travel graph description

Travel Graphs

This section covers travel graphs, speed, distance, time, trapezium rule and velocity.

Speed, Distance and Time

The following is a basic but important formula which applies when speed is constant (in other words the speed doesn't change):

Remember, when using any formula, the units must all be consistent. For example speed could be measured in m/s, distance in metres and time in seconds.

If speed does change, the average (mean) speed can be calculated: Average speed = total distance travelled total time taken

In calculations, units must be consistent, so if the units in the question are not all the same (e.g. m/s, m and s or km/h, km and h), change the units before starting, as above.

The following is an example of how to change the units:

Change 15km/h into m/s. 15km/h = 15/60 km/min (1) = 15/3600 km/s = 1/240 km/s (2) = 1000/240 m/s = 4.167 m/s (3)

In line (1), we divide by 60 because there are 60 minutes in an hour. Often people have problems working out whether they need to divide or multiply by a certain number to change the units. If you think about it, in 1 minute, the object is going to travel less distance than in an hour. So we divide by 60, not multiply to get a smaller number.

If a car travels at a speed of 10m/s for 3 minutes, how far will it travel? Firstly, change the 3 minutes into 180 seconds, so that the units are consistent. Now rearrange the first equation to get distance = speed × time. Therefore distance travelled = 10 × 180 = 1800m = 1.8km

Velocity and Acceleration

Velocity is the speed of a particle and its direction of motion (therefore velocity is a vector quantity, whereas speed is a scalar quantity).

When the velocity (speed) of a moving object is increasing we say that the object is accelerating . If the velocity decreases it is said to be decelerating. Acceleration is therefore the rate of change of velocity (change in velocity / time) and is measured in m/s².

A car starts from rest and within 10 seconds is travelling at 10m/s. What is its acceleration?

Distance-Time Graphs

These have the distance from a certain point on the vertical axis and the time on the horizontal axis. The velocity can be calculated by finding the gradient of the graph. If the graph is curved, this can be done by drawing a chord and finding its gradient (this will give average velocity) or by finding the gradient of a tangent to the graph (this will give the velocity at the instant where the tangent is drawn).

Velocity-Time Graphs/ Speed-Time Graphs

A velocity-time graph has the velocity or speed of an object on the vertical axis and time on the horizontal axis. The distance travelled can be calculated by finding the area under a velocity-time graph. If the graph is curved, there are a number of ways of estimating the area (see trapezium rule below). Acceleration is the gradient of a velocity-time graph and on curves can be calculated using chords or tangents, as above.

The distance travelled is the area under the graph. The acceleration and deceleration can be found by finding the gradient of the lines.

On travel graphs, time always goes on the horizontal axis (because it is the independent variable).

Trapezium Rule

This is a useful method of estimating the area under a graph. You often need to find the area under a velocity-time graph since this is the distance travelled.

Area under a curved graph = ½ × d × (first + last + 2(sum of rest))

d is the distance between the values from where you will take your readings. In the above example, d = 1. Every 1 unit on the horizontal axis, we draw a line to the graph and across to the y axis. 'first' refers to the first value on the vertical axis, which is about 4 here. 'last' refers to the last value, which is about 5 (green line).] 'sum of rest' refers to the sum of the values on the vertical axis where the yellow lines meet it. Therefore area is roughly: ½ × 1 × (4 + 5 + 2(8 + 8.8 + 10.1 + 10.8 + 11.9 + 12 + 12.7 + 12.9 + 13 + 13.2 + 13.4)) = ½ × (9 + 2(126.8)) = ½ × 262.6 = 131.3 units²

Ultimate Guide to Travel Graphs| Cambridge IGCSE Mathematics

Kulni gamage.

Created on July 20, 2023

[Please watch the video attached at the end of this blog for a visual explanation on the Ultimate Guide to Travel Graphs]

Travel graphs fall under the bigger topic of Algebra, but unlike the other lessons under this topic, this is really simple and easy to understand… as long as you read the question carefully and pay attention to the readings of the graphs ?!

What is a travel graph?

A travel graph is a line graph that tells us how much distance has been covered by the object in question within a particular time. We can use travel graphs to represent the motion of cars, people, trains, buses, etc. This means travel graphs can tell us whether an object is still (stationary), moving at a constant speed, accelerating, or decelerating. The y -axis represents the distance travelled while the x -axis shows the time it takes to travel a distance such as that.

