Uniformly Accelerated (NonUniform) Motion
Important Formula
What do the formula symbols mean?
Position
$$ x(t) $$ Unit $$ \mathrm{m} $$Initial position
$$ x_0 $$ Unit $$ \mathrm{m} $$Initial velocity
$$ \class{blue}{v_0} $$ Unit $$ \frac{\mathrm m}{\mathrm s} $$Time
$$ t $$ Unit $$ \mathrm{s} $$Acceleration
$$ \class{red}{\boldsymbol a} $$ Unit $$ \frac{\mathrm m}{\mathrm{s}^2} $$Table of contents
Nonuniform 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 nonuniform 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 onedimensional. The position \( x(t) \) of the nonuniformly moving object on this path at time \( t \) is given by the following formula:
The object has started accelerating (or decelerating) at the
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 nonuniformly moving object as a function of time \(t\) in a diagram, that is, the function \(x(t)\) from Eq. 1
, you get a positiontime graph of an accelerated motion, as shown in Illustration 5
. The position function \( x(t) \) is a parabola opened upwards or downwards.
In the positiontime diagram, two possible cases of nonuniform 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 velocitytime 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.