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Grade 9 Maths Book An Easy Way to Learn Maths

INNODEMS

Section 3.4 Time, Distance and Speed

Subsection 3.4.1 Speed in km/hr

Activity 3.4.1. Matching Question.

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Definition 3.4.1.

Speed refers to the rate at which an object changes its position, typically measured in units of distance per time.
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Speed \(= \frac{\text{Distance covered}}{\text{Time taken}}\)
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Example 3.4.2.

A bus travels 120 kilometers in 2 hours. What is its speed in km/h?
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Solution.
Recall that, Speed \(= \frac{\text{Distance covered}}{\text{Time taken}}\)
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Here, the distance covered is \(120\) kilometers and the time taken is \(2\) hours.
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Therefore, Speed \(= \frac{120 \text{ km}}{2 \text{ h}} = 60 \text{ km/h}\text{.}\)
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Example 3.4.3.

David covered \(450\) meters in \(15\) minutes. Find his speed in km/h.
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Solution.
Convert time in minutes to hours.
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\begin{align*} 1 \text{ minute} \amp = (1 \div 60) \text{ hours} \\ \amp = \frac{15}{60} \text{ hours} \\ \amp = \frac{1}{4} \text{ hours} \end{align*}
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Convert distance in metres to kilometers.
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\begin{align*} 450 \text{ m} \amp = (450 \div 1000) \text{ km} \\ \amp = \frac{450}{1000} \\ \amp = 0.45 \text{ km} \end{align*}
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\begin{align*} \text{ Speed } \amp = \frac{\text{Distance Covered}}{\text{Time taken}} \\ \amp = 0.45 \text{km} \div \frac{1}{60} \text{h}\\ \amp = (0.45 \div \frac{6}{1}) \text{km/h} \\ \amp = 9 \text{km/h} \end{align*}
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Example 3.4.4.

Samuel left his house at \(6:10\) a.m. and arrived at the library at \(6:40\) a.m. If the total distance he walked was \(7.5\) kilometers, calculate his average speed in kilometers per hour.
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Solution.
First, we need to find the total time taken in hours.
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The time taken is from \(6:10\) a.m. to \(6:40\) a.m., which is \(30\) minutes.
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Convert minutes to hours:
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\begin{align*} 30 \text{ minutes} \amp = (30 \div 60) \text{ hours} \\ \amp = \frac{1}{2} \text{ hours} \end{align*}
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Now, we can calculate the average speed using the formula:
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\begin{align*} \text{ Speed } \amp = \frac{\text{Distance Covered}}{\text{Time taken}} \\ \amp = 7.5 \text{km} \div \frac{1}{2} \text{h}\\ \amp = (7.5 \div \frac{1}{2}) \text{km/h} \\ \amp = 15 \text{km/h} \end{align*}
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Subsection 3.4.2 Speed in m/s

Activity 3.4.2.

Work in groups
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  1. Go outside to the school field. Measure and mark a straight distance of 40 meters using a tape measure or meter wheel.
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  2. Choose one student from your group to run the \(40\)-meter distance while another student uses a stopwatch to time how long it takes in seconds.
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  3. Record the distance (in meters) and the time taken (in seconds) in your notebook.
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  4. Use the formula \(Speed = \frac{\text{Distance}}{\text{Time}}\) to calculate the running speed in meters per second (m/s).
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  5. Repeat the activity with different group members and compare your results. Discuss why the speeds may differ among individuals.
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Example 3.4.6.

Onyango walked \(3\) km from Lubao to Kakamega town in \(90\) minutes. Determine his speed in metres per second (m/s).
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Solution.
Speed \(= \frac{\text{Distance covered}}{\text{Time taken}}\)
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Convert the distance from kilometres to metres:
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\(3\) km = (\(3\) \(\times\) \(1000\)) m = \(3000\) m
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Convert the time from minutes to seconds:
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\(90\) minutes = (\(90\) \(\times\) \(60\)) s = \(5400\) s
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Speed \(= \frac{3000 \text{ m}}{5400 \text{ s}} = \frac{5}{9} \text{ m/s}\)
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Speed \(= \frac{5}{9} \text{ m/s}\)
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Example 3.4.7.

A car travels \(150\) m in \(10\) seconds. What is its speed in metres per second (m/s)?
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Solution.
Speed \(= \frac{\text{Distance covered}}{\text{Time taken}}\)
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Speed \(= \frac{150 \text{ m}}{10 \text{ s}} = 15 \text{ m/s}\)
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Subsection 3.4.3 Average Speed

Activity 3.4.3.

Work in groups
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  1. Identify a measured path around the school where a "journey" can take place, such as walking 100 meters from the gate to a classroom block, and another 50 meters from the classroom block to the school canteen.
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  2. One group member acts as the "traveler" and walks the first 100 meters while another times the walk using a stopwatch. Record the time taken for this segment.
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  3. After finishing the first part of the journey, the "traveler" takes a timed rest break of 5 minutes to simulate a lunch break. Record this break time.
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  4. The traveler then walks the second 50-meter stretch. Record the time taken for this second part of the walk.
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  5. Now calculate the total distance walked and the total time taken (including the 5-minute break).
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  6. Use the formula \(\frac{\text{Total distance}}{\text{Total time}}\) to compute the average speed in meters per second.
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  7. Discuss in your group: How does the break time affect the average speed? What would the speed be without including the break time?
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Remark 3.4.9.

Average speed is defined as the total distance traveled divided by the total time taken. It gives us an overall idea of how fast an object is moving, regardless of any variations in speed during the journey.
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Consider a scenario where a car travels \(100\) km in \(1\) hour, then \(50\) km in \(30\) minutes. What is the average speed of the car for the entire journey?
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Example 3.4.10.

A cyclist cycles \(30\) km in \(2\) hours and then \(20\) km in \(1\) hour. What is the average speed of the cyclist for the entire trip?
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Solution.
Total distance = \(30\) km + \(20\) km = \(50\) km
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Total time = \(2\) hours + \(1\) hour = \(3\) hours
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Average speed = \(\frac{50 \text{ km}}{3 \text{ hours}} \approx 16.67 \text{ km/h}\)
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Example 3.4.11.

A car travels \(150\) km in \(2\) hours, then takes a \(30\)-minute break, and later travels another \(100\) km in \(1.5\) hours. What is the average speed of the car for the entire journey?
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Solution.
Total distance = \(150\) km + \(100\) km = \(250\) km
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Total time = \(2\) hours + \(0.5\) hour (break) + \(1.5\) hours = \(4\) hours
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Average speed = \(\frac{250 \text{ km}}{4 \text{ hours}} = 62.5 \text{ km/h}\)
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Subsection 3.4.4 Velocity

\(\textbf{Key Takeaway}\)
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Definition 3.4.14.

Velocity is refered to as distance covered in a specific direction per unit time. It is a vector quantity, meaning it has both magnitude and direction.
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Example 3.4.15.

A car travels \(100\) meters to the east in \(20\) seconds. What is its velocity in m/s?
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Solution.
Velocity = \(\frac{\text{Total Distance Covered in a Specified Direction}}{\text{Total Time Taken}}\)
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\(\frac{100 \text{ m}}{20 \text{ s}} = 5 \text{ m/s}\text{.}\)
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Example 3.4.16.

A Spacecraft travels \(300\) kilometers to the north in \(5\) minutes. Calculate its velocity in km/h.
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Solution.
First, convert the time from minutes to hours: \(5 \text{ minutes} = \frac{5}{60} \text{ hours} = \frac{1}{12} \text{ hours}\text{.}\)
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Velocity = \(\frac{\text{Total Distance Covered in a Specified Direction}}{\text{Total Time Taken}}\)
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\(\frac{300 \text{ km}}{\frac{1}{12} \text{ hours}} = 300 \text{ km} \times \frac{12}{1} = 3600 \text{ km/h}\text{.}\)
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Subsection 3.4.5 Acceleration

Activity 3.4.5.

Work in groups.
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  1. From the left handside of Figure 3.4.18 below, click to check the Road, Photos, Markers and Graphs boxes.
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  2. Click the Play button to start the simulation.
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  3. Observe the motion of the two objects as they accelerate.
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  4. Click the Reset button to reset the simulation and Pause button to pause the simulation.
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  5. Discuss and share with the rest of the class your observations about the acceleration of the objects.
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Figure 3.4.18. Acceleration of an object
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\(\textbf{Key Takeaway}\)
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Definition 3.4.19.

Acceleration is defined as the rate of change of velocity per unit time. It is a vector quantity, meaning it has both magnitude and direction.
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Acceleration can be positive (speeding up), negative (slowing down), or zero (constant speed).
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The formula for acceleration is given by: \(\text{Acceleration} = \frac{\text{Change in Velocity}}{\text{Time Taken}}\)
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Example 3.4.20.

A Jet fighter accelerates from \(0\) m/s to \(100\) m/s in \(10\) seconds. What is its acceleration?
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Solution.
Initial velocity \(= 0\) m/s
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Final velocity \(= 100\) m/s
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Time taken \(= 10\) seconds
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Acceleration \(= \frac{\text{Final Velocity} - \text{Initial Velocity}}{\text{Time Taken}}\)
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\begin{align*} \amp = \frac{(100 - 0) \text{m/s}}{10 \text{s}}\\ \amp = \frac{100 \text{m/s}}{10 \text{s}} \\ \amp = 10 \text{m/s}^2 \end{align*}
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Example 3.4.21.

