Uniform (unaccelerated) Motion in Physics
Table of contents
Imagine you are standing on the side of the road watching an object, for example a car, pass you at a constant speed on a straight road. With physics you can easily predict WHERE the object will be, after 10 seconds, 20 seconds or 100 seconds. For this purpose there is the following law:
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You started to observe the car at the position \(x_0\). \(x_0\) is the initial position of the object. Usually the initial position is set to zero: \(x_0 = 0\).
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With a velocimeter you found out that the object moves with a constant velocity \( \class{blue}{v} \). For example with 10 meters per second: \( \class{blue}{v} = 10 \, \mathrm{m}/\mathrm{s} \). Every second the object travels 10 meters.
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\( x(t) \) is the position of the object at the time \( t \). For example, the position \(x\) after 10 seconds we write as \( x(10) \).
A vehicle is traveling at 10 meters per second. What is its position \( x \) after 5 seconds? Multiply the velocity \( \class{blue}{v} \) by the time \( t = 5 \, \mathrm{s} \).
~&=~ \class{blue}{10 \, \frac{\mathrm{m}}{\mathrm{s}}} \cdot 5 \, \mathrm{s} \\
~&=~ 50 \, \mathrm{m}
\end{align} $$
The object is 50 meters away from the initial position x₀ = 0 after 5 seconds.
Note that only if the initial position \(x_0\) is zero, \(x(t)\) yields the distance the object will travel after a given time \( t \). If the initial position is not zero, \(x(t)\) is NOT the distance traveled!
Distance traveled
How do we determine the covered distance of the object within a given time? The covered distance is written as \( \class{red}{\Delta x} \). This is the distance of two positions \( x_1 \) and \( x_2 \) of the object at two different times \( t_1 \) and \( t_2 \).\( x_2 - x_1 \) is the distance covered \( \class{red}{\Delta x} \) by the object within the time \( t_2 - t_1 \). The time difference \( t_2 - t_1 \) is abbreviated as \( \Delta t \).
You get the distance covered \( \class{red}{\Delta x} \), that is the distance between the positions \( x_2 \) and \( x_1 \), by multiplying the speed of the object by the time \( \Delta t \):
The object moves at 10 meters per second.
~&=~ \class{blue}{10 \, \frac{\mathrm{m}}{\mathrm{s}}} \cdot 7 \, \mathrm{s} \\
~&=~ \class{red}{70 \, \mathrm{m}}
\end{align} $$
Uniform motion illustrated in a diagram
You can illustrate the position \(x(t)\) of an object at time \(t\) in a diagram. The position \(x\) is plotted on the vertical axis and the time \(t\) on the horizontal axis. This results in a position-time graph, as shown in Illustration 3.
The position-time graph during a uniform movement is a straight line. The slope of the straight line is the ratio of the traveled distance \( \class{green}{\Delta x} \) to the time \( \class{brown}{\Delta t} \) required for it. The slope therefore corresponds to the speed of the object! The steeper the line, the faster the object moves.
On the other hand, if you plot the velocity \( v \) on the vertical axis and the time \(t\) on the horizontal axis, you get a velocity-time graph of uniform motion, as shown in Illustration 4.
Since the velocity does NOT change with time in a uniform motion, the result is a horizontal straight line. This means: At any time \(t\), for example at time \(t_1\) or at time \(t_2\) on the horizontal axis, the velocity on the vertical axis always has the same value. This horizontal straight line is a constant velocity function \( \class{blue}{v(}t\class{blue}{)} \).
In the next lesson, we will look at a non-uniform motion, that is, an accelerated motion.
