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Subsection 2.9.5 Velocity Time Graph
Activity 2.9.6 .
\(\textbf{Work in groups}\)
A train moving at
\(40 \text{ m/s}\) along a North-South railway track passes through a station R at
\(5:30 \textbf{ PM}\text{.}\) The train is decelerating at
\(4 \textbf{ m/s²}\) northward.
Find the velocity of the train:
\(3 \textbf{ seconds}\) after
\(5:30 \textbf{ PM}\)
\(6 \textbf{ seconds}\) after
\(5:30 \textbf{ PM}\)
\(2 \textbf{ seconds}\) after
\(5:30 \textbf{ PM}\)
Determine the average velocity of the train:
In the first
\(5 \textbf{ seconds}\) after
\(5:30 \textbf{ PM}\)
In the first
\(10 \textbf{ seconds}\) after
\(5:30 \textbf{ PM}\)
Draw a velocity-time graph, find the distance of the train from R at
\(12\) seconds past
\(5:30 \textbf{ PM}\)
\(\textbf{Key Takeaway}\)
\(\textbf{A velocity-time graph}\) is a graph that shows how an object’s 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.
\(\textbf{Acceleration}\) 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*}
\textbf{Acceleration} = \amp \frac{\textbf{Change in velocity}}{\textbf{Corresponding change in time}}\\
\textbf{a} = \amp \frac{\textbf{Δv}}{\textbf{Δt}}\\
\textbf{a} = \amp \frac{\textbf{v - u}}{\textbf{t}}
\end{align*}
\(\textbf{a}\) = Acceleration which is measured in
\(\textbf{m/s²}\)
\(\textbf{Δv}\) =
\(\textbf{v - u}\) = change in velocity which is measured in metres per second
\(\textbf{m/s}\)
\(\textbf{v}\) is the final velocity and
\(\textbf{u}\) is the initial velocity.
\(\textbf{Δt}\) =
\(\textbf{t}\) = time taken which is measured in seconds(s)
\(\textbf{Deceleration}\) is negative acceleration, meaning the object is slowing down. It happens when the velocity decreases over time.
In a velocity-time graph, deceleration appears as a downward-sloping line.
Example 2.9.15 .
Table 2.9.16. Draw the velocity-time graph to represent the data.
Solution .
This is a velocity-time graph that shows Acceleration, Constant Speed, and Deceleration.
Example 2.9.17 .
A car starts from rest and accelerates to a velocity of
\(40 \textbf{ m/s}\) in
\(10 \textbf{ seconds}\) . It then maintains this velocity for
\(15 \textbf{ seconds}\) before decelerating to rest, with the total time of motion being
\(45 \textbf{ seconds}\)
Draw the velocity-time graph to represent its motion.
Find the total distance covered.
Determine the average velocity.
Calculate the acceleration and deceleration.
Solution .
The total distance is the area under the velocity-time graph, which consists of a trapezium.
Distance during acceleration (triangle).
\begin{align*}
\textbf{Area} = \amp \frac{1}{2}\times \textbf{ base}\times \textbf{ height}\\
\amp \frac{1}{2} \times 10 \times 40\\
= \amp 200\textbf{ m}
\end{align*}
Distance during constant velocity (rectangle)
\begin{align*}
\textbf{Area} = \textbf{base} \times \textbf{ height}\amp \\
\amp 15 \times 40\\
= \amp 600\textbf{ m}
\end{align*}
Distance during deceleration (triangle)
\begin{align*}
\textbf{Area} = \amp \frac{1}{2}\times \textbf{ base}\times \textbf{ height}\\
\amp 20 \times 40\\
= \amp 400\textbf{ m}
\end{align*}
\(200+600+400\) =
\(1200\textbf{ m}\)
Therefore, the total distance is
\(1200\textbf{ m}\)
\begin{align*}
\textbf{Average velocity} = \amp \frac{\textbf{Total distance}}{\textbf{ Total time}}\\
= \amp \frac{1200}{45}\\
\amp = 26.67\textbf{ m/s}
\end{align*}
Therefore, the average velocity is
\(26.67\textbf{ m/s}\)
Acceleration and Deceleration
\begin{align*}
\textbf{a} = \amp \frac{\textbf{v - u}}{\textbf{t}}
\end{align*}
\(\textbf{v}\) = is the final velocity (
\(40 \textbf{ m/s}\text{,}\) from the graph)
\(\textbf{u}\) = is the initial velocity (
\(0 \textbf{ m/s}\) )
\(\textbf{t}\) = is the time taken during acceleration (
\(10 \textbf{ seconds}\) ).
\begin{align*}
\amp \frac{40 - 0}{10}\\
= \amp 4\textbf{ m/s²}
\end{align*}
Therefore, acceleration is
\(4\textbf{ m/s²}\)
\(\textbf{v}\) is the final velocity (
\(0 \textbf{ m/s}\text{,}\) since the car comes to rest),
\(\textbf{u}\) is the initial velocity (
\(40 \textbf{ m/s}\text{,}\) from the graph)
\(\textbf{t}\) is the time taken during deceleration (
\(10 \textbf{ seconds}\) ).
\begin{align*}
\amp \frac{0 - 40}{10}\\
= \amp -4\textbf{ m/s²}
\end{align*}
Therefore, the object is decelerating at
\(4\textbf{ m/s²}\)
Exercises Exercises
1. Table 2.9.18.
Draw the velocity time graph to represent the data
Use the graph to describe the motion of the vehicle. Is the velocity constant, increasing, or decreasing? Explain your answer.
Calculate the average velocity of the vehicle between
\(\textbf{t}\) =
\(0 \textbf{ s}\) and
\(\textbf{t}\) =
\(3 \textbf{ s}\)
2. A motorcycle starts from rest and accelerates uniformly to a speed of
\(30 \textbf{ m/s}\) in
\(8 \textbf{ seconds}\text{.}\) It then continues at this speed for
\(12 \textbf{ seconds}\) before decelerating uniformly to rest over
\(10 \textbf{ seconds}\text{.}\) The total time of motion is
\(30 \textbf{ seconds}\)
Draw the velocity-time graph to represent its motion.
Find the total distance covered.
Determine the average velocity.
Calculate the acceleration and deceleration.
3. After takeoff, an airplane reaches a cruising speed of
\(250 \textbf{ m/s}\) and maintains it for
\(30 \textbf{ minutes}\text{.}\) Draw a velocity-time graph representing the motion of the airplane from
\(\textbf{t}\) =
\(0\) to
\(\textbf{t}\) =
\(1800 \textbf{ seconds}\text{.}\)
4. A cyclist starts at
\(5 \textbf{ m/s}\) and increases speed to
\(20 \textbf{ m/s}\) in
\(15 \textbf{ sconds}\text{.}\)
Draw the Velocity-Time Graph.
Calculate the acceleration of the cyclist.
What type of motion is represented by the velocity-time graph you drew?
Checkpoint 2.9.19 .