James went on a cycling ride. The travel graph shows James’s distance from home on this cycle ride. Find James’s speed for the first 60 minutes.

We are told that James is going on a cycling ride and that the distance he travelled is what is shown here. What is required of us is to find James’s speed for the first 60 minutes. This means you need to mark the point at 60 minutes on the x -axis as shown in the picture below.

We have learnt when learning about Speed, Time and Distance that speed = distance/ time.

Therefore, in a distance-time graph, the slope that we see (also called the gradient) is what will give us the speed.

We can see that this particular cycle has travelled 12 km in 60 minutes.

When we substitute these values into the formula above, it would look like this:

When this formula is solved, then we get James’s speed for the first 60 minutes , which is 0.2 km/min .

A train journey takes one hour. The diagram shows the speed-time graph for this journey. Calculate the total distance of this journey.

We need to be careful here because this time what we have in this question is a speed-time graph and not a distance-time graph.

In a speed-time graph, the gradient of a slope represents either acceleration or deceleration, which will be concepts touched upon in later lessons. The distance travelled is found by calculating the area under a speed-time graph.

In order to calculate the distance travelled by the train therefore, we need to calculate the area under the graph that is drawn. To make things easier, we can separate this graph into shapes:

2 triangles and 1 rectangle

Triangle 1: Area of a triangle = ½ × base of triangle × height of triangle Area = ½ × 4 km × 3 km/min Area = 6 km2

Rectangle: Length of the rectangle can be found by finding out the difference between 50 and 4, which equals to 46 min. The height of the rectangle is 3 km/min. Area = length × base Area = 46 min × 3 km/min Area = 138 km

Triangle 2: Area of a triangle = ½ × base of triangle × height of triangle Area = ½ × 10 km × 3 km/min Area = 15 km

Total Area under the graph = Total distance travelled by the train Therefore, Total Distance = Triangle 1 + Rectangle + Triangle 2 Total Distance = 6 km + 138 km + 15 km Total Distance = 159 km

Those are the basic tips and tricks you would need to learn in this particular chapter!

Travel Graphs at Exams

Pay attention to what sort of graph is given to you. Is it a distance-time graph? A speed-time graph? Depending on the type of graph, remember, the value represented by the gradient changes, and so does the value represented by the area under the graph.

Practice as many questions as you can on this topic. They are quite simple once you get the hang of them. You can find some questions in this quiz to check where you stand!

If you are struggling with IGCSE revision or the Mathematics subject in particular, you can reach out to us at Tutopiya to join revision sessions or find yourself the right tutor for you.

Watch the video below for a visual explanation of the lesson on the ultimate guide to travel graphs!

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Travel Graphs GCSE Maths Revision Questions

Travel Graphs GCSE Questions PDF

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This set of questions provides travel graphs in the form of distance-time graphs. Learners will need to read information from the graphs, such as distances at certain times and periods of rest. They will also need to calculate the speed between different times.

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Maths, but not as you know it…, gcse maths: travel graphs (speed-distance-time).

Learn how to use travel graphs relating speed, distance and time to solve problems. Tom introduces the topic, Bobby applies it to Usain Bolt’s 100-metre world record, and Susan uses an interactive tool on GeoGebra – go to www.geogebra.org/m/ABuBYdX7 to join in at home.

This is the second lesson in a new series with the Maths Appeal duo – Bobby Seagull and Susan Okereke – and Tom Rocks Maths where we’ll be exploring the GCSE Maths syllabus to show the world that maths is accessible to everyone!

Watch Bobby and Susan answer YOUR questions here .

Watch Tom’s livestream here .

Menu Level 1 Level 2 Level 3 Level 4 Exam Help More Graphs

This is level 4; Draw a travel graph from the given description. Click on the points at the ends of where the straight line segments of the graphs should be.

This is Travel Graphs level 4. You can also try: Level 1 Level 2 Level 3

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Description of Levels

Level 1 - Reading information from distance-time graphs

Level 2 - Matching distance-time graphs with their descriptions

Level 3 - Reading information from speed-time graphs

Level 4 - Draw a travel graph from the given description

Exam Style Questions - A collection of problems in the style of GCSE or IB/A-level exam paper questions (worked solutions are available for Transum subscribers).

