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Summary

Displacement-time graphs

In a nutshell

Displacement-time graphs shows the displacement of an object as it moves, in a straight line, towards or away from an origin point. The graph displays displacement in metres on the y-axis and time in seconds on the x-axis. Displacement-time graphs can go below the x-axis, unlike distance-time graphs.

In graphs and equations, displacement is symbolised with an $s$ , measured in metres.

Equations

description	equation
Velocity of an object at a particular displacement over a certain time.	$v=\dfrac{s}{t}$
Average velocity of an object.	$v=\dfrac{s}{t}$
Average speed of an object.	$s=\dfrac{d}{t}$

Variable definitions

QUANTITY NAME	SYMBOL	UNIT NAME	UNIT
$Velocity$	$v$	$metres\ per\ second$	$ms^{-1}$
$Displacement$	$s$	$metre$	$m$
$Time$	$t$	$seconds$	$s$
$Speed$	$s$	$metres\ per\ second$	$ms^{-1}$

Features of the graph

Displacement is a vector which shows the direction an object moves in, and it is negative when an object is moving backwards from the point of origin. In the graph, the gradient of a slope indicates the velocity of the object.

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

The gradient of a displacement-time graph can indicate a number of other things:

A positive gradient means that the object is travelling forwards.
A negative gradient means that the object is moving backwards.
A horizontal line shows that the object is stationary.
A straight diagonal line shows that the object is moving at a constant velocity.
A curved line shows that the object is either accelerating or decelerating.

Any time that the line touches the x-axis, the object is at the origin point. Whenever the line moves toward the x-axis, it means the object is coming back towards the origin point. Whenever the line goes below the x-axis, it means the object is moving away from the origin point again, but this time in the opposite/backward direction.

Finding velocity and speed

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

At any point on the graph, the velocity can be worked out using the gradient:

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

The velocity of an object can be negative, however the speed cannot be. The speed is the absolute value of the velocity. This means if the velocity is $-8\ ms^{-1}$ , then the speed is $8\ ms^{-1}$ . The negative velocity tells us that the object is moving backwards.

Working out average velocity and average speed is slightly different.

For average velocity, you take the value of the final displacement of the object from the point of origin, and divide that by the total time:

$\boxed{average\ velocity=\dfrac{displacement\ from\ origin}{time\ taken}}$

For average speed, you use the total distance travelled over the whole journey and divide that by the total time:

$\boxed{average\ speed = \dfrac{total\ distance\ travelled}{time\ taken}}$

Example 1

This graph shows Alex's sprints during training for 8 seconds.

a: Jess looks at the graph and says Alex was moving at a constant velocity between points C and D.

b: Work out his velocity between points A and B.

c: Work out his average speed and average velocity for the whole sprint.

a:The graph shows a horizontal line between points C and D. This means Alex had zero displacement during this time, so he was not moving. His velocity was zero.

No, Jess is not correct.

b: Velocity can be worked out using the gradient. From point A to B, Alex moves $2m$ in $3 seconds$ . His velocity is:

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

$v=\dfrac{2}{3}\ ms^{-1}$ 

His velocity from point $A$ to $B$ is $\underline{\dfrac 2 3 \ m.s^{-1}}$ .

c: Average speed:

$average\ speed=\dfrac{total\ distance\ trav elled}{time\ taken}$

His total distance travelled was $2\ m+5\ m+3\ m=10\ m$ . His total time taken was $8s$ . Therefore:

$average\ speed=\dfrac{10}{8}=\dfrac{5}{4}\ ms^{-1}$ 

His average speed is $\underline{1.25 \ ms^{-1}}$ .

Average velocity:

$average\ velocity=\dfrac{displacement\ from\ origin}{time\ taken}$ 

His total displacement was $0m$ , as he returned to his starting point. Therefore:

$average\ velocity=\dfrac{0}{8}=0\ ms^{-1}$ 

His average velocity was $\underline{ 0 \ ms^{-1}}$ .

Learn with Basics

Length:

Unit 1

Motion and speed

Unit 2

Distance-time and velocity-time graphs

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Unit 3