Linear motion

Linear motion is how an object moves in a straight line. To describe these motions, we need to know some technical terms like displacement, distance, velocity, speed and acceleration, and how to represent how they change over time. Use this resource to learn about the terms and their relationships to one another.

Before you begin, make sure you understand the basics of scalar and vector quantities.

Describing linear motion

In linear motion, these quantities help us describe not just how far something moves, but how fast it moves and how its motion changes over time. Let's define each quantity and show how they are connected when an object travels in a straight line.

Displacement \(\left(x\right)\)

When an object is moved from one point to another, it is said to be displaced. If we start in Melbourne and travel to Dandenong, we can say that we have been displaced \(32\,\text{km}\) from Melbourne. To then define our position, we need to specify a direction: we are \(32\,\text{km}\) South East of Melbourne. That is, we give a magnitude \(\left(32\,\text{km}\right)\) and a direction (South East). This makes position, or displacement, a vector quantity. Displacement is often used to show the position of a something relative to a specific point.

In physics, we assign a positive sign to a specific direction. For example, if East is the positive direction, then travelling \(2\,\text{km}\) East is \(+2\,\text{km}\). This would make travelling \(2\,\text{km}\) West, \(-2\,\text{km}\).

Distance \(\left(d\right)\)

Distance is the magnitude (or size) of the displacement. It is the length of the path taken by something. It has no direction, so it is a scalar quantity. For example, you can say your distance is \(60\,\text{km}\) away, but your displacement would be \(60\,\text{km}\) in a direction, such as \(60\,\text{km}\) West.

Velocity \(\left(v\right)\)

Velocity is the rate of change of displacement, \(x\). That is, for an object moving at constant velocity:

\[v=\frac{x}{t}\]

where \(t\) is the time taken.

Since displacement is a vector quantity, so is velocity.

The units of velocity depend on the units of displacement and time. If displacement is in metres \(\left(\text{m}\right)\) and time is in seconds \(\left(\text{s}\right)\), the unit of velocity is metres per second \((\text{m/s}\) or \(\text{m}\,\text{s}^{-1})\).

Speed

The speed of an object is the magnitude of its velocity. It can also be calculated by:

\[\text{speed}=\frac{d}{t}\]

where \(t\) is the time taken.

It has no direction, making it a scalar quantity.

Acceleration \(\left(a\right)\)

Acceleration is the rate of change of velocity:

\[a=\frac{v}{t}\]

If \(v\) is in metres per second, the unit of acceleration is metres per second per second: \(\text{m/s}^{2}\) or \(\text{m}\,\text{s}^{-2}\). Since velocity is a vector quantity, so is acceleration. If acceleration is constant, we use the expression 'the acceleration is uniform'.

Representing linear motion

We can describe linear motion using graphs that show how displacement, velocity and acceleration change over time.

Distance vs time graphs

Let’s suppose a person, starting from rest, runs a distance \(d\) of \(12\,\text{m}\) in a time \(t\) of \(2\,\text{s}\). Assuming that their speed is constant, the distance–time graph would look like this:

The gradient of a distance (a scalar quantity) versus time graph gives the speed (also a scalar quantity). In this case, the speed is given by:

\[\begin{align*} \text{speed} & = \frac{d}{t} \\[6pt]
& = \frac{12}{2}\\[6pt]
& = 6\text{ m/s}
\end{align*}\]

Displacement vs time graphs

Consider the displacement–time graph:

The gradient of a displacement (a vector quantity) versus time graph gives velocity (also a vector quantity). Expand the accordions to learn what is happening between each point on the graph.

Velocity vs time graphs

The velocity–time graph shows a car accelerating uniformly from rest to \(60\,\text{m/s}\) in \(20\,\text{s}\), then travels at a constant velocity of \(60\,\text{m/s}\) for the next \(20\,\text{s}\), then decelerates uniformly to rest in \(20\,\text{s}\). The total journey takes \(60\,\text{s}\).

From A to B, the object accelerates from \(0\,\text{m/s}\) to \(60\,\text{m/s}\). From B to C, the object moves at a constant velocity of \(60\,\text{m/s}\) and from C to D, the object slows to a stop from \(60\,\text{m/s}\).

Let's look at some features of the graph in detail: the area under the graph and its gradient.

Area under velocity–time graph

The area is the product of velocity and time, so the units are \(\text{m}/\text{s}\,\times \text{s}=\text{m}\). That is, the area under the graph gives displacement.

The area under a velocity–time graph gives displacement.

The area under a speed–time graph gives distance.

Consider the velocity–time graph again. The area has been divided into three regions: \(1\), \(2\) and \(3\). Let the area of each region be \(A_{1}\), \(A_{2}\) and \(A_{3}\), respectively. We have:

\[\begin{align*} A_{1} & = \frac{1}{2}\times b\times h \\[6pt]
& = \frac{1}{2}\times20\times60 \\[6pt]
& = 600\,\text{m} \end{align*}\] \[\begin{align*} A_{2} & = b\times h \\[6pt]
& = 20\times60\\[6pt]
& = 1200\,\text{m} \end{align*}\] \[\begin{align*} A_{3} & = \frac{1}{2}\times b\times h \\[6pt]
& = \frac{1}{2}\times20\times60\\[6pt]
& = 600\,\text{m} \end{align*}\]

The total displacement is therefore: \(A_{1}+A_{2}+A_{3}=600+1200+600=2400\,\text{m}\).