More Graphs including lesson Starters, visual aids, investigations and self-marking exercises.

Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.

Distance-Time Graphs

For a basic introduction to distance-time graphs see Hurdles Race . For more details play the video below.

Don't wait until you have finished the exercise before you click on the 'Check' button. Click it often as you work through the questions to see if you are answering them correctly. You can double-click the 'Check' button to make it float at the bottom of your screen.

NESA is regularly updating its advice as the coronavirus outbreak unfolds. Get our latest COVID-19 advice

This webpage has been archived to prepare for transfer to the new NESA website . Reference to syllabus outcomes and content on this webpage may not be current. Teachers are encouraged to visit the Key Learning Area page for recent student work samples on the NESA website .

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Areas for Assessment

Describing travel graphs, description of activity.

For each of the six travel graphs, students will be asked to write a brief description of the journey that the graph might represent.

Students will complete the accompanying worksheet.

Patterns and Algebra, Working Mathematically

This assessment activity could be presented to students during, or at the end of, a unit on Graphs of Physical Phenomena, where students have been drawing and interpreting a variety of travel graphs, as well as providing descriptions of the meanings of different gradients for graphs of particular events.

Working mathematically Knowledge, skills and understanding developed through inquiry, application of problem-solving strategies, communication, reasoning and reflection
Patterns and algebra Knowledge, skills and understanding in patterning, generalisation and algebraic reasoning

Graphs of Physical Phenomena (PAS5.2.5) Draws and interprets graphs of physical phenomena

Applying Strategies (WMS5.2.2) Selects and uses appropriate problem-solving strategies that include selecting and organising key information and identifying and working on related problems

Communicating (WMS5.2.3) Uses appropriate mathematical language and algebraic, statistical and other notations and conventions in written, oral or graphical form

Reasoning (WMS5.2.4) Uses mathematical arguments to reach and justify conclusions

Criteria for assessing learning

(These criteria would normally be communicated to students with the activity.)

Students will be assessed on their ability to:

describe journeys represented in travel graphs
use mathematical terminology

Possible prompts to assist student engagement

Where does the graph begin? What does this mean?
Consider some specific examples.
Can you explain why you think the graph has that shape?
How would you describe the slope of the line? What does this mean?
What occurs when the graph changes slope? How would you describe this in terms of a journey?
Is it possible to be in two places at the same time? Does this occur in any of the graphs?

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4.2: Hamiltonian Circuits and the Traveling Salesman Problem

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David Lippman
Pierce College via The OpenTextBookStore

In the last section, we considered optimizing a walking route for a postal carrier. How is this different than the requirements of a package delivery driver? While the postal carrier needed to walk down every street (edge) to deliver the mail, the package delivery driver instead needs to visit every one of a set of delivery locations. Instead of looking for a circuit that covers every edge once, the package deliverer is interested in a circuit that visits every vertex once.

Hamilton Circuits and Paths

A Hamiltonian circuit is a circuit that visits every vertex once with no repeats. Being a circuit, it must start and end at the same vertex. A Hamiltonian path also visits every vertex once with no repeats, but does not have to start and end at the same vertex.

Hamiltonian circuits are named for William Rowan Hamilton who studied them in the 1800’s.

One Hamiltonian circuit is shown on the graph below. There are several other Hamiltonian circuits possible on this graph. Notice that the circuit only has to visit every vertex once; it does not need to use every edge.

This circuit could be notated by the sequence of vertices visited, starting and ending at the same vertex: ABFGCDHMLKJEA. Notice that the same circuit could be written in reverse order, or starting and ending at a different vertex.

Unlike with Euler circuits, there is no nice theorem that allows us to instantly determine whether or not a Hamiltonian circuit exists for all graphs.[1]

Does a Hamiltonian path or circuit exist on the graph below?

We can see that once we travel to vertex E there is no way to leave without returning to C, so there is no possibility of a Hamiltonian circuit. If we start at vertex E we can find several Hamiltonian paths, such as ECDAB and ECABD.

