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Grade 10 Mathematics Teachers’ Book

INNODEMS

Subsection 2.10.6 Velocity-Time Graphs

Learner Experience 2.10.15.

Work in groups
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  1. 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 \(4 \,\text{ m/s²}\) northward.
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    Find the velocity of the train:
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    • \(3\) seconds after \(5:30\, \text{ PM}\)
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    • \(6\) seconds after \(5:30\, \text{ PM}\)
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    • \(2\) seconds after \(5:30\, \text{ PM}\)
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  2. Determine the average velocity of the train:
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    • In the first \(5 \,\text{ seconds}\) after \(5:30\, \text{ PM}\)
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    • In the first \(10 \,\text{ seconds}\) after \(5:30\, \text{ PM}\)
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  3. Draw a velocity-time graph, find the distance of the train from R at \(12\) seconds past \(5:30\, \text{ PM}\)
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Key Takeaway 2.10.56.

A velocity-time graph is a graph that shows how an object’s velocity (speed with direction) changes over time.
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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.
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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.
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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.
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\begin{align*} \text{Acceleration} = \amp \frac{\text{Change in velocity}}{\text{Corresponding change in time}}\\ \text{a} = \amp \frac{\text{Δv}}{\text{Δt}}\\ \text{a} = \amp \frac{\text{v - u}}{\text{t}} \end{align*}
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Where;
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  • \(\text{a}\) = Acceleration which is measured in \(\text{m/s²}\)
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  • \(\text{Δv}\) = \(\text{v - u}\) = change in velocity which is measured in metres per second \(\text{m/s}\)
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    \(\text{v}\) is the final velocity and \(\text{u}\) is the initial velocity.
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  • \(Δt\) = \(\text{t}\) = time taken which is measured in seconds(s)
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Deceleration is negative acceleration, meaning the object is slowing down. It happens when the velocity decreases over time.
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  • In a velocity-time graph, deceleration appears as a downward-sloping line.
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Example 2.10.57.

Given the table below
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Table 2.10.58.
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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.
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Solution.
This is a velocity-time graph that shows Acceleration, Constant Speed, and Deceleration.
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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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  1. Draw the velocity-time graph to represent its motion.
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  2. Find the total distance covered.
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  3. Determine the average velocity.
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  4. Calculate the acceleration and deceleration.
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Solution.
  1. The velocity-time graph;
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  2. The total distance is the area under the velocity-time graph, which consists of a trapezium.
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    • Distance during acceleration (triangle).
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      \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)
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      \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)
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      \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:
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      \(200+600+400\) = \(1200\,\text{ m}\)
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      Therefore, the total distance is \(1200\,\text{ m}\)
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  3. Average Velocity
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    \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*}
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    Therefore, the average velocity is \(26.67\,\text{ m/s}\)
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  4. Acceleration and Deceleration
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    Using the equation:
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    \begin{align*} \text{a} = \amp \frac{\text{v - u}}{\text{t}} \end{align*}
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    Where;
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    \(\text{v}\) = is the final velocity (\(40 \, \text{ m/s}\text{,}\) from the graph)
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    \(\text{u}\) = is the initial velocity (\(0 \, \text{ m/s}\))
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    \(\text{t}\) = is the time taken during acceleration (\(10 \, \text{ seconds}\)).
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    \begin{align*} \amp \frac{40 - 0}{10}\\ = \amp 4 \,\text{ m/s²} \end{align*}
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    Therefore, acceleration is \(4 \,\text{ m/s²}\)
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    Under deceleration;
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    \(\text{v}\) is the final velocity (\(0 \, \text{ m/s}\text{,}\) since the car comes to rest),
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    \(\text{u}\) is the initial velocity (\(40 \, \text{ m/s}\text{,}\) from the graph)
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    \(\text{t}\) is the time taken during deceleration (\(10 \, \text{ seconds}\)).
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    \begin{align*} \amp \frac{0 - 40}{10}\\ = \amp -4 \,\text{ m/s²} \end{align*}
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    Therefore, the object is decelerating at \(4 \,\text{ m/s²}\)
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Checkpoint 2.10.60.

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Checkpoint 2.10.61.

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Exercises Exercises

1.

Given the table below
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Table 2.10.62.
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Time \(0\) \(1\) \(2\) \(3\) \(4\) \(5\) \(6\) \(7\)
Velocity (m/s) \(0\) \(5\) \(15\) \(30\) \(45\) \(50\) \(55\) \(60\)
  1. Draw the velocity time graph to represent the data
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  2. Use the graph to describe the motion of the vehicle. Is the velocity constant, increasing, or decreasing? Explain your answer.
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  3. Calculate the average velocity of the vehicle between \(\text{t}\) = \(0 \text{ s}\) and \(\text{t}\) = \(3 \text{ s}\)
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Answer.
  1. The velocity-time graph;
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  2. The motion of the vehicle is increasing as the velocity is increasing with time.
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  3. Average Velocity from t = \(0 \text{ s}\) to \(\text{t}\) = \(3 \text{ s}\)
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    \begin{align*} = \amp \frac{(2.5+10+22.5)}{3}\\ \amp = 11.67 \,\text{ m/s} \end{align*}
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    Therefore, the average velocity is \(11.67\,\text{ m/s}\)
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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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  1. Draw the velocity-time graph to represent its motion.
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  2. Find the total distance covered.
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  3. Determine the average velocity.
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  4. Calculate the acceleration and deceleration.
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Answer.
  1. The velocity-time graph;
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  2. The total distance is the area under the velocity-time graph, which consists of a trapezium.
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    \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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    Total distance = \(120+360+150\) = \(630\,\text{ m}\)
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  3. Average Velocity \(= 21 \, \text{ m/s}\)
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  4. The acceleration is calculated as: \(a = \frac{30 - 0}{8} = 3.75 \, \text{ m/s}^2\text{.}\)
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    The deceleration is calculated as: \(\frac{0 - 30}{10} = -3 \, \text{ m/s}^2\text{.}\)
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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 \(\text{t}\) = \(1800 \,\text{ seconds}\text{.}\)
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Answer.
The velocity-time graph;
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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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  1. Draw the Velocity-Time Graph.
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  2. Calculate the acceleration of the cyclist.
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  3. What type of motion is represented by the velocity-time graph you drew?
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Answer.
  1. The velocity-time graph;
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  2. The acceleration is calculated as: \(a = \frac{20 - 5}{15} = 1 \, \text{ m/s}^2\text{.}\)
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  3. The motion represented by the velocity-time graph is uniformly accelerated motion since the cyclist’s velocity increases at a constant rate over time.
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