Uniform Rectilinear Motion

Lessons

Senior 1 · Physics

Uniform Rectilinear Motion

5 lessons

Lesson 4 of 5

Uniform Rectilinear Motion: Equations and Graphs

25 min

The needle that does not move

On a long, straight, level stretch of the road east of Kigali, a bus driver holds the speedometer needle steady at $60\,\mathrm{km\,h^{-1}}$ for 5 minutes. The bus is moving in a straight line and its velocity is not changing. How far does it travel in those 5 minutes, and what would a graph of its journey look like?

Definition

Uniform rectilinear motion (URM) is motion in a straight line at constant velocity. In URM:

the trajectory is a straight line;
the speed and the direction never change;
equal distances are covered in equal times;
the acceleration is zero: $a = 0$ .

Experiment: a toy car at steady speed

Figure 4.1: A battery-powered toy car runs along a straight track marked every 0.5 m.

Switch the car on a little behind the zero mark so that it is already at a steady speed when it passes 0. One pupil starts a stopwatch as it passes the zero mark and calls out every 2 s; at each call, partners note the car's position.

Time $t$ (s)	0	2	4	6	8	10
Distance $d$ (m)	0	0.50	1.00	1.49	2.01	2.50

In every 2 s the car covers about 0.50 m: equal distances in equal times. Its speed is $\frac{0.50\,\mathrm{m}}{2\,\mathrm{s}} = 0.25\,\mathrm{m\,s^{-1}}$ , and it does not change. The small differences (1.49 m, 2.01 m) are measuring errors.

Important

Safety. Run the track along a wall, away from doorways, so that nobody steps on the car. Switch the car off before picking it up.

The equations of uniform rectilinear motion

Because the velocity $v$ is constant, the distance covered grows steadily with time:

d = v\,t

If the object starts at position $x_{0}$ (measured from the reference point) when $t = 0$ , its position at time $t$ is

x = x_0 + v\,t

and at all times $v = \text{constant}$ and $a = 0$ .

Worked example 1: the bus in the opening

How far does a bus moving at a steady $60\,\mathrm{km\,h^{-1}}$ travel in 5.0 minutes?

Change the time to hours: $5.0\,\mathrm{min} = \frac{5.0}{60}\,\mathrm{h}$ .

d = v\,t = 60\,\mathrm{km\,h^{-1}} \times \frac{5.0}{60}\,\mathrm{h} = 5.0\,\mathrm{km}

Worked example 2: finding a position

At noon a bus is 12 km east of Kigali, moving east at a steady $60\,\mathrm{km\,h^{-1}}$ on a straight road. Where is it at 12:30?

$x_0 = 12\,\mathrm{km}$ , $v = 60\,\mathrm{km\,h^{-1}}$ , $t = 0.50\,\mathrm{h}$ :

x = x_0 + v\,t = 12\,\mathrm{km} + 60\,\mathrm{km\,h^{-1}} \times 0.50\,\mathrm{h} = 12 + 30 = 42\,\mathrm{km}

It is 42 km east of Kigali.

The distance–time graph

Plot the toy car's results with time on the horizontal axis and distance on the vertical axis. The points lie on a straight line through the origin.

Figure 4.2: Distance against time. Line A is the toy car. The steeper line B is a faster object, and the flat line C is an object standing still 1.0 m from the start.

The gradient (steepness) of a distance–time graph is the speed:

v = \text{gradient} = \frac{\Delta d}{\Delta t} = \frac{1.0\,\mathrm{m}}{4\,\mathrm{s}} = 0.25\,\mathrm{m\,s^{-1}}

A steeper straight line means a greater constant speed.
A horizontal line means the object is at rest: its distance does not change.
A curved line would mean the speed is changing, so the motion is not uniform.

The speed–time graph

For uniform motion the speed is the same at every moment, so a speed–time graph is a horizontal line. The area under a speed–time graph is the distance travelled: for a rectangle,

d = \text{area} = v \times t = 0.25\,\mathrm{m\,s^{-1}} \times 10\,\mathrm{s} = 2.5\,\mathrm{m}

Figure 4.3: Speed against time. The toy car's graph is flat; the shaded area is the 2.5 m it travels in 10 s. The dashed sloping line is an object speeding up, which is not uniform motion.

Worked example 3: reading a graph

(a) A cyclist's distance–time graph is a straight line from $(0\,\mathrm{s}, 0\,\mathrm{m})$ to $(60\,\mathrm{s}, 300\,\mathrm{m})$ . Find her speed. (b) A car's speed–time graph is flat at $20\,\mathrm{m\,s^{-1}}$ for 45 s. How far does it travel?