Depending on the problem being solved, sometimes weights are assigned to the edges. The weights could represent the distance between two locations, the travel time, or the travel cost.

With Hamiltonian circuits, our focus will not be on existence, but on the question of optimization; given a graph where the edges have weights, can we find the optimal Hamiltonian circuit; the one with lowest total weight.

To answer this question of how to find the lowest cost Hamiltonian circuit, we will consider some possible approaches. The first option that might come to mind is to just try all different possible circuits.

Brute Force Algorithm (a.k.a. exhaustive search)

List all possible Hamiltonian circuits
Find the length of each circuit by adding the edge weights
Select the circuit with minimal total weight.

Apply the Brute force algorithm to find the minimum cost Hamiltonian circuit on the graph below.

To apply the Brute force algorithm, we list all possible Hamiltonian circuits and calculate their weight:

$\begin{array}{|l|l|} \hline \textbf { Circuit } & \textbf { Weight } \\ \hline \text { ABCDA } & 4+13+8+1=26 \\ \hline \text { ABDCA } & 4+9+8+2=23 \\ \hline \text { ACBDA } & 2+13+9+1=25 \\ \hline \end{array}$

Note: These are the unique circuits on this graph. All other possible circuits are the reverse of the listed ones or start at a different vertex, but result in the same weights.

From this we can see that the second circuit, ABDCA, is the optimal circuit.

The Brute force algorithm is optimal; it will always produce the Hamiltonian circuit with minimum weight. Is it efficient? To answer that question, we need to consider how many Hamiltonian circuits a graph could have. For simplicity, let’s look at the worst-case possibility, where every vertex is connected to every other vertex. This is called a complete graph. In figure A, there are examples of complete graphs with different numbers of vertices.

$Complete graphs. One has two vertices, one has 3 vertices, one has 5 vertices, and the last one has 9 vertices$

Out of convenience, mathematicians sometimes use specific notation for complete graphs based on the number of vertices. A complete graph with $n$ vertices can be represented by K n . For example, the graphs represented above are K 2 , K 3 , K 5 and K 9 .

Suppose we had a complete graph with five vertices like the air travel graph above. From Seattle there are four cities we can visit first. From each of those, there are three choices. From each of those cities, there are two possible cities to visit next. There is then only one choice for the last city before returning home.

This can be shown visually:

Counting the number of routes, we can see there are $4 \cdot 3 \cdot 2 \cdot 1=24$ routes. For six cities there would be $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120$ routes.

Number of Possible Circuits

For $n$ vertices in a complete graph, there will be $(n-1) !=(n-1)(n-2)(n-3) \cdots 3 \cdot 2 \cdot 1$ routes. Half of these are duplicates in reverse order, so there are $\frac{(n-1) !}{2}$ unique circuits.

The exclamation symbol, !, is read “factorial” and is shorthand for the product shown.

How many circuits would a complete graph with 8 vertices have?

A complete graph with 8 vertices would have $(8-1) !=7 !=7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=5040$ possible Hamiltonian circuits. Half of the circuits are duplicates of other circuits but in reverse order, leaving 2520 unique routes.

While this is a lot, it doesn’t seem unreasonably huge. But consider what happens as the number of cities increase:

$\begin{array}{|l|l|} \hline \textbf { Cities } & \textbf { Unique Hamiltonian Circuits } \\ \hline 9 & 8 ! / 2=20,160 \\ \hline 10 & 9 ! / 2=181,440 \\ \hline 11 & 10 ! / 2=1,814,400 \\ \hline 15 & 14 ! / 2=43,589,145,600 \\ \hline 20 & 19 ! / 2=60,822,550,204,416,000 \\ \hline \end{array}$

As you can see the number of circuits is growing extremely quickly. If a computer looked at one billion circuits a second, it would still take almost two years to examine all the possible circuits with only 20 cities! Certainly Brute Force is not an efficient algorithm.

Unfortunately, no one has yet found an efficient and optimal algorithm to solve the TSP, and it is very unlikely anyone ever will. Since it is not practical to use brute force to solve the problem, we turn instead to heuristic algorithms ; efficient algorithms that give approximate solutions. In other words, heuristic algorithms are fast, but may or may not produce the optimal circuit.

