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

Uniformly Accelerated (Non-Uniform) Motion

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.

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:

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0

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:

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Position of the accelerated object with starting point x0 = 0
x(t) ~=~ \class{blue}{v_0} \, t ~+~ \frac{1}{2} \, \class{red}{a} \, t^2
0

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:

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Position of the accelerated object without initial velocity
x(t) ~=~ \frac{1}{2} \, \class{red}{a} \, t^2
0

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

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Distance covered by an accelerated object with initial velocity
\class{green}{\Delta x} ~=~ \class{blue}{v_0} \, (t_2 - t_1) ~+~ \frac{1}{2} \, \class{red}{a} \, (t_2 - t_1)^2
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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?

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Calculation example: Accelerated motion with initial velocity
\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}
0

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:

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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:

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Velocity as a function of time
\class{blue}{v(}t\class{blue}{)} ~=~ {\class{blue}{v_0}} ~+~ \class{red}{a} \, t
0

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:

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Velocity as a function of position
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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:

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How to calculate: Speed after covered distance
\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} }}
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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 \(x_1\):

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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?

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How to calculate: Average velocity during accelerated motion
\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}}
0

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.

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.