My name is Alexander FufaeV and here I will explain the following topic:

# Uniformly Accelerated (Non-Uniform) Motion

## Important Formula

What do the formula symbols mean?

## Position

Unit
Current position $$x(t)$$ of an accelerated object at time $$t$$ on the $$x$$ axis.

## Initial position

Unit
Position of the object at the time $$t = 0$$. It is usual to set the initial position $$x_0 = 0$$. Then the position $$x(t)$$ corresponds to the distance covered by the body after the time $$t$$.

## Initial velocity

Unit
Velocity with which the object starts the acceleration process. If the object starts from rest, then the initial velocity is $$v_0 = 0$$.

## Time

Unit
This is the time during which the object accelerates.

## Acceleration

Unit
Uniform acceleration of the object. That means: Every second the current velocity of the body increases by the value $$a$$.
Explanation

Non-uniform motion of an object (such as a car) means that the object is moving with a changing velocity. In contrast to uniform motion, where the velocity remains constant, the object accelerates or decelerates, or changes its direction in the case of non-uniform motion. Often, the term "accelerated motion" is used to refer to any change in velocity, whether it's acceleration, deceleration, or change in direction.

A crucial physical quantity of this lesson is the acceleration $$\class{red}{a}$$. Acceleration is measured in units of $$\mathrm{m}/\mathrm{s}^2$$ (meters per square second) and indicates how fast the velocity of an object changes with time.

In this lesson we consider an accelerated motion of the object, where the acceleration $$\class{red}{a}$$ does not change in time, so it is constant. The motion with a constant acceleration is called uniformly accelerated motion.

In the following, we assume that the object is moving on a straight path. In other words: The movement is one-dimensional. The position $$x(t)$$ of the non-uniformly moving object on this path at time $$t$$ is given by the following formula:

The object has started accelerating (or decelerating) at the $$x_0$$ with the acceleration $$\class{red}{a}$$. This was at time $$t = 0$$. At this initial position, the object was not necessarily stationary, but already had a constant initial velocity $$\class{blue}{v_0}$$. This velocity will continue to increase or decrease during the acceleration process.

Usually the initial position during the acceleration is set to zero: $$x_0 = 0$$. Then the formula 1 simplifies to the following formula:

Of course, it is also possible that the object was stationary before acceleration. So it had no initial velocity: $$\class{blue}{v_0} = 0$$. Then formula 2 simplifies further to the following formula:

## Distance covered during accelerated motion

The object accelerating with the acceleration $$\class{red}{a}$$ covers the distance $$\class{brown}{\Delta t}$$ within the time span $$\class{green}{\Delta x}$$:

If the object started the acceleration from standstill ( $$\class{blue}{v_0} = 0$$ ), then you can calculate the distance traveled by the object as follows:

## Velocity as a function of time or position

The velocity $$\class{blue}{v(}t\class{blue}{)}$$ of an accelerating object can be calculated as a function of time $$t$$ as follows:

The velocity $$\class{blue}{v(}x\class{blue}{)}$$ of an accelerating object can also be calculated as a function of the current position $$x$$ of the object:

## Average velocity

In accelerated motion, of course, the velocity of the object changes from one point in time to the next. Or in other words: At the position $$x_1$$ at the time $$t_1$$ it has the velocity $$\class{blue}{v_1}$$ and at the position $$x_2$$ at the time $$t_2$$ the object has another velocity $$\class{blue}{v_2}$$. You can calculate a constant average velocity $$\class{blue}{\bar v}$$ that the object had between the positions $$x_2$$ and $$x_1$$:

The average velocity is the distance $$\class{green}{\Delta x} = x_2 - x_1$$ covered within the time span $$\class{brown}{\Delta t} = t_2 - t_1$$.

## Accelerated motion on a diagram

If you plot the position $$x$$ of the non-uniformly moving object as a function of time $$t$$ in a diagram, that is, the function $$x(t)$$ from Eq. 1, you get a position-time graph of an accelerated motion, as shown in Illustration 5. The position function $$x(t)$$ is a parabola opened upwards or downwards.

In the position-time diagram, two possible cases of non-uniform motion are shown, depending on whether the acceleration $$a$$ is positive or negative:

• If the acceleration is positive: $$a > 0$$, then the object accelerates. Its velocity increases. For an accelerating object, the parabola is open upwards.

• If the acceleration is negative: $$a < 0$$, then the object decelerates. Its velocity $$v$$ decreases and becomes zero at a certain time: $$v=0$$. The acceleration is also zero at this time: $$a = 0$$. In the case of a decelerating object, the parabola is open downwards.

What does the graph look like when you plot the velocity $$\class{blue}{v(}t\class{blue}{)}$$ of a nonuniformly moving object as a function of time $$t$$? Then you get a velocity-time diagram with a straight line, as shown in illustration 6. The constant slope of this straight line is the ratio of the velocity change $$\Delta v$$ to the time span $$\class{brown}{\Delta t}$$, which was needed for this velocity change. The slope therefore corresponds to the constant acceleration.