Nearest Neighbor Algorithm (NNA)

Select a starting point.
Move to the nearest unvisited vertex (the edge with smallest weight).
Repeat until the circuit is complete.

Consider our earlier graph, shown to the right.

From D, the nearest neighbor is C, with a weight of 8.

From C, our only option is to move to vertex B, the only unvisited vertex, with a cost of 13.

From B we return to A with a weight of 4.

The resulting circuit is ADCBA with a total weight of $1+8+13+4 = 26$.

We ended up finding the worst circuit in the graph! What happened? Unfortunately, while it is very easy to implement, the NNA is a greedy algorithm, meaning it only looks at the immediate decision without considering the consequences in the future. In this case, following the edge AD forced us to use the very expensive edge BC later.

LA to Chicago: $100
Chicago to Atlanta: $75
Atlanta to Dallas: $85
Dallas to Seattle: $120

Total cost: $450

In this case, nearest neighbor did find the optimal circuit.

Going back to our first example, how could we improve the outcome? One option would be to redo the nearest neighbor algorithm with a different starting point to see if the result changed. Since nearest neighbor is so fast, doing it several times isn’t a big deal.

Repeated Nearest Neighbor Algorithm (RNNA)

Do the Nearest Neighbor Algorithm starting at each vertex
Choose the circuit produced with minimal total weight

Starting at vertex A resulted in a circuit with weight 26.

Starting at vertex B, the nearest neighbor circuit is BADCB with a weight of 4+1+8+13 = 26. This is the same circuit we found starting at vertex A. No better.

Starting at vertex C, the nearest neighbor circuit is CADBC with a weight of 2+1+9+13 = 25. Better!

Starting at vertex D, the nearest neighbor circuit is DACBA. Notice that this is actually the same circuit we found starting at C, just written with a different starting vertex.

The RNNA was able to produce a slightly better circuit with a weight of 25, but still not the optimal circuit in this case. Notice that even though we found the circuit by starting at vertex C, we could still write the circuit starting at A: ADBCA or ACBDA.

Try it Now 1

The table below shows the time, in milliseconds, it takes to send a packet of data between computers on a network. If data needed to be sent in sequence to each computer, then notification needed to come back to the original computer, we would be solving the TSP. The computers are labeled A-F for convenience.

$\begin{array}{|l|l|l|l|l|l|l|} \hline & \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D} & \mathrm{E} & \mathrm{F} \\ \hline \mathrm{A} & \_ \_ & 44 & 34 & 12 & 40 & 41 \\ \hline \mathrm{B} & 44 & \_ \_ & 31 & 43 & 24 & 50 \\ \hline \mathrm{C} & 34 & 31 & \_ \_ & 20 & 39 & 27 \\ \hline \mathrm{D} & 12 & 43 & 20 & \_ \_ & 11 & 17 \\ \hline \mathrm{E} & 40 & 24 & 39 & 11 & \_ \_ & 42 \\ \hline \mathrm{F} & 41 & 50 & 27 & 17 & 42 & \_ \_ \\ \hline \end{array}$

a. Find the circuit generated by the NNA starting at vertex B.

b. Find the circuit generated by the RNNA.

At each step, we look for the nearest location we haven’t already visited.

From B the nearest computer is E with time 24.

From E, the nearest computer is D with time 11.

From D the nearest is A with time 12.

From A the nearest is C with time 34.

From C, the only computer we haven’t visited is F with time 27

From F, we return back to B with time 50.

The NNA circuit from B is BEDACFB with time 158 milliseconds.

While certainly better than the basic NNA, unfortunately, the RNNA is still greedy and will produce very bad results for some graphs. As an alternative, our next approach will step back and look at the “big picture” – it will select first the edges that are shortest, and then fill in the gaps.

Cheapest Edge Algorithm (Best Edge/Greedy Algorithm)

1. Select the cheapest unused edge in the graph.

2. Repeat step 1, adding the cheapest unused edge to the circuit, unless:

a. adding the edge would create a circuit that doesn’t contain all vertices, or

b. adding the edge would give a vertex degree 3.

3. Repeat until a circuit containing all vertices is formed.

Using the four vertex graph from earlier, we can use the Sorted Edges algorithm.

The cheapest edge is AD, with a cost of 1. We highlight that edge to mark it selected.