Check out the further resources at the end of this page if you need a refresher on calculating the area of triangles and rectangles.

Gradient of a velocity–time graph

The gradient (or slope) of a velocity–time graph gives the acceleration of the object. It is calculated using:

\[m=\frac{\text{rise}}{\text{run}}\]

Expand the accordions to learn what is happening between each point on the graph.

What the gradient and area under each graph represents is summarised in the following table.

Example 1 – sketching and describing linear motion graphs

A car accelerates from a stationary position. The rate of acceleration is constant. Sketch graphs of the acceleration versus time, the velocity versus time and the displacement versus time.

Since the acceleration is constant, the acceleration–time graph will simply be a horizontal line.

As the car is accelerating, this means that its velocity is increasing over time. The velocity–time graph is a linear graph with a gradient equal to acceleration.

Displacement is the total distance travelled, so we get it by looking at how fast the car is going during each time interval. At the start, the car is moving slowly, so it does not cover much distance in the first few seconds. Later, it is moving faster, so it covers more distance in each second than before.

This means the displacement does not increase at a steady rate – it increases more and more quickly as time goes on, creating a curve.

The gradient (slope) at any point on the curve is the velocity at that time, or the instantaneous velocity. You can find the approximate value of the gradient by drawing a tangent at the point of interest and then calculating the gradient of the tangent.

Exercise – sketching and describing linear motion graphs

1. The graph shows the position (displacement) of a dancer moving in a straight line across a stage. The dancer’s movements are designated by sections A to D.
   

   

   1. Determine the dancer's starting position.
   2. Determine in which section (A—D) the dancer is at rest.
   3. Determine in which section the dancer is moving in a positive direction.
   4. Determine in which sections the dancer is moving with a negative velocity.

2. The graph represents the straight line motion of a radio-controlled toy car.
   

   

   1. Describe, in words, the motion of the car.
   2. Determine the position of the car at:
      1. \(2\,\text{s}\)
      2. \(4\,\text{s}\)
      3. \(6\,\text{s}\)
      4. \(10\,\text{s}\).
   3. Determine at what time the car returned to its starting point.
   4. Calculate the velocity of the car:
      1. during the first \(2\,\text{s}\)
      2. at \(3\,\text{s}\)
      3. from \(4\,\text{s}\) to \(6\,\text{s}\)
      4. at \(8\,\text{s}\)
      5. from \(8\,\text{s}\) to \(9\,\text{s}\).
   5. During its \(10\,\text{s}\) motion, calculate the car's:
      1. distance travelled
      2. displacement.

3. The following position–time graph is for a cyclist travelling along a straight road.
   

   

   1. Describe, in words, the motion of the cyclist.
   2. Calculate the velocity of the cyclist during the first \(30\,\text{s}\).
   3. Calculate the cyclist’s velocity during the final \(10\,\text{s}\).
   4. Estimate the cyclist’s instantaneous velocity at \(35\,\text{s}\).
   5. Determine the average velocity of the cyclist between \(30\,\text{s}\) and \(40\,\text{s}\).

4. Consider the velocity–time graphs, A to E.
   

   

   Identify which velocity-time graph best represents the motion of:
   1. a car coming to a stop at a traffic light
   2. a swimmer moving at a constant speed
   3. a cyclist accelerating from rest with constant acceleration
   4. a car accelerating from rest and changing through its gears.

5. The following graph shows the motion of a dog running along a footpath in a northerly direction.
   

   

   1. Describe the motion of the dog during each section of the graph (A to F).
   2. Calculate the displacement of the dog after:
      1. \(2\,\text{s}\)
      2. \(7\,\text{s}\)
      3. \(10\,\text{s}\)
   3. Plot a position–time graph for the dog’s motion.

6. The straight line motion of a train is shown in the following velocity–time graph.
   

   

   1. Determine how long the train takes to reach its cruising speed.
   2. Calculate the acceleration of the train \(10\,\text{s}\) after starting. (Hint: You can only approximate this. Take a tangent to the graph at \(10\,\text{s}\) and calculate its gradient.)
   3. Calculate the acceleration of the train \(40\,\text{s}\) after starting
   4. Calculate the approximate displacement of the train after \(120\,\text{s}\).

7. The velocity–time graphs for a bus and a bicycle travelling along the same straight stretch of road are shown in the graph. The bus is initially at rest and starts moving as the bicycle passes it.
   

   

   1. Calculate the initial acceleration of the bus.
   2. Determine at what time the bus first starts gaining ground on the bicycle.
   3. Determine at what time the bus overtakes the bicycle.
   4. Determine how far the bicycle has traveled before the bus catches up to it.
   5. Determine the average velocity of the bus during the first \(8\,\text{s}\).
   6. Draw an acceleration–time graph for the bus.
   7. Use your acceleration–time graph to determine the change in velocity of the bus over the first \(8\,\text{s}\).

Images on this page by RMIT, licensed under CC BY-NC 4.0