Lesson 2 of 5Kinematics

Unit 1 · Lesson 2 of 5

Uniform Motion and Motion Graphs

Constant-velocity motion, and how to pull velocity out of the slope of a position-time graph and displacement out of the area under a velocity-time graph.

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Key things to remember

1Uniform motion means constant velocity: equal displacements in equal times, same direction throughout.
2 $v = \frac{Δ d}{Δ t}$ , rearranged as $Δ d = v Δ t$ or $Δ t = \frac{Δ d}{v}$ .
3Slope of a position-time graph is velocity. Slope of a velocity-time graph is acceleration.
4Area under a velocity-time graph is displacement; area below the axis counts as negative.
5Slope of a chord gives average velocity; slope of a tangent gives instantaneous velocity.
6A horizontal line on a $d$ - $t$ graph means at rest; a horizontal line on a $v$ - $t$ graph means constant velocity.

Uniform motion means constant velocity: equal displacements in equal time intervals, in an unchanging direction. It is the simplest kind of motion there is, and it is worth taking seriously because graphs of uniform motion are where the two most useful graph skills in the whole course — reading a slope and reading an area — are easiest to see.

v = \frac{Δ d}{Δ t} ⟹ Δ d = v Δ t Δ t = \frac{Δ d}{v}

One relationship, rearranged three ways. Which form you use depends only on what the question is missing.

Average or instantaneous?

For genuinely uniform motion there is no difference — the velocity at every instant equals the average. The distinction only starts to matter in the next topic, once the velocity is allowed to change.

Position-time graphs

Fig 1The same journey shown three ways. Position-time, velocity-time and acceleration-time graphs each answer a different question.

On a position-time ( $d$ - $t$ ) graph, time runs along the horizontal axis and position up the vertical one. The slope is the velocity, because slope is rise over run, and here that is $\frac{Δ d}{Δ t}$ — the definition of velocity. This is the single most useful sentence in the unit.

Shape of the $d$ - $t$ graph	What the object is doing
Horizontal line	At rest — position is not changing
Straight line, positive slope	Uniform motion in the positive direction
Straight line, negative slope	Uniform motion in the negative direction
Steeper straight line	Same kind of motion, but faster
Curve bending upward	Speeding up (slope is increasing)
Curve flattening off	Slowing down (slope is decreasing)

A graph is not a picture of the path

d

t

graph that rises and then falls does not mean the object went up a hill and came down. It means the object moved in the positive direction and then came back toward the origin. The vertical axis is position along a line, not height — unless the question has specifically defined 'position' as height.

Velocity-time graphs

Fig 2On a velocity-time graph the slope gives acceleration and the shaded area gives displacement.

On a velocity-time ( $v$ - $t$ ) graph the area between the line and the time axis is the displacement. For uniform motion the shape is a rectangle, and its area is height $\times$ width $= v \times Δ t$ , which is exactly $Δ d$ . Area below the time axis counts as negative displacement, because a negative velocity moves the object backwards.

The two rules worth memorizing

Slope of a

d

t

graph is velocity. Slope of a

v

t

graph is acceleration. Area under a

v

t

graph is displacement. Area under an

a

t

graph is the change in velocity. Going up the list means taking slopes; going down means taking areas.

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Set an initial velocity and an acceleration, then watch all three graphs draw themselves in step with the object.

Reading a graph like a physicist

Check the axes first. A velocity-time graph and a position-time graph of the same motion look nothing alike, and mistaking one for the other makes every subsequent answer wrong.
Use two widely separated points when you calculate a slope. Picking two points close together magnifies your reading error.
Carry the units through the slope. Metres divided by seconds gives m/s, which is how you know the slope is a velocity rather than something else.
Split an area into rectangles and triangles. Any $v$ - $t$ graph in this course decomposes into those two shapes, and the area of a triangle is $\frac{1}{2} bh$ .

One last piece of vocabulary. The slope of a chord between two points on a curved $d$ - $t$ graph gives the average velocity over that interval. The slope of a tangent drawn at a single point gives the instantaneous velocity at that moment. Drawing a good tangent by hand is a skill worth practising: extend the line well past the point of contact so the rise and the run are both large enough to read accurately.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

A position-time graph passes through

(2.0 s, 12 m)

and

(8.0 s, 42 m)

as a straight line. Find the velocity.

Example 2

A cyclist travels at a constant

6.5 m/s [E]

for

2.0 min

. How far does she travel, and what does the velocity-time graph look like?

Example 3

A velocity-time graph shows an object at a constant

+ 4.0 m/s

for the first

5.0 s

, then a constant

- 2.0 m/s

for the next

5.0 s

. Find the total displacement and the total distance.

Where marks get lost

Reading a $d$ - $t$ graph as a picture of the journey, so a rising then falling line is described as going over a hill instead of moving out and coming back.
Taking the value on a $d$ - $t$ graph as the velocity. The value is a position; the slope is the velocity.
Forgetting that area below the time axis on a $v$ - $t$ graph is negative displacement, so a there-and-back trip comes out with far too large a displacement.
Calculating a slope from two points that are almost on top of each other, which turns a small reading error into a large slope error.
Leaving time in minutes or hours when the velocity is in m/s. Convert before substituting, every time.
Adding magnitudes to get displacement, or adding signed values to get distance — they are computed differently and questions often ask for both.

Flashcards 8

Uniform motion

Slope of a position-time graph

Slope of a velocity-time graph

Area under a velocity-time graph

+ 4 more

Practice

A position-time graph is a horizontal line. What is the object doing?

A straight-line

d

t

graph goes from

(1.0 s, 5.0 m)

(6.0 s, 30.0 m)

. What is the velocity in m/s?

m/s

A car drives at a constant

22 m/s

for

3.0 min

. How far does it travel, in metres?

On a

v

t

graph an object holds

+ 8.0 m/s

for

4.0 s

, then

- 3.0 m/s

for

4.0 s

. What is the total displacement in metres?

Which quantity is given by the area under a velocity-time graph?

What do you call the line drawn touching a curve at a single point, whose slope gives the instantaneous velocity?

d

t

graph curves upward, getting steeper as time goes on. What is happening?

Order these

d

t

graph slopes from the fastest motion to the slowest, ignoring direction.

1A line rising $30 m$ in $3.0 s$
2A horizontal line
3A line rising $10 m$ in $4.0 s$
4A steep line rising $40 m$ in $2.0 s$

Match each graph feature to what it tells you.

Slope of a

d

t

graph

Slope of a

v

t

graph

Area under a

v

t

graph

Area under an

a

t

graph

A runner covers

150 m

at a constant

5.0 m/s

. How many seconds does it take?

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