The next shortest edge is AC, with a weight of 2, so we highlight that edge.

For the third edge, we’d like to add AB, but that would give vertex A degree 3, which is not allowed in a Hamiltonian circuit. The next shortest edge is CD, but that edge would create a circuit ACDA that does not include vertex B, so we reject that edge. The next shortest edge is BD, so we add that edge to the graph.

Notice that the algorithm did not produce the optimal circuit in this case; the optimal circuit is ACDBA with weight 23.

While the Sorted Edge algorithm overcomes some of the shortcomings of NNA, it is still only a heuristic algorithm, and does not guarantee the optimal circuit.

Your teacher’s band, Derivative Work , is doing a bar tour in Oregon. The driving distances are shown below. Plan an efficient route for your teacher to visit all the cities and return to the starting location. Use NNA starting at Portland, and then use Sorted Edges.

\( \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline & & & & & & & & & & \\ & \text { Ashland } & \text { Astoria } & \text { Bend } & \text { Corvallis } & \text { Crater Lake } & \text { Eugene } & \text { Newport } & \text { Portland } & \text { Salem } & \text { Seaside } \\ \hline \text { Ashland } & \_ & 374 & 200 & 223 & 108 & 178 & 252 & 285 & 240 & 356 \\ \hline \text { Astoria } & 374 & \_ & 255 & 166 & 433 & 199 & 135 & 95 & 136 & 17 \\ \hline \text { Bend } & 200 & 255 & \_ & 128 & 277 & 128 & 180 & 160 & 131 & 247 \\ \hline \text { Corvallis } & 223 & 166 & 128 & \_ & 430 & 47 & 52 & 84 & 40 & 155 \\ \hline \text { Crater Lake } & 108 & 433 & 277 & 430 & \_ & 453 & 478 & 344 & 389 & 423 \\ \hline \text { Eugene } & 178 & 199 & 128 & 47 & 453 & \_ & 91 & 110 & 64 & 181 \\ \hline \text { Newport } & 252 & 135 & 180 & 52 & 478 & 91 & \_ & 114 & 83 & 117 \\ \hline \text { Portland } & 285 & 95 & 160 & 84 & 344 & 110 & 114 & \_ & 47 & 78 \\ \hline \text { Salem } & 240 & 136 & 131 & 40 & 389 & 64 & 83 & 47 & \_ & 118 \\ \hline \text { Seaside } & 356 & 17 & 247 & 155 & 423 & 181 & 117 & 78 & 118 & \_ \\ \hline \end{array}\)

Using NNA with a large number of cities, you might find it helpful to mark off the cities as they’re visited to keep from accidently visiting them again. Looking in the row for Portland, the smallest distance is 47, to Salem. Following that idea, our circuit will be:

Using Sorted Edges, you might find it helpful to draw an empty graph, perhaps by drawing vertices in a circular pattern. Adding edges to the graph as you select them will help you visualize any circuits or vertices with degree 3.

We start adding the shortest edges:

The graph after adding these edges is shown to the right. The next shortest edge is from Corvallis to Newport at 52 miles, but adding that edge would give Corvallis degree 3.

Continuing on, we can skip over any edge pair that contains Salem or Corvallis, since they both already have degree 2.

The graph after adding these edges is shown to the right. At this point, we can skip over any edge pair that contains Salem, Seaside, Eugene, Portland, or Corvallis since they already have degree 2.

At this point the only way to complete the circuit is to add:

Crater Lk to Astoria 433 miles

The final circuit, written to start at Portland, is:

Portland, Salem, Corvallis, Eugene, Newport, Bend, Ashland, Crater Lake, Astoria, Seaside, Portland.

Total trip length: 1241 miles.

While better than the NNA route, neither algorithm produced the optimal route. The following route can make the tour in 1069 miles:

Portland, Astoria, Seaside, Newport, Corvallis, Eugene, Ashland, Crater Lake, Bend, Salem, Portland

Try it Now 2

Find the circuit produced by the Sorted Edges algorithm using the graph below.