(a) $v = \frac{\Delta d}{\Delta t} = \frac{300\,\mathrm{m}}{60\,\mathrm{s}} = 5.0\,\mathrm{m\,s^{-1}}$

(b) $d = \text{area} = 20\,\mathrm{m\,s^{-1}} \times 45\,\mathrm{s} = 900\,\mathrm{m}$

Warning

Read the axes first. A horizontal line means "at rest" on a distance–time graph but "constant speed" on a speed–time graph. And a graph is not a picture of the road: a steep line on a distance–time graph means fast motion, not a steep hill.

Tip

Distance–time: gradient gives speed. Speed–time: area gives distance. Say it with the axes: "up over along" for the gradient, "height times width" for the area.

Uniform motion at work: conveyor belts

Tea is one of Rwanda's main exports. In a tea factory, freshly picked leaves ride along conveyor belts that move at a steady speed, so each leaf spends exactly the right time in each stage of processing. If a belt 12 m long moves at $0.40\,\mathrm{m\,s^{-1}}$ , a leaf takes $t = \frac{d}{v} = \frac{12}{0.40} = 30\,\mathrm{s}$ to ride its full length. Bottling lines in drinks factories and airport luggage belts use the same idea.

Distance-time graphs (Cognito). Watch how the gradient gives the speed, and what flat and curved lines mean.

✦

Practice

Which of these are examples of uniform rectilinear motion? (a) a car at a steady $50\,\mathrm{km\,h^{-1}}$ on a straight road; (b) a moto going round a roundabout at a steady speed; (c) a ball rolling faster and faster down a slope.
A bus moves at a steady $15\,\mathrm{m\,s^{-1}}$ on a straight road. How far does it go in 2.0 minutes?
A cyclist is 3.0 km from a school at $t = 0$ and rides away from it along a straight road at a steady $6.0\,\mathrm{m\,s^{-1}}$ . How far from the school is she after 10 minutes?
A distance–time graph is a straight line from $(0, 0)$ to $(8\,\mathrm{s}, 40\,\mathrm{m})$ . Find the speed.
A speed–time graph is a horizontal line at $12\,\mathrm{m\,s^{-1}}$ from 0 to 25 s. Find the distance travelled.
Describe the distance–time graph of a person who walks at a steady speed for 5 minutes, stands still for 2 minutes, then walks on at a greater steady speed.
The results for a trolley are: $t = 0, 1, 2, 3, 4\,\mathrm{s}$ and $d = 0, 0.3, 0.6, 1.2, 1.5\,\mathrm{m}$ . A pupil says the motion is uniform. Is she right? Explain.

Show answersHide answers

Only (a). In (b) the direction keeps changing, and in (c) the speed keeps changing.
$d = vt = 15 \times 120 = 1800\,\mathrm{m} = 1.8\,\mathrm{km}$
$vt = 6.0 \times 600 = 3600\,\mathrm{m} = 3.6\,\mathrm{km}$ ; $x = x_0 + vt = 3.0 + 3.6 = 6.6\,\mathrm{km}$
$v = \frac{40\,\mathrm{m}}{8\,\mathrm{s}} = 5.0\,\mathrm{m\,s^{-1}}$
$d = 12 \times 25 = 300\,\mathrm{m}$
A straight sloping line, then a horizontal line (at rest), then a steeper straight sloping line.
No. The trolley covers 0.3 m in each of the first two seconds, then 0.6 m, then 0.3 m. The distances in equal times are not equal, so its speed was not constant (or one reading is wrong and should be repeated).

Key points

Uniform rectilinear motion: a straight path, constant velocity, equal distances in equal times, $a = 0$ .
Equations: $d = v t$ and $x = x_{0} + v t$ .
Distance–time graph: a straight line whose gradient is the speed; flat means at rest.
Speed–time graph: a horizontal line; the area under it is the distance.

Sources and credits

2.8: Graphical Analysis of One-Dimensional Motion, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Abasoromyi bicyayi.jpg by Wood Wildlife Photography, Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: GCSE Physics - Distance-Time Graphs, Cognito, embedded from YouTube.

The worked examples, diagrams, Rwandan applications and practice exercises are written by Ganzaa.

1The needle that does not move

2Experiment: a toy car at steady speed

3The equations of uniform rectilinear motion

4The distance–time graph

5The speed–time graph

6Uniform motion at work: conveyor belts