One of the most common types of movement of objects in space with which a person meets every day is uniformly accelerated rectilinear movement. In the 9th grade of comprehensive schools, in the course of physics they study this type of movement in detail. Let's consider it in the article.
Kinematic characteristics of the movement
Before giving formulas describing uniformly accelerated rectilinear motion in physics, we 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 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 a straight segment on this line. It is measured in meters (m) in the SI system of physical units.
Speed, or how often it is called linear speed, is the speed of a change in the position of a body in space along its trajectory of movement. Denote the speed by v. It is measured in meters per second (m / s).
Acceleration is the third important value for describing a rectilinear uniformly accelerated motion. It shows how quickly the speed of the body changes over time. Acceleration is denoted by a and determined in meters per square second (m / s 2 ).
The path S and speed v are variable characteristics in a rectilinear uniformly accelerated motion. Acceleration is a constant value.
The relationship of speed and acceleration
Imagine that a 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 put pressure on the gas pedal, and the car began to increase its speed, acquiring acceleration a. If you start the countdown 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:
v = v 0 + a * t.
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 axis at the point (0; v 0 ) and having a certain angle of inclination to the abscissa axis (the tangent of this angle is acceleration value a).
The figure shows two graphs. The difference between them is only 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 started to accelerate from the rest state (for example, a starting car).
Note that if in the example above the driver would depress the brake pedal instead of the gas pedal, then the braking movement would be described by the following formula:
v = v 0 - a * t.
This type of movement is called rectilinear, equally slow.
Formulas traveled
In practice, it is often important to know not only acceleration, but also the significance 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 = v 0 * t + a * t 2/2.
The first term corresponds to uniform motion without acceleration. The second term is a contribution to the path traveled by pure accelerated motion.
In case of braking of a moving object, the expression for the path will take the form:
S = v 0 * t - a * t 2/2.
Unlike 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 branches of the parabola will be the graphs of the functions S (t). The figure below shows these graphs in a schematic form.
Parabolas 1 and 3 correspond to the accelerated movement of the body, parabola 2 describes the process of inhibition. 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 its movement.
Further in the article we will solve three different problems on the use of the above formulas.
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 for 20 seconds with an acceleration of 1 m / s 2 . Then its speed does not change. How long does the car take the passenger to point B?
The distance that a car will travel in 20 seconds will be equal to:
S 1 = a * t 1 2/2.
At the same time, the speed that he will gain in 20 seconds is equal to:
v = a * t 1 .
Then the desired travel time t can be calculated by the following formula:
t = (S - S 1 ) / v + t 1 = (S - a * t 1 2/2) / (a * t 1 ) + t 1 .
Here S is the distance between A and B.
We transfer all the known data to the SI system and substitute it into the recorded expression. We get the answer: t = 1510 seconds or about 25 minutes.
The task of calculating the braking distance
Now we solve the problem of equally slow motion. Suppose a truck was moving at a speed of 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 stops in 15 seconds.
The braking distance S can be calculated using the following formula:
S = v 0 * t - a * t 2/2.
The braking time t and the initial speed v 0 we know. The acceleration a can be found from the expression for speed, given that its final value is zero. We have:
v 0 - a * t = 0;
a = v 0 / t.
Substituting the obtained expression into the equation, we arrive at the final formula for the path S:
S = v 0 * t - v 0 * t / 2 = v 0 * t / 2.
We substitute the values from the condition and record the answer: S = 145.8 meters.
Free fall speed problem
Perhaps the most common rectilinear uniformly accelerated motion in nature is the free fall of bodies in the field of planetary gravity. We will solve the following problem: they released the body from a height of 30 meters. What speed will it have when it falls to the surface of the earth?
The desired speed can be calculated by the formula:
v = g * t.
Where g = 9.81 m / s 2 .
The time of the fall of the body is determined from the corresponding expression for the path S:
S = g * t 2/2;
t = √ (2 * S / g).
We substitute the time t in the formula for v, we obtain:
v = g * √ (2 * S / g) = √ (2 * S * g).
The value of the path S traveled by the body is known from the condition, substituting it into equality, we obtain: v = 24.26 m / s or about 87 km / h.