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Grade 10 Mathematics Teachers’ Book

INNODEMS

Subsection 2.10.2 Speed and Velocity

Learner Experience 2.10.3.

Work in groups
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Materials: Stop watch, Measuring tape and ball/round object
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  1. Roll a ball or any round object on the table
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  2. Measure the rolling distance using the tape measure
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  3. Record the time when rolling
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  4. Find the speed of the rolling ball
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Learner Experience 2.10.4.

Work in groups
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What you need
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What to do
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Mark a central point and label it C (origin).
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Measure and mark:
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Two students stand at point C.
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The two groups of students to set the stopwatch and shout "go":
  • Student 1 walks to Point A (East)
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  • Student 2 walks to Point B (West)
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Record the time taken for each student to reach their respective points.
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(b)

Compare the average velocity of student 1 and student 2.
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Exploration 2.10.5. Speed vs. Velocity: The Round Trip.

While we often use the words "speed" and "velocity" interchangeably in daily life, in physics, they describe two very different things.
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Start the car’s journey below and use the questions in the instructions below to uncover the difference between velocity and speed.
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Instructions.

Press the Start Trip button and watch the data counters carefully as the car reaches the "Turn" and begins to come back:
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  • The First Leg: During the first \(4\) seconds, as the car moves toward the right, how do the values for Average Speed and Average Velocity compare? Why are they identical during this part of the trip?
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  • The Turn-Around: After the car passes the 10m mark and starts moving back toward Home, what happens to the Average Velocity? Why does it start to decrease even though the car is still moving?
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  • Distance vs. Displacement: Notice the Total Distance never stops growing. Since Speed is calculated as \(\text{speed} = \frac{\text{distance}}{\text{time}}\text{,}\) how does this explain why the Speed stays high while Velocity drops?
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  • Critical Thinking: If the car drove all the way back to "Home" (\(0\) m), what would its final Average Velocity be? What would its Average Speed be?
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Figure 2.10.10. Interactive: Comparing Average Speed and Average Velocity
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Key Takeaway 2.10.11.

Speed is the rate of change of distance with time. SI unit: \(km/h\) .
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Speed is how fast something moves.
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Speed normally varies over time, so the average speed is often used
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\begin{gather*} \text{Average speed} = \frac{\text{Total distance covered}}{\text{Total time taken}} \end{gather*}
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Velocity is Speed in a specified direction or the rate of change of displacement with time. SI unit: \(km/h\text{.}\)
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Symbol of velocity is given as v while speed is given as s.
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\begin{gather*} \text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time taken}} \end{gather*}
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or \(v = \frac{d}{t} \)
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where \(d\) represents the distance and \(t\) represents time
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For motion with constant velocity, the equation is
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\begin{gather*} s = vt \end{gather*}
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where \(v\) is velocity , \(t\) is time and \(s\) is displacement
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Example 2.10.12.

A car travels 100 km in 2 hours. Find its speed.
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Solution.
\(\text{Speed} = \frac{\text{distance}}{\text{time}}\)
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\(\text{Speed} = \frac{100}{2} = 50 \text{ km/h} \)
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Example 2.10.13.

Find the velocity of a car that travels 150 km north in 3 hours.
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Solution.
Displacement = 150 km (in the north direction)
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Time taken = 3 hours
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Average velocity = Displacement / Time taken = 150 km / 3 h = 50 km/h (in the north direction)
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\begin{align*} \text{Average velocity} =\amp \frac{\text{Displacement}}{\text{Time taken}} \\ =\amp \frac{150 \text{ km}}{3 \text{ h}} \\ =\amp 50 \text{ km/h (north direction)} \end{align*}
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Example 2.10.14.

A motorist covers a distance of 360 km in 3 hours travelling from Busia to Bahangongo. Calculate the average speed.
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Solution.
\(\text{Speed} = \frac{\text{distance}}{\text{time}}\)
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\begin{gather*} \text{Average speed} = \frac{\text{360 km}}{\text{3 h}} \end{gather*}
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\begin{gather*} \text{120 km/h} \end{gather*}
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Example 2.10.15.

A cyclist travels \(10 \, \text{km}\) in the positive direction in \(40 \, \text{minutes}\text{,}\) and then reverses, riding \(6 \, \text{km}\) in the negative direction in \(20 \, \text{minutes}\text{.}\) Calculate the average velocity.
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Solution.
Displacement = \(10 \text{ km (positive)}\) + \(-6 \text{ km (negative)}\) = \(4 \text{ km (positive)}\)
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Time taken = \(40 \text{ minutes} + 20 \text{ minutes} = 60 \text{ minutes} = 1 \text{ hour}\)
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\begin{gather*} \text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}} \\ = \frac{4 \text{ km}}{1 \text{ h}} \\ = 4 \text{ km/h (positive direction)} \end{gather*}
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Example 2.10.16.