A Bus slows down from \(60\) m/s to \(30\) m/s in \(5\) seconds. Calculate its acceleration?
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Solution.
Initial velocity \(= 60\) m/s
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Final velocity \(= 30\) m/s
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Time taken \(= 5\) seconds
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Acceleration \(= \frac{\text{Final Velocity} - \text{Initial Velocity}}{\text{Time Taken}}\)
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\begin{align*} \amp = \frac{(30 - 60) \text{m/s}}{5 \text{s}}\\ \amp = \frac{-30 \text{m/s}}{5 \text{s}} \\ \amp = -6 \text{m/s}^2 \end{align*}
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Checkpoint 3.4.22.

A car accelerates from \(20\) m/s to \(50\) m/s in \(3\) seconds. Calculate its acceleration.
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Subsection 3.4.6 Identifying Longitudes

Activity 3.4.7.

Work in pairs or small groups.
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In this activity, you will use an orange to explore the concept of longitudes, including the Prime Meridian and other meridians on a globe.
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Materials Needed:
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  • 1 orange (per group)
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  • Marker pen (preferably black or dark color)
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  • Ruler (optional)
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  • Toothpicks or a small container (optional, to hold the orange steady)
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Instructions:
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  1. Take the orange and place it on a flat surface. If it keeps rolling, use a small container or toothpicks to hold it steady.
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  2. Use a marker to carefully draw a dotted line from the top of the orange (where the stem scar is) to the bottom. This line should pass through the center and divide the orange vertically into two equal parts.
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  3. Discuss with your group:
    • What do we call this vertical line on the real Earth?
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    • What are the names of the two parts it divides the globe into?
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  4. On either side of the dotted line, draw 2 or 3 more curved lines from top to bottom of the orange (north to south), equally spaced. These lines should bend slightly as if wrapping around the globe.
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  5. These curved lines represent other meridians. What name do we give to all such vertical lines running from the North Pole to the South Pole?
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  6. Label the dotted vertical line as the Prime Meridian and one of the other lines as 60° East or 30° West (your choice).
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  7. Now, use your finger to point to a location between two of the meridian lines. Pretend it is a city on Earth.
    • Can you estimate its longitude based on your lines?
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    • What would you call its location: East or West of the Prime Meridian?
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  8. Share your orange globe and observations with the rest of the class. Explain what the lines mean and how they help us locate places on Earth.
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\(\textbf{Key Takeaway}\)
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Definition 3.4.24.

Longitude are imaginary lines that run from the North Pole to the South Pole, measuring the distance east or west of the prime meridian.
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Any north-sourth line can be meridian. A prime meridian is the meridian chosen to be zero degrees longitude for common reference.
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Prime meridian is also known as the Greenwich Meridian because it passes through the Greenwich Observatory in London, England.
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Figure 3.4.25. Identifying Longitudes
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Longitude is the measurement east or west of the prime meridian. In Figure 3.4.25 we can see the prime meridian at \(0^\circ\) and other longitudes marked. The longitude \(28^\circ E\) is 28 degrees east of the prime meridian, while \(65^\circ W\) is 65 degrees west of the prime meridian.
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Example 3.4.26.

In the diagram below, identify points that are on the same longitude.
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Solution.
Points E, F and G are on the same longitude.
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Points B and C are on the same longitude.
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Example 3.4.27.

Identify the longitudes of the points A, B, C, D, E and F in the diagram below.
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Solution.
  1. B is \(38^\circ E\) of the prime meridian.
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  2. C is \(32^\circ W\) of the prime meridian.
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  3. D is \(60^\circ E\) of the prime meridian.
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  4. E is \(75^\circ W\) of the prime meridian.
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  5. F is \(38^\circ E\) of the prime meridian.
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Subsection 3.4.7 Relating Longitudes to Time

Activity 3.4.8.

Work in groups.
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Use the diagram in Figure 3.4.31 and follow the steps below:
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  1. Identify pairs of cities that lie on the same longitude. Record their names and longitudes.
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  2. Use the rule that places on the same longitude experience the same local time to explain which cities share the same time of day.
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  3. Identify pairs of cities that are equidistant from the Prime Meridian but on opposite sides (e.g., 53°E and 53°W). Discuss whether they experience the same time of day and explain your reasoning.
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  4. Discuss: If it is 10:00 AM at City B, what is the time at City E? Use the 4 minutes per degree rule to support your answer.
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  5. Present your findings to the class and explain how Earth’s rotation affects the relationship between longitude and time.
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\(\textbf{Key Takeaway}\)
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Remark 3.4.29.

The earth rotates on its axis, completing one full rotation every 24 hours. This rotation causes the cycle of day and night.
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The earth rotates from west to east, which means that as the earth rotates, different parts of the world experience daylight and darkness at different times.
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Places that are located at the same longitude will experience the same time of day, while places that are located at different longitudes will experience different times of day.
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Example 3.4.30.

