Curriculum Alignment
- Strand
- 2.0 Measurements and Geometry
- Sub-Strand
- 2.10 Linear Motion
- Specific Learning Outcomes
- Draw a velocity-time graph from given tables in different situations and interpret velocity-time graphs in different situations.
Subsection 2.10.6 Velocity-Time Graphs
Teacher Resource 2.10.55.
To assist your teaching, we have prepared lesson resources, aligned with this textbook and the CBC. The Lesson Plan links to syllabus learning outcomes and provides suggest time allocations. The Step-by-Step Guide provides more detailed guidance on how to teach the content, including suggested questions to ask learners, and possible answers.
Learner Experience 2.10.15.
Work in groups
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A train moving at \(40 \,\text{ m/s}\) along a North-South railway track passes through a station R at \(5:30\, \text{ PM}\text{.}\) The train is decelerating at \(4 \,\text{ m/sΒ²}\) northward.Find the velocity of the train:
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Determine the average velocity of the train:
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Draw a velocity-time graph, find the distance of the train from R at \(12\) seconds past \(5:30\, \text{ PM}\)
Key Takeaway 2.10.56.
A velocity-time graph is a graph that shows how an objectβs velocity (speed with direction) changes over time.
A velocity-time graph displays velocity on the y-axis and time on the x-axis, where the slope indicates acceleration and the area under the graph represents displacement.
A horizontal line signifies constant velocity, an upward slope indicates acceleration, a downward slope indicates deceleration, and a line at zero velocity means the object is stationary.
Acceleration is the rate of change of velocity of an object over time. It occurs when an object speeds up, slows down, or changes direction.
\begin{align*}
\text{Acceleration} = \amp \frac{\text{Change in velocity}}{\text{Corresponding change in time}}\\
\text{a} = \amp \frac{\text{Ξv}}{\text{Ξt}}\\
\text{a} = \amp \frac{\text{v - u}}{\text{t}}
\end{align*}
Where;
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\(\text{Ξv}\) = \(\text{v - u}\) = change in velocity which is measured in metres per second \(\text{m/s}\)
Deceleration is negative acceleration, meaning the object is slowing down. It happens when the velocity decreases over time.
Example 2.10.57.
Given the table below
| Time | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) |
|---|---|---|---|---|---|---|---|---|
| Velocity (m/s) | \(0\) | \(10\) | \(20\) | \(30\) | \(40\) | \(40\) | \(30\) | \(10\) |
Draw the velocity-time graph to represent the data.
Example 2.10.59.
A car starts from rest and accelerates to a velocity of \(40 \, \text{ m/s}\) in \(10 \,\text{ seconds}\) . It then maintains this velocity for \(15 \,\text{ seconds}\) before decelerating to rest, with the total time of motion being \(45 \,\text{ seconds}\)
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Draw the velocity-time graph to represent its motion.
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Find the total distance covered.
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Determine the average velocity.
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Calculate the acceleration and deceleration.
Solution.
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The velocity-time graph;
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The total distance is the area under the velocity-time graph, which consists of a trapezium.
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Distance during acceleration (triangle).\begin{align*} \text{Area} = \amp \frac{1}{2}\times \text{ base}\times \text{ height}\\ \amp \frac{1}{2} \times 10 \times 40\\ = \amp 200 \,\text{ m} \end{align*}
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Distance during constant velocity (rectangle)\begin{align*} \text{Area} = \text{base} \times \text{ height}\amp \\ \amp 15 \times 40\\ = \amp 600 \,\text{ m} \end{align*}
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Distance during deceleration (triangle)\begin{align*} \text{Area} = \amp \frac{1}{2}\times \text{ base}\times \text{ height}\\ \amp 20 \times 40\\ = \amp 400\,\text{ m} \end{align*}
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Total distance:Therefore, the total distance is \(1200\,\text{ m}\)
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Average Velocity\begin{align*} \text{Average velocity} = \amp \frac{\text{Total distance}}{\text{ Total time}}\\ = \amp \frac{1200}{45}\\ \amp = 26.67\,\text{ m/s} \end{align*}Therefore, the average velocity is \(26.67\,\text{ m/s}\)
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Acceleration and DecelerationUsing the equation:\begin{align*} \text{a} = \amp \frac{\text{v - u}}{\text{t}} \end{align*}Where;\begin{align*} \amp \frac{40 - 0}{10}\\ = \amp 4 \,\text{ m/sΒ²} \end{align*}Therefore, acceleration is \(4 \,\text{ m/sΒ²}\)Under deceleration;\begin{align*} \amp \frac{0 - 40}{10}\\ = \amp -4 \,\text{ m/sΒ²} \end{align*}Therefore, the object is decelerating at \(4 \,\text{ m/sΒ²}\)
Checkpoint 2.10.60.
