Of the various types of uneven motion, we have singled out the simplest - motion with constant acceleration. However, there is no movement with a strictly constant acceleration, just as there is no movement with a strictly constant speed. All these are the simplest models of real movements.
Do the exercises
- The point moves along a curvilinear trajectory with acceleration, the modulus of which is constant and equal to 2 m/s 2 . Does this mean that in 1 s the modulus of the point's velocity changes by 2 m/s?
- The point moves with variable acceleration, the module of which at some point in time is 3 m/s 2 . How to interpret this value of the acceleration of the moving point?
Acceleration - the amount of change in the speed of a body per unit time. In other words, acceleration is the rate of change of speed.
A - acceleration, m/s 2
t - speed change interval, s
V 0 - initial speed of the body, m / s
V - final speed of the body, m/s
An example of using a formula.
The car accelerates from 0 to 108km/h (30m/s) in 3 seconds.
The acceleration with which the car accelerates is:
a \u003d (V-V o) / t \u003d (30m / s - 0) / 3c \u003d 10m / s 2
Another, more precise, formulation reads: acceleration is equal to the derivative of the body's velocity: a=dV/dt
The term acceleration is one of the most important in physics. Acceleration is used in tasks for acceleration, braking, throws, shots, falls. But, at the same time, this term is one of the most difficult to understand, first of all, because the unit of measurement m/s 2(meter per second per second) is not used in everyday life.
The device for measuring acceleration is called an accelerometer. Accelerometers, in the form of miniature microchips, are used in many smartphones and allow you to determine the force with which the user acts on the phone. Data on the force of impact on the device, allow you to create mobile applications, which react to screen rotation and shake.
Reaction mobile devices to rotate the screen, it is provided precisely by the accelerometer - a microchip that measures the acceleration of the device.
An exemplary accelerometer circuit is shown in the figure. A massive weight, with sudden movements, deforms the springs. Deformation measurement using capacitors (or piezoelectric elements) allows you to calculate the force on the weight and acceleration.
Knowing the deformation of the spring, using Hooke's law (F=k∙Δx), you can find the force acting on the weight, and knowing the weight of the weight, using Newton's second law (F=m∙a), you can find the acceleration of the weight.
On the circuit board of the iPhone 6, the accelerometer fits into a microchip measuring just 3mm by 3mm.
