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Velocity-time graphs

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Summary

Velocity-time graphs

In a nutshell

Velocity-time graphs display the movement of an object over time. In these graphs, velocity ( $v$ ) is plotted on the y-axis and time ( $t$ ) is plotted on the x-axis. Velocity is the speed of an object in a given direction. When the object is moving backwards, the velocity is negative. This means that the line on a velocity-time graph can cross the x-axis.

Equations

description	equation
Acceleration of an object at a given time.	$a = \dfrac{v}{t}$
Displacement of an object.	$s=v\times t$

Variable definitions

quantity name	symbol	unit name	unit
$Velocity$	$v$	$Metres\ per\ second$	$ms^{-1}$
$Displacement$	$s$	$Metre$	$m$
$Time$	$t$	$Seconds$	$s$
$Acceleration$	$a$	$Metres\ per\ second\ squared$	$ms^{-2}$

Features of the graph

Maths; Constant acceleration; KS5 Year 12; Velocity-time graphs

A positive gradient indicates a constant acceleration.
A horizontal line indicates no acceleration, and means the object is moving at a constant velocity.
A negative gradient shows negative acceleration, which is described as deceleration.
If the line touches the x-axis, it means the object is stationary - the velocity is $0\ ms^{-1}$ .

Calculating acceleration and displacement

Using the velocity-time graph, you can work out both acceleration and displacement.

The acceleration of an object at any point is the gradient of the slope:

$\boxed{a = \dfrac{v}{t}}$

The displacement of an object is equal to the area under the curve:

$\boxed{s=v\times t}$

This equation can also be used to work out the total distance travelled, and if the graph is positive, then the displacement of the object will be equal to the distance travelled. However, the difference occurs when the velocity becomes negative and crosses the x-axis. When this happens, the displacement is equal to the area under the graph above the x-axis, minus the absolute value of the area above the graph below the x-axis. The distance, however, is equal to the area under the graph above the x-axis, plus the absolute value of the area above the graph below the x-axis.

Note: The 'absolute' value refers to the scalar value of a number - the number without a direction. This means an absolute number can never be negative.

Example 1

A man drives from his house towards his office. This graph represents the first 3 hours of his 5 hour commute.

a) Work out his initial acceleration before he reaches a constant velocity. Give your answer in $ms^{-1}$ .

b) Work out his total displacement over the 3 hours. Give your answer in $m$ .

a: The man reaches a constant velocity when the graph becomes horizontal, after 1 hour. His velocity at this hour mark is $30\ kmh^{-1}$ . The question asks for the answer in $ms^{-2}$ , so convert this velocity and time:

$30\ kmh^{-1}= 8.333\ ms^{-1}\ and \ 1\ hour=3600\ seconds$ 

Work out the man's initial acceleration:

$a=\dfrac{v}{t}$

$a = \dfrac{8.333}{3600}=0.0023\ ms^{-2}$

His initial acceleration was $\underline{0.0023 \ ms^{-2}}$ .

b: The total displacement is equal to the area under the graph. This can be worked out by splitting the graph into four sections: $0 \ to\ 1\ hour,\ 1\ to\ 1.5\ hours,\ 1.5\ to\ 2\ hours\ and\ 2\ to\ 3\ hours.$

The displacement from $0\ to\ 1\ hour$ :

$area\ of\ a\ triangle = \dfrac{1}{2}(b\times h)$

$s=\dfrac{1}{2}(30\times 1)=15\ km$ 

The displacement from $1\ to\ 1.5\ hours$ :

$s=v\times t$

$s=0.5\times 30 = 15\ km$ 

The displacement from $1.5\ to\ 2\ hours$ :

$area\ of\ a\ trapezium = \dfrac{1}{2}(a + b )\times h$ 

$s=\dfrac{1}{2}(30+20)\times 0.5 = 12.5\ km$ 

The displacement from $2\ to\ 3\ hours$ :

$area\ of\ a\ trapezium = \dfrac{1}{2}(a + b )\times h$

 $s=\dfrac{1}{2}(20+50)\times 1 = 35\ km$ 

Total displacement:

$s=15+15+12.5+35=77.5\ km=77,500\ m$

His total displacement over $3$ hours was $\underline{77 \ 500 \ m}$ .

Learn with Basics

Length:

Unit 1

Motion and speed

Unit 2

Distance-time and velocity-time graphs

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Unit 3