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Data Handling - Travel and Conversion Graphs

Grade 6IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A distance-time graph illustrates how far an object has traveled over a period of time. In these graphs, time is plotted on the horizontal x-axis and distance from a starting point is plotted on the vertical y-axis. A horizontal line segment indicates that the object is stationary (distance is not changing).

A distance-time graph showing a trip, a stationary period, and a return journey.
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The gradient (slope) of a distance-time graph represents the speed of the object. A steeper line indicates a higher speed, while a gentler slope indicates a slower speed. The gradient can be calculated using m=change in distancechange in timem = \frac{\text{change in distance}}{\text{change in time}}.

Comparison of steep and gentle gradients on a graph.
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Conversion graphs are straight-line graphs used to convert one unit of measurement to another (e.g., Celsius to Fahrenheit or Pounds to Kilograms). If the conversion is directly proportional, the line will pass through the origin (0,0)(0, 0).

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To read a conversion graph, find the given value on one axis, move vertically or horizontally to meet the line, and then move to the other axis to find the corresponding converted value.

📐Formulae

Speed=DistanceTimeSpeed = \frac{Distance}{Time}

Average Speed=Total Distance TraveledTotal Time TakenAverage\ Speed = \frac{Total\ Distance\ Traveled}{Total\ Time\ Taken}

Distance=Speed×TimeDistance = Speed \times Time

Time=DistanceSpeedTime = \frac{Distance}{Speed}

Gradient (m)=y2−y1x2−x1Gradient\ (m) = \frac{y_2 - y_1}{x_2 - x_1}

💡Examples

Problem 1:

A cyclist travels at a constant speed for 2 hours2\text{ hours} and covers a distance of 30 km30\text{ km}. He then rests for 1 hour1\text{ hour} and finally travels back to the start point in 1.5 hours1.5\text{ hours} at a constant speed. Calculate the speed for the first part of the journey and the speed for the return journey.

Solution:

For the first part: Speed=30 km2 h=15 km/hSpeed = \frac{30\text{ km}}{2\text{ h}} = 15\text{ km/h}. For the return journey: Speed=30 km1.5 h=20 km/hSpeed = \frac{30\text{ km}}{1.5\text{ h}} = 20\text{ km/h}.

Explanation:

In the first part, the cyclist covers 30 km30\text{ km} in 2 hours2\text{ hours}. During the rest period, the distance remains constant at 30 km30\text{ km} (horizontal line). On the return journey, the cyclist covers the same 30 km30\text{ km} back to the start (0 km0\text{ km}) in 1.5 hours1.5\text{ hours}.

Problem 2:

A conversion graph for miles and kilometers passes through the origin (0,0)(0,0) and the point (5,8)(5, 8), where xx is miles and yy is kilometers. Use this relationship to convert 25 miles25\text{ miles} to kilometers.

Solution:

Given the ratio 5 miles=8 km5\text{ miles} = 8\text{ km}, for 25 miles25\text{ miles}: 25 miles=5×5 miles25\text{ miles} = 5 \times 5\text{ miles} Distance in km=5×8 km=40 kmDistance\ in\ km = 5 \times 8\text{ km} = 40\text{ km}

Explanation:

Since the graph is a straight line through the origin, the units are directly proportional. We find the scale factor by dividing the target value by the known value (25÷5=525 \div 5 = 5) and then multiply the corresponding kilometer value by that scale factor.

Problem 3:

A car travels 120 km120\text{ km} in 2.5 hours2.5\text{ hours}. Calculate its average speed.

Solution:

Average Speed=1202.5=48 km/hAverage\ Speed = \frac{120}{2.5} = 48\text{ km/h}

Explanation:

Divide the total distance by the total time taken to find the average speed. If the graph of this journey was a straight line, 48 km/h48\text{ km/h} would be the constant gradient of that line.

Problem 4:

A car's journey is recorded on the distance-time graph below. The car travels 60 km60\text{ km} in 1.5 hours1.5\text{ hours}, stops for 0.5 hours0.5\text{ hours}, and then continues another 40 km40\text{ km} in 1 hour1\text{ hour}. Calculate the speed for the first part of the journey and the average speed for the entire 3 hour3\text{ hour} trip.

Distance-time graph showing a car journey with two travel segments and one stop.

Solution:

  1. Speed for the first part: Speed=DistanceTime=60 km1.5 h=40 km/h\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{60\text{ km}}{1.5\text{ h}} = 40\text{ km/h}

  2. Average speed for the whole trip: Total Distance=60+0+40=100 km\text{Total Distance} = 60 + 0 + 40 = 100\text{ km} Total Time=1.5+0.5+1.0=3 hours\text{Total Time} = 1.5 + 0.5 + 1.0 = 3\text{ hours} Average Speed=100 km3 h≈33.33 km/h\text{Average Speed} = \frac{100\text{ km}}{3\text{ h}} \approx 33.33\text{ km/h}

Explanation:

Speed is the gradient of the distance-time graph. Average speed considers the total distance divided by the total time, including the time spent stationary.

Problem 5:

The conversion graph below shows the relationship between Gallons and Liters. Use the graph to convert 4 gallons4\text{ gallons} to Liters. Then, determine how many Gallons are approximately equal to 36 Liters36\text{ Liters} by using the same ratio.

Conversion graph for Gallons to Liters passing through (0,0) and (4,18).

Solution:

  1. From the graph, find 44 on the Gallons (x-axis). Move up to the line and then left to the Liters (y-axis). The value is 18 Liters18\text{ Liters}.

  2. To find the ratio: 1 Gallon=184=4.5 Liters1\text{ Gallon} = \frac{18}{4} = 4.5\text{ Liters}

  3. To convert 36 Liters36\text{ Liters} to Gallons: Gallons=364.5=8 Gallons\text{Gallons} = \frac{36}{4.5} = 8\text{ Gallons}

Explanation:

Conversion graphs represent linear relationships. By finding the value on one axis, we can map it to the other. Alternatively, we can find the unit rate (gradient) to solve via calculation.