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Subsection 2.10.4 Acceleration

Learner Experience 2.10.10.

Work in groups
  1. Define
    1. Velocity
    2. Velocity
  2. A car travels \(200\) meters in \(10\) seconds. What is its average velocity?
  3. A car accelerates from \(75\) km/h to \(90\) km/h in \(10\) seconds.
    Find it’s acceleration
  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?
    2. Is the acceleration positive or negative? Why?

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.
Use the controls to investigate how acceleration changes the motion of the car and how the equations of motion describe this behaviour.

Instructions.

  1. Use the slider labeled β€œInitial velocity” to set the starting speed of the car.
  2. Use the slider labeled β€œAcceleration” to control how quickly the car’s speed changes.
  3. Click Start to begin the animation.
  4. Observe the arrow above the car. This arrow represents the velocity of the car and changes length as the speed increases.
  5. Watch the dotted line and the label above the car to see how the distance traveled changes with time.
  6. Click Pause at any time to examine the car’s position, velocity, and time.
  7. Click Resume to continue the motion.
  8. Click Reset to restart the simulation and try a different combination of initial velocity and acceleration.
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?
  2. Keep the initial velocity the same but increase the acceleration. How does the speed of the car change over time?
  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?
  4. Try a larger acceleration value. How does this affect how quickly the car moves across the screen?
  5. Based on your observations, explain why the distance traveled by the car depends on both the initial velocity and the acceleration.
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.

Key Takeaway 2.10.37.

Acceleration - The rate of change of velocity with time. SI unit is \(m/sΒ²\)
Acceleration is given by
\begin{gather*} \text{Acceleration} \, = \frac{\text{Change in velocity}}{\text{Time Taken}} \end{gather*}
\begin{gather*} a = \frac{\Delta v}{\Delta t} \end{gather*}
Negative acceleration is called deceleration or retardation
For motion with constant acceleration, the three equations of motion are:
  1. \(\text{final velocity}\) = Initial velocity + (acceleration \(\times \) time)
    \(v = u + at\)
    where;
  2. Displacement = \((\text{initial velocity} \times \text{time}) + (\frac{1}{2} \times \text{acceleration} \times \text{time}^2) \)
    \(s = ut + \frac{1}{2} at^2 \)
  3. \(\text{Final velocity}^2 = \text{Initial velocity}^2\) + (2 \(\times\) acceleration \(\times\) displacement)
    \(v^2 = u^2 + \, (2 \times \textbf{acceleration}) \times s^2\)

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 \(-5 \, \text{m/s}^2\text{.}\)

Example 2.10.39.

A car starts from rest and accelerates at 2 m/sΒ² for 5 seconds. Find its final velocity.
Solution.
\(v = u + at \)
\(= 0 + (2 \times 5) = 10 \, \text{ m/s}\)
The final velocity is \(10 \, \text{ m/s}\)

Example 2.10.40.

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

Example 2.10.41.

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

Exercises Exercises

1.

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

2.

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

3.

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

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?
Answer.
\(a = \frac{(17 \, \text{m/s} - 5 \, \text{m/s})}{6 \, \text{s}} = 2 \, \text{m/s}^2 \)

5.

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