1. Title
The Study of Motion
Kinematics (kinema-tics rhymes with cinema-tics) is the study of Motion (without worrying about the force that might have caused the motion)
There movie explained that there three cases we need to consider and the methods we use depend very much on the type of motion we are investigating:
1) Steady-speed
Referring specifically to a single period of steady speed motion
If any change occurs to the speed during the period being investigated, then revert to CASE 2
2) Phases of Acceleration, Steady-speed and/or Deceleration
Any situation involving non-uniform acceleration (or if the acceleration ceases/changes)
When using CASE 3, it doesn't matter if the acceleration changes during the period under investigation
3) Uniform Acceleration
Referring specifically to a single period of constant acceleration or deceleration
If any ''change' in the acceleration occurs during the period under consideration, we must revert to CASE 2
NOTE: I've swapped CASE 2 and CASE 3 around
(compared to the order in the movie)...
Each of these 3 situations needs to be dealt with differently:
Case 1: Steady Speed
For a single period of steady speed motion, use: \(speed\;=\;\frac{distance}{time}\)
Allow the units of the SPEED to dictate the units of distance and time
Let's check your understanding of CASE 1...
Question 1
A train travels at 12 m/s for 1 hour and 15 minutes. Find the distance travelled. Give your answer in km.
I know it says, "Give your answer in km", but we must let the units of the speed dictate the units of the ‘distance’ and the units of the ‘time’
So the units of distance will be ‘metres’ ─────┐ ▼ SPEED = 12 m/s ▲ └──── and the units of time will be ‘seconds’
As the speed is in metres per second, so our units for distance must be metres and our units for time must be seconds:
╒══════════════════════════════════╕ │ time = 1 hour 15 minutes │ │ └───┬──┘ │ │ ┌───────┴──┐ │ │ = …… minutes + 15 minutes │ │ └───────┬───────────────┘ │ │ = ……… minutes │ │ = ……… × 60 seconds │ ╖ │ = ………… seconds ├───► time = ………… s ║ ╘══════════════════════════════════╛ speed = 12 m/s ╟ speed = distance distance = ??? ║ time ╜ 12 = distance ……… ………… m = distance Now convert this into kilometres as required: …… km = distance
Question 2
Question 2: a) Apply the method explained in the movie to write \(y\; = \;{x}^{2} \, \color{#bd398c}{+\,10}x \, \color{#3C7850}{+\,12}\) as a completed square
b) Use the GEOGEBRA APPLET (above) to produce a diagram which explains your completed square geometrically
<p>Question 2: a) Apply the method explained in the movie to write \(y\; = \;{x}^{2} \, \color{#bd398c}{+\,10}x \, \color{#3C7850}{+\,12}\) as a completed square<br>
b) Use the GEOGEBRA APPLET <i><font color="#9499B1">(above)</font></i> to produce a diagram which explains your completed square geometrically</p>
<pre>Clue: <font color="#B0DEA9">╔════════════════════════════════════════╗</font>
<font color="#B0DEA9">╒══</font><font color="#B0DEA9">╩═╕</font> <font color="#B0DEA9">║</font>
x² + 10x <font color="#62BE53">+</font><font color="#62BE53"> 12</font> <font color="#B0DEA9">║</font>
<font color="#E8B0D3">╘═╦══╛</font> