Alexander Fufaev
My name is Alexander FufaeV and here I write about:

Uniformly Accelerated (Non-Uniform) Motion

Important Formula

Formula: Uniformly Accelerated Motion
What do the formula symbols mean?


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

Initial position

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

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


This is the time during which the object accelerates.


Uniform acceleration of the object. That means: Every second the current velocity of the body increases by the value \(a\).
Change of Position During Accelerated Motion WITH Initial Position and Velocity
Table of contents
  1. Important Formula
  2. Distance covered during accelerated motion
  3. Velocity as a function of time or position
  4. Average velocity
  5. Accelerated motion on a diagram

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.

Example: What does an acceleration of 5 m/s² mean?

A vehicle accelerates with \( \class{red}{a} = 5 \, \mathrm{m}/\mathrm{s}^2 \). This means: Every second the speed of the vehicle increases by 5 meters per second. For example, if at the beginning the vehicle had the speed \( 10 \, \mathrm{m}/\mathrm{s} \), then after one second it has a speed of \( 15 \, \mathrm{m}/\mathrm{s} \). After another second \( 20 \, \mathrm{m}/\mathrm{s} \) and so on. The vehicle accelerates!

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.

Change of Position During Accelerated Motion WITH Initial Position and Velocity

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 \( \class{green}{\Delta x} \):

Distance covered by an accelerating Tesla

A Tesla vehicle starts to accelerate with an initial velocity of \( 10 \, \text{m}/\text{s} \) and a uniform acceleration of \( 25 \, \text{m}/\text{s}^2 \). What distance has the vehicle covered after 5 seconds?

Here, the start time \( t_1 = 0 \) was set and the end time \( t_2 = 5 \, \mathrm{s} \), so that \( \class{brown}{\Delta t} = 5 \, \mathrm{s} \).

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:

Distance Covered by an Object During Uniformly Accelerated Motion

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:

Different Velocities at Different Positions During Accelerated Motion
Example: Speed after a certain distance

A vehicle has accelerated from 0 to \( 27.8 \, \mathrm{m}/\mathrm{s} \) (100 km/h) within 200 meters in 4 seconds. What is the speed of the vehicle after these 200 meters?

The distance covered between \(x_0\) and \(x\) is: \( x - x_0 = 200 \, \mathrm{m} \). An increase in speed from 0 to \( 27.8 \, \mathrm{m}/\mathrm{s} \) corresponds to an acceleration \( \class{red}{a} = \frac{ 27.8 \, \mathrm{m}/\mathrm{s} }{ 4 \, \mathrm{s} } = 6.95 \, \mathrm{m}/\mathrm{s}^2 \). The initial velocity in this case is \( \class{blue}{v_0} = 0 \). Inserting the values results in:

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 \(x_1\):

Position-Time Graph and Average Velocity as Slope

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

Example: Average velocity of a Tesla vehicle

A Tesla vehicle drives 300 kilometers (300,000 meters) in a total time of 5 hours (18,000 seconds). At what average velocity did the vehicle travel?

In SI units, this corresponds to an average velocity of \( \class{blue}{16.7 \, \mathrm{m} / \mathrm{s}} \).

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.

Position-Time Graph During an Accelerated Motion

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.

Velocity-Time Graph During Uniformly Accelerated Motion