In physics, in the sections of kinematics and dynamics, various types of mechanical motion are studied along different types of trajectories. This article is devoted to the consideration of graphs and formulas of uniformly accelerated motion of bodies along a straight path and around a circle.
The concept of acceleration
Before we proceed to the analysis of the formulas of uniformly accelerated motion, we must define acceleration itself. Under it, in physics, a vector quantity is assumed, which describes the change in speed over time. The mathematical formulation of this definition is as follows:
a¯ = dv¯ / dt.
For example, a change in speed of 1 m / s in one second is characterized by an acceleration of 1 m / s 2 .
The recorded expression allows you to calculate the so-called instantaneous speed. In practice, it is often necessary to know not a value a¯ at a given point in time, but some average value a cp ¯ over a certain period of time. In this case, apply the following formula:
a¯ = Δv¯ / Δt.
Here Δv¯ is the velocity change vector over time Δt.
Note that the acceleration vector is always directed towards the change in speed, therefore, it does not depend directly on the speed vector. In turn, the speed is always directed tangentially to the trajectory at a given point.
Uniformly accelerated rectilinear motion
This type of movement often appears in physical problems. In practice, it is also realized, for example, during acceleration of a vehicle from a standstill, during a free fall of a heavy body or during braking of a vehicle. In all these cases, we are talking about moving objects with constant acceleration. That is why the movement itself is called uniformly accelerated (a = const).
The speed and acceleration of uniformly accelerated motion are related by the following expression:
v = v 0 + a * t.
Here v 0 is the speed that the body had before the appearance of acceleration a. At the beginning of movement with acceleration from a state of relative rest, the value of v 0 can be omitted. Then we get:
v = a * t.
As you can see, the graphs of uniformly accelerated motion for the function v (t) will be straight lines that start either from the point (0; v 0 ) or from the point (0; 0). The angle between the abscissa and the straight line is the arctangent of the acceleration value.
If the initial velocity v 0 is present, the acceleration a can be negative, which in practice corresponds to the braking of the body. The graph v (t) will also be a straight line, but it will tend to a zero value of speed. The corresponding formula takes the form:
v = v 0 - a * t.
Since the acceleration of uniformly accelerated motion does not depend on time, the graph of the function a (t) is a straight line parallel to the time axis t.
Moving with uniformly accelerated motion
Above were given three formulas of uniformly accelerated motion in a straight line that relate speed and time (acceleration is a constant value). In order to calculate the path that the body travels in time t with this type of displacement, the recorded expressions should be integrated over time. As a result of the integration operation, we obtain the following three formulas for the path S:
1) S = a * t 2/2;
2) S = v 0 * t + a * t 2/2;
3) S = v 0 * t - a * t 2/2.
All three expressions show that for the path, the graphs of uniformly accelerated motion are parabolas, or rather, its right branch. For formulas 1) and 2) we are talking about the increasing branch of the parabola, since the acceleration vector coincides with the velocity vector. For the third expression, the right branch of the parabola tends to some constant positive value S 0 corresponding to the path that the body will go before it stops completely.
Equally accelerated circumferential movement
This type of movement is very different from straightforward. Firstly, with accelerated rotation, the velocity changes its modulus and its vector, which leads to the appearance of two components of acceleration: tangent and centripetal. Secondly, during rotation, it makes no sense to evaluate how far the body has traveled, because it moves under the same circle.
In connection with the foregoing, angular velocities and accelerations are used to describe circular motion. Angular acceleration shows how quickly the angular velocity changes in radians per second. With linear acceleration a, the angular α is associated with the following expression:
a = α * r.
Where r is the radius of the trajectory of rotation.
For uniformly accelerated movement along a circular path, the following kinematic formulas are valid:
θ = ω 0 * t ± α * t 2/2;
ω = ω 0 ± α * t.
Here θ is the rotation angle in radians over time t. It can be used to calculate the linear distance L that the body will go along the circle:
L = θ * r.
Freefall Challenge
Having considered all the important formulas of uniformly accelerated motion, we will solve the following problem: the body is thrown vertically upwards with an initial speed of 35 m / s. It is necessary to determine to what height it can rise and after what time it will fall to the ground. The forces of friction can be neglected.
During the rise, the acceleration of gravity g, directed against the speed, acts on the body, that is, the rise time will be equal to:
v = v 0 - g * t =>
t = v 0 / g.
Neglecting the forces of friction, it is safe to say that the rise time will be equal to the fall time, so the total time of the body’s movement is:
t tot = 2 * t = 2 * v 0 / g = 7.14 seconds.
The lifting height h can be calculated by the following formula:
h = v 0 * t - g * t 2/2 = v 0 2 / (2 * g) = 62.44 meters.
Thus, the body after throwing up will reach a height of 62.4 meters, and will fall to the surface of the earth 7.1 seconds after the start of movement.