In Figure 3.4.31 below, cities A, B, C, D, E, and F are located at various longitudes. Determine which of these cities experience the same time of day.
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A diagram showing the relationship between longitudes and time zones.
Figure 3.4.31.
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Solution.
Cities B and C lie on the same longitude (\(22^\circ\) E), which means they experience the same local time.
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Cities E and F are at \(53^\circ\) W, and City D is at \(53^\circ\) E. Since they are equally distant from the Prime Meridian, they experience the same local time.
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Subsection 3.4.8 Local time of places on the Earth along different Longitudes

Activity 3.4.10.

Work in groups.
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Use an atlas or a map of the world showing longitudes and the Prime Meridian. Then complete the following tasks:
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  1. Identify three towns or cities located at different longitudes (one in the Eastern Hemisphere, one near the Prime Meridian, and one in the Western Hemisphere). Record their longitudes.
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  2. Assume the local time in the town closest to the Prime Meridian is \(12:00\) noon. Using the concept that \(1^\circ\) of longitude equals \(4\) minutes, calculate the local time in the other two towns.
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  3. Share your group’s findings with the class. Explain how the direction (east or west) of a town’s longitude affected its local time compared to the Prime Meridian.
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  4. Discuss: What challenges might arise in global communication or travel due to differences in local time?
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\(\textbf{Key Takeaway}\)
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Remark 3.4.33.

The earth rotates on its axis from west to east, which means that places to the east of a given location will experience sunrise earlier than those to the west. This results in different local times for places along different longitudes.
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The earth takes \(24\) hours to complete one full rotation covering \(360\) degrees of longitude, which means that each hour corresponds to \(15\) degrees of longitude (360 degrees \(\div\) \(24\) hours \(=\) \(15\) degrees/hour). Therefore, for every \(15\) degrees of longitude difference between two places, there is a one-hour difference in local time.
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Therefore, the earth covers \(1^\circ\) of longitude in \(4\) minutes. We gain time as we move eastward and lose time as we move westward.
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Example 3.4.34.

Town A is situated at \(40^\circ\) E longitude. What will be the local time in Town A when it is \(12:00\) noon at the Prime Meridian?
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Solution.
Town A is located \(40^\circ\) east of the Prime Meridian. To find the time difference, we use the fact that the Earth rotates through \(1^\circ\) of longitude every \(4\) minutes.
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Therefore, the time difference between Town A and the Prime Meridian is:
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\(40^\circ \times 4 = 160\) minutes
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Converting \(160\) minutes into hours and minutes:
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\(160 \text{ minutes} = 2 \text{ hours and } 40 \text{ minutes}\)
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Since Town A is east of the Prime Meridian, its local time is ahead.
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So, when it is \(12:00\) noon at the Prime Meridian, the local time in Town A is \(2:40\) PM.
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Example 3.4.35.

If it is 10:00 am at \(50^\circ\) E, what time is it at \(30^\circ\) W?
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Solution.
The total difference in longitude between \(50^\circ\) E and \(30^\circ\) W is:
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\(50 + 30 = 80\) degrees
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Since the Earth rotates \(1^\circ\) every \(4\) minutes, the time difference is:
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\(80 \times 4 = 320\) minutes
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Converting \(320\) minutes into hours and minutes:
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\(320 \text{ minutes} = 5 \text{ hours and } 20 \text{ minutes}\)
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Since \(50^\circ\) E is east of \(30^\circ\) W, we need to subtract the time difference from the time at \(50^\circ\) E.
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Therefore, the time at \(30^\circ\) W is:
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\(10:00 \text{ am} - 5 \text{ hours and } 20 \text{ minutes} = 4:40 \text{ am}\)
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Example 3.4.36.

If it is 3:00 pm at \(90^\circ\) W, what time is it at \(60^\circ\) E?
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Solution.
The total difference in longitude between \(90^\circ\) W and \(60^\circ\) E is:
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\(90 + 60 = 150\) degrees
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Since the Earth rotates \(1^\circ\) every \(4\) minutes, the time difference is:
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\(150 \times 4 = 600\) minutes
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Converting \(600\) minutes into hours and minutes:
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\(600 \text{ minutes} = 10 \text{ hours}\)
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Since \(90^\circ\) W is west of \(60^\circ\) E, we need to add the time difference to the time at \(90^\circ\) W.
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Therefore, the time at \(60^\circ\) E is:
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\(3:00 \text{ pm} + 10 \text{ hours} = 1:00 \text{ am}\)
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Technology 3.4.38. Explore more on Time, Distance, and Speed.

  1. Explore the concept of average speed and how it relates to time and distance with this interactive video.
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  2. Explore more on the concept of Acceleration, which is the rate of change of velocity with respect to time.
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