Uniformly Accelerated (Non-Uniform) Motion

by Alexander FufaeV

Table of contents

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.

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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:

Uniformly Accelerated Motion (Position, Speed, Time)

$$ \begin{align} x(t) ~=~ x_0 ~+~ \class{blue}{v_0} \, t ~+~ \frac{1}{2} \, \class{red}{a} \, t^2 \end{align} $$

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:

$$ \begin{align} x(t) ~=~ \class{blue}{v_0} \, t ~+~ \frac{1}{2} \, \class{red}{a} \, t^2 \end{align} $$

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:

$$ \begin{align} x(t) ~=~ \frac{1}{2} \, \class{red}{a} \, t^2 \end{align} $$

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 accelerated object with initial velocity

$$ \begin{align} \class{green}{\Delta x} ~=~ \class{blue}{v_0} \, (t_2 - t_1) ~+~ \frac{1}{2} \, \class{red}{a} \, (t_2 - t_1)^2 \end{align} $$

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?

$$ \begin{align} \class{green}{\Delta x} ~&=~ \class{blue}{v_0} \, (t_2 - t_1) ~+~ \frac{1}{2} \, \class{red}{a} \, (t_2 - t_1)^2 \\\\
~&=~ \class{blue}{10 \, \frac{\text m}{\text s}} \cdot \class{brown}{5 \, \text{s}} ~+~ \frac{1}{2} \cdot \class{red}{25 \, \text{m}/\text{s}^2} \cdot (\class{brown}{5 \, \text{s}} )^2 \\\\
~&=~ 362.5 \, \text{m} \end{align} $$

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:

Uniformly Accelerated Motion (Distance, Time, Acceleration)

$$ \begin{align} \class{green}{\Delta x} ~=~ \frac{1}{2} \, \class{red}{a} \, ( t_2 ~-~ t_1 )^2 \end{align} $$

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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:

Velocity as a function of time

$$ \begin{align} \class{blue}{v(}t\class{blue}{)} ~=~ {\class{blue}{v_0}} ~+~ \class{red}{a} \, t \end{align} $$

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:

Velocity as a function of position

$$ \begin{align} \class{blue}{v(}x\class{blue}{)} ~=~ \sqrt{ {\class{blue}{v_0}}^2 ~+~ 2 \class{red}{a} \,(x - x_0) } \end{align} $$

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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:

$$ \begin{align} \class{blue}{v(}x\class{blue}{)} ~&=~ \sqrt{ {\class{blue}{v_0}}^2 ~+~ 2 \class{red}{a} \,(x - x_0) } \\\\
~&=~ \sqrt{ 2 \cdot \class{red}{6.95 \, \mathrm{m}/\mathrm{s}^2} \cdot 200 \, \mathrm{m} } \\\\
~&=~ \class{blue}{52.7 \, \frac{ \mathrm{m} }{ \mathrm{s} }} \end{align} $$

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 $x_1$:

Average Velocity (Distance, Time)

$$ \begin{align} \class{blue}{\bar v} ~=~ \frac{x_2 - x_1}{t_2 - t_1} \end{align} $$

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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?

$$ \begin{align} \class{blue}{\bar v} ~&=~ \frac{ \class{green}{\Delta x} }{ \class{brown}{\Delta t} } \\\\
~&=~ \frac{ \class{green}{ \class{green}{300 \, \mathrm{km}}} }{ \class{brown}{5 \, \mathrm{h}} } \\\\
~&=~ \class{blue}{60 \, \mathrm{km} / \mathrm{h}} \end{align} $$

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.

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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.

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Velocity-Time Graph During Uniformly Accelerated Motion