AB: Add, cost 11

BG: Add, cost 13

AE: Add, cost 14

EC: Skip (degree 3 at E)

FG: Skip (would create a circuit not including C)

BF, BC, AG, AC: Skip (would cause a vertex to have degree 3)

GC: Add, cost 36

CF: Add, cost 37, completes the circuit

Final circuit: ABGCFEA

[1] There are some theorems that can be used in specific circumstances, such as Dirac’s theorem, which says that a Hamiltonian circuit must exist on a graph with n vertices if each vertex has degree n /2 or greater.

Traveltime curves how they are created
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Travel-time Curves: How they are created

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Where do travel-time graphs come from?

A travel time curve is a graph of the time that it takes for seismic waves to travel from the epicenter of an earthquake to seismograph stations at varying distances away. The velocity of seismic waves through different materials yield information about Earth's deep interior. IRIS' travel times graphic for the 1994 Northridge, CA earthquake (described in No.5. Exploring the Earth Using Seismology) is animated to show how travel times are determined. Seismic waves "bounce" the buildings to merely illustrate arrival times and wave behavior, not to depict reality. The resultant seismograms show that stations around the world record somewhat predictable arrival times.

CLOSED CAPTIONING: A .srt file is included with the download. Use an appropriate media player to utilize captioning.

Time vs. distance graph of P and S seismic waves
Seismic waves recorded at different points are a function of distance from the earthquake
Seismic velocities within the earth have been calculated from these data

Related Animations

Seismic waves travel through the earth to a single seismic station. Scale and movement of the seismic station are greatly exaggerated to depict the relative motion recorded by the seismogram as P, S, and surface waves arrive.

We use exaggerated motion of a building (seismic station) to show how the ground moves during an earthquake, and why it is important to measure seismic waves using 3 components: vertical, N-S, and E-W. Before showing an actual distant earthquake, we break down the three axes of movement to clarify the 3 seismograms.

A cow and a tree in this narrated cartoon for fun and to emphasize that seismic waves traveling away from an earthquake occur everywhere, not just at seismic stations A, B, C, and D. A person would feel a large earthquake only at station A near the epicenter. Stations B, C, D, and the cow are too far from the earthquake to feel the seismic waves though sensitive equipment records their arrival.

A gridded sphere is used to show a single station recording five equidistant earthquakes.

A gridded sphere is used to show: 1) the seismic stations don't need to be lined up longitudinally to create travel-time curves, as they appear in the first animation, and 2) a single station records widely separated earthquakes that plot on the travel-time curves.

The shadow zone is the area of the earth from angular distances of 104 to 140 degrees from a given earthquake that does not receive any direct P waves. The different phases show how the initial P wave changes when encountering boundaries in the Earth.

The shadow zone results from S waves being stopped entirely by the liquid core. Three different S-wave phases show how the initial S wave is stopped (damped), or how it changes when encountering boundaries in the Earth.

Related Fact-Sheets

Earthquakes create seismic waves that travel through the Earth. By analyzing these seismic waves, seismologists can explore the Earth's deep interior. This fact sheet uses data from the 1994 magnitude 6.9 earthquake near Northridge, California to illustrate both this process and Earth's interior structure.

NOTE: Out of Stock; self-printing only.

Related Software-Web-Apps

jAmaSeis is a free, java-based program that allows users to obtain and display seismic data in real-time from either a local instrument or from remote stations.

Seismic Waves is a browser-based tool to visualize the propagation of seismic waves from historic earthquakes through Earth’s interior and around its surface. Easy-to-use controls speed-up, slow-down, or reverse the wave propagation. By carefully examining these seismic wave fronts and their propagation, the Seismic Waves tool illustrates how earthquakes can provide evidence that allows us to infer Earth’s interior structure.

Related Lessons

Students work first in small groups, and then as a whole class to compare predicted seismic wave travel times, generated by students from a scaled Earth model, to observed seismic data from a recent earthquakes. This activity uses models, real data and emphasizes the process of science.

To understand plate tectonic processes and hazards, and to better understand where future earthquakes are likely to occur, it is important to locate earthquakes as they occur. In this activity students use three-component seismic data from recent earthquakes to locate a global earthquake.

In this multi-step lab, students explore the concepts of seismic wave propagation through materials with different mechanical properties, and examine seismic evidence from a recent earthquake to infer Earth’s internal structure and composition. This lab is designed to be done with an instructor present to answer questions and guide students to conclusions

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