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.

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:

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

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:

Position of the accelerated object with starting point x0 = 0

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

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:

Position of the accelerated object without initial velocity

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

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:

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

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

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:

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 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.

+ Perfect for high school and undergraduate physics students + Contains over 500 illustrated formulas on just 140 pages + Contains tables with examples and measured constants + Easy for everyone because without vectors and integrals