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Grade 10 Mathematics Teachers’ Book

INNODEMS

Subsection 2.10.4 Acceleration

Learner Experience 2.10.10.

Work in groups
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  1. Define
    1. Velocity
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    2. Velocity
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  2. A car travels \(200\) meters in \(10\) seconds. What is its average velocity?
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  3. A car accelerates from \(75\) km/h to \(90\) km/h in \(10\) seconds.
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    Find it’s acceleration
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  4. A vehicle moving at \(\text{25 m/s}\) applies brakes and comes to a stop in \(\text{5 seconds}\text{.}\)
    1. What is the acceleration of the vehicle?
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    2. Is the acceleration positive or negative? Why?
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Exploration 2.10.11. Motion with Constant Acceleration.

In this activity, you will explore how the motion of an object changes when it moves with constant acceleration. The simulation shows a car moving along a straight road. You can adjust the initial velocity (\(u\)) and the acceleration (\(a\)) using the sliders. As the car moves, the simulation displays its position, velocity, and distance traveled.
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Use the controls to investigate how acceleration changes the motion of the car and how the equations of motion describe this behaviour.
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Instructions.
  1. Use the slider labeled “Initial velocity” to set the starting speed of the car.
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  2. Use the slider labeled “Acceleration” to control how quickly the car’s speed changes.
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  3. Click Start to begin the animation.
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  4. Observe the arrow above the car. This arrow represents the velocity of the car and changes length as the speed increases.
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  5. Watch the dotted line and the label above the car to see how the distance traveled changes with time.
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  6. Click Pause at any time to examine the car’s position, velocity, and time.
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  7. Click Resume to continue the motion.
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  8. Click Reset to restart the simulation and try a different combination of initial velocity and acceleration.
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Use the simulation to investigate the following questions:
  1. Set the acceleration to \(0\text{.}\) Start the animation. How does the car move when there is no acceleration? What happens to the velocity arrow?
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  2. Keep the initial velocity the same but increase the acceleration. How does the speed of the car change over time?
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  3. Pause the animation after a few seconds. Compare the values of velocity, time, and position. How does acceleration influence how quickly the velocity increases?
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  4. Try a larger acceleration value. How does this affect how quickly the car moves across the screen?
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  5. Based on your observations, explain why the distance traveled by the car depends on both the initial velocity and the acceleration.
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Figure 2.10.36. Interactive simulation of motion with constant acceleration. Adjust the initial velocity and acceleration to observe how the car’s position and velocity change over time.
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Key Takeaway 2.10.37.

Acceleration - The rate of change of velocity with time. SI unit is \(m/s²\)
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Acceleration is given by
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\begin{gather*} \text{Acceleration} \, = \frac{\text{Change in velocity}}{\text{Time Taken}} \end{gather*}
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\begin{gather*} a = \frac{\Delta v}{\Delta t} \end{gather*}
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Negative acceleration is called deceleration or retardation
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For motion with constant acceleration, the three equations of motion are:
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  1. \(\text{final velocity}\) = Initial velocity + (acceleration \(\times \) time)
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    \(v = u + at\)
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    where;
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  2. Displacement = \((\text{initial velocity} \times \text{time}) + (\frac{1}{2} \times \text{acceleration} \times \text{time}^2) \)
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    \(s = ut + \frac{1}{2} at^2 \)
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  3. \(\text{Final velocity}^2 = \text{Initial velocity}^2\) + (2 \(\times\) acceleration \(\times\) displacement)
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    \(v^2 = u^2 + \, (2 \times \textbf{acceleration}) \times s^2\)
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Note 2.10.38.

Deceleration is donated by a negative sign in the acceleration value. For example: if the deceleration is \(5 \, \text{m/s}^2\text{,}\) then it’s acceleration is \(-5 \, \text{m/s}^2\text{.}\)
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Example 2.10.39.

A car starts from rest and accelerates at 2 m/s² for 5 seconds. Find its final velocity.
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Solution.
\(v = u + at \)
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\(= 0 + (2 \times 5) = 10 \, \text{ m/s}\)
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The final velocity is \(10 \, \text{ m/s}\)
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Example 2.10.40.

A rocket accelerates from rest to \(250\) m/s in \(10\) seconds. Calculate the rocket’s acceleration.
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Solution.
Initial velocity \(u = 0 \)m/s
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Final velocity \(v = 250 \)m/s
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Time (t) = \(10 \)seconds
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Acceleration (a) = (v - u) / t
\begin{gather*} \frac{(v - u)}{t} \end{gather*}
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\begin{gather*} a = \frac{(250- 0 )\,\text{m/s}}{10 s} \end{gather*}
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\begin{gather*} a = \frac{(250 \, \text{m/s})}{10 \, \text{s}} \end{gather*}
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\begin{gather*} a = 25 \, \text{m/s}^2 \end{gather*}
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The rocket’s acceleration is \(25 \, \text{m/s}^2\)
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Example 2.10.41.

An object decelerates from 20 m/s to 5 m/s in 3 seconds. What is its acceleration?
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Solution.
Initial velocity \(u = 20 \)m/s
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Final velocity \(v = 5 \)m/s
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Time (t) = \(3 \)seconds
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Acceleration (a) = (v - u) / t
\begin{gather*} \frac{(v - u)}{t} \end{gather*}
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\begin{gather*} a = \frac{(5 - 20 ) \, \text{m/s}}{3 \, \text{seconds}} \end{gather*}
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\begin{gather*} a = \frac{(-15 \, \text{m/s})}{3 \, \text{seconds}} \end{gather*}
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\begin{gather*} a = -5 \, \text{m/s}^2 \end{gather*}
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The object’s acceleration is \(-5 \, \text{m/s}^2\text{.}\) The negative sign indicates deceleration (slowing down).
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Exercises Exercises

1.

A car’s velocity changes from \(10 \, \text{m/s}\) to \(30 \, \text{m/s}\) in \(4\) seconds. Find its acceleration.
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Answer.
\(a = \frac{(30 \, \text{m/s} - 10 \, \text{m/s})}{4 \, \text{s}} = 5 \, \text{m/s}^2 \)
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2.

A car moves with \(8 \, \text{m/s}^2\) acceleration for \(5\) seconds, reaching \(40 \, \text{m/s}\text{.}\) Find its initial velocity.
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Answer.
\(u = 40 \, \text{m/s} - (8 \, \text{m/s}^2 \times 5 \, \text{s}) = 0 \, \text{m/s} \)
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3.

A train moving at \(40 \, \text{km/h} \) decelerates at \(0.5 \, \text{m/s}^2\text{.}\) Find the time taken to stop
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Answer.
\(t = \frac{(0 \, \text{m/s} - 11.11 \, \text{m/s})}{-0.5 \, \text{m/s}^2} = 22.22 \, s \)
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4.

A cyclist’s speed increases from \(5\) m/s to \(17\) m/s over a period of \(6\) seconds. What is the cyclist’s average acceleration?
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Answer.
\(a = \frac{(17 \, \text{m/s} - 5 \, \text{m/s})}{6 \, \text{s}} = 2 \, \text{m/s}^2 \)
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5.

A runner’s velocity changes from \(3\) m/s to \(7\) m/s. What is the runner’s average velocity?
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Answer.
\(\text{Average velocity} = \frac{(3 \, \text{m/s} + 7 \, \text{m/s})}{2} = 5 \, \text{m/s} \)
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Checkpoint 2.10.42.

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Checkpoint 2.10.43.

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