Alex is cycling from home to the market. He rides the first 8 km at a speed of 16 km/h. After reaching a park, Alex takes a 20-minute break to play football with his friends. Then, he cycles the remaining 6 km to the market at a speed of 12 km/h. What was Alex’s average speed for the entire journey?
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Solution.
  1. Calculating the time for each cycling part.
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      Distance: \(8 km\)
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      Speed: \(16 km/h\)
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      \begin{gather*} \text{Time}\, = \, \frac{\text{Distance}}{\text{Speed}} \, = \, \frac{\text{8 km}}{\text{16 km/h}} \, = \text{0.5 hours} \end{gather*}
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      Distance: \(6 km\)
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      Speed: \(12 km/h\)
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      \begin{gather*} \text{Time}\, = \, \frac{\text{Distance}}{\text{Speed}} \, = \, \frac{\text{6 km}}{\text{12 km/h}} \, = \text{0.5 hours} \end{gather*}
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  2. Convert the break time to hours
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    \(\frac{\text{20 minutes}}{\text{60 minutes}} = \frac{1}{3} \, \text{hours}\)
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  3. Calculating the total distance gives
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    \begin{gather*} \text {Total distance = 8 km + 6 km = 14 km } \end{gather*}
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  4. Calculating the total time gives
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    Total time =
    \begin{gather*} \frac{1}{2} \, \text{hours} + \frac{1}{3} \, \text{hours} + \frac{1}{2} \, \text{hours} = 1\frac{1}{3} \, \text{hours} \end{gather*}
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  5. Now Calculating the average speed gives
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    \(\text{Average speed} \, = \, \frac{\text{Total distance}}{\text{Total time}}\)
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    \(\text{Average speed} \, = \, \frac{\text{14 km}}{\text{1.33 hours}} \, = \, \text{10.53 km/h} \)
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Alex’s average speed for the entire journey was approximately \(10.53\) km/h.
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Checkpoint 2.10.17.

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Checkpoint 2.10.18.

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Checkpoint 2.10.19.

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Exercises Exercises

1.

Juma cycled for \(3\) hours to a trading centre \(60\) km away, then drove \(250 \,\)km in a van at \(50 \,\) km/h and finally cycled home at \(20 \,\) km/h. Find the average speed for the whole journey.
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Answer.
\(\text{Average speed} = \frac{60 + 250 + 60}{3 + \frac{250}{50} + \frac{60}{20}} \)
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\(= \frac{370}{11} \, \text{km/h}\)
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2.

If a sprinter runs \(100 \text{ m}\) in \(10 \text{ seconds}\text{,}\) what is his average velocity?
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Answer.
\(10 \, \text{m/s}\)
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3.

Sarah is rushing to school. She walks the first \(500 \text{ meters} \,\) at \(5 \, \text{km/h}\text{,}\) realizing she’s late, she then sprints the next \(300 \text{ meters} \,\) at \(10 \, \text{km/h}\) What was her average speed for the entire trip to school?
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Answer.
\(\text{Average speed} = \frac{0.5 + 0.3}{\frac{0.5}{5} + \frac{0.3}{10}} \)
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\(= \frac{80}{13} \ \, \text{km/h}\)
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4.

The world record for the men’s marathon is \(2:03:38\text{.}\) If the distance is \(42.195 \, \text{km}\text{,}\) what is the average velocity during the run?
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Answer.
\(5.69 \, \text{m/s}\)
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5.

John walks to the store, a distance of \(1 \, \text{km}\text{,}\) at a speed of \(4 \, \text{km/hr}\text{.}\) He spends \(15 \, \text{minutes}\) at the store, and then walks back at \(5 \, \text{km/hr}\text{.}\) What was John’s average speed for the entire trip, including the time spent at the store
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Answer.
\(\text{Average speed} = \frac{1 + 1}{\frac{1}{4} + \frac{1}{5} + \frac{15}{60}} \)
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\(= \frac{20}{7} \ \, \text{km/h}\)
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6.

A hiker covered \(12 \,\text{ km in}\, 3\, \text{ hours}\text{.}\) After taking a \(15 \, \text{ minute}\) break, what speed must the hiker maintain to reach their destination within a total travel time of \(4 \, \text{ hours}\text{?}\)
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Answer.
\(\text{Required speed} = \frac{12 \text{ km}}{4 \text{ hours} - 3 \text{ hours} - \frac{15 }{60} \text{hours}} \)
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\(= 16 \, \text{km/h}\)
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7.

A runner completed \(10 \, \text{ km in} 1.5 \, \text{ hours}\text{.}\) After a \(\text{5 minute}\) rest, what pace does the runner need to maintain to finish a total distance within \(\text{2 hours}\text{?}\)
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Answer.
\(\text{Required pace} = \frac{10 \text{ km}}{2 \text{ hours} - 1.5 \text{ hours} - \frac{5}{60} \text{hours}}\)
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\(= 24 \, \text{km/h}\)
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8.

A train journey consisted of three segments. The train traveled for \(3 \text{ hours at} \, 90 \text{ km/h}\text{,}\) then paused for \(0.75 \text{ hours}\) at a station, and finally continued for \(2 \text{ hours at} \, 70 \text{ km/h}\text{.}\) What was the average speed of the train for the entire journey?
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Answer.
\(\text{Average speed} = \frac{(3 \times 90) + (2 \times 70)}{3 + 0.75 + 2} \)
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\(= \frac{1640}{23} \, \text{km/h}\)
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9.

A sailboat is traveling north at \(10 \, \text{km/h}\text{,}\) relative to the water. The water is flowing north at \(5 \, \text{km/h}\text{.}\) What is the velocity of the boat relative to ground?
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Answer.
\(15 \, \text{km/h}\)
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