Checkpoint 2.10.61.
Exercises Exercises
1.
Given the table below
| Time | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) |
|---|---|---|---|---|---|---|---|---|
| Velocity (m/s) | \(0\) | \(5\) | \(15\) | \(30\) | \(45\) | \(50\) | \(55\) | \(60\) |
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Draw the velocity time graph to represent the data
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Use the graph to describe the motion of the vehicle. Is the velocity constant, increasing, or decreasing? Explain your answer.
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Calculate the average velocity of the vehicle between \(\text{t}\) = \(0 \text{ s}\) and \(\text{t}\) = \(3 \text{ s}\)
Answer.
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The velocity-time graph;
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The motion of the vehicle is increasing as the velocity is increasing with time.
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\begin{align*} = \amp \frac{(2.5+10+22.5)}{3}\\ \amp = 11.67 \,\text{ m/s} \end{align*}Therefore, the average velocity is \(11.67\,\text{ m/s}\)
2.
A motorcycle starts from rest and accelerates uniformly to a speed of \(30 \,\text{ m/s}\) in \(8 \, \text{ seconds}\text{.}\) It then continues at this speed for \(12 \, \text{ seconds}\) before decelerating uniformly to rest in \(10 \, \text{ seconds}\text{.}\)
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Draw the velocity-time graph to represent its motion.
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Find the total distance covered.
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Determine the average velocity.
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Calculate the acceleration and deceleration.
Answer.
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The velocity-time graph;
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The total distance is the area under the velocity-time graph, which consists of a trapezium.\begin{gather*} = \frac{1}{2} \times 8 \times 30\\ = 120 \,\text{m}\\ = 30 \times 12\\ = 360 \,\text{ m}\\ = \frac{1}{2} \times 10 \times 30\\ = 150\,\text{ m} \end{gather*}
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Average Velocity \(= 21 \, \text{ m/s}\)
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The acceleration is calculated as: \(a = \frac{30 - 0}{8} = 3.75 \, \text{ m/s}^2\text{.}\)The deceleration is calculated as: \(\frac{0 - 30}{10} = -3 \, \text{ m/s}^2\text{.}\)
3.
After takeoff, an airplane reaches a cruising speed of \(250 \,\text{ m/s}\) and maintains it for \(30 \, \text{ minutes}\text{.}\) Draw a velocity-time graph representing the motion of the airplane from \(\text{t}\) = \(0\) to \(\text{t}\) = \(1800 \,\text{ seconds}\text{.}\)
4.
A cyclist starts at \(5 \,\text{ m/s}\) and increases speed to \(20 \,\text{ m/s}\) in \(15 \,\text{ seconds}\text{.}\)
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Draw the Velocity-Time Graph.
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Calculate the acceleration of the cyclist.
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What type of motion is represented by the velocity-time graph you drew?
Answer.
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The velocity-time graph;
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The acceleration is calculated as: \(a = \frac{20 - 5}{15} = 1 \, \text{ m/s}^2\text{.}\)
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The motion represented by the velocity-time graph is uniformly accelerated motion since the cyclistβs velocity increases at a constant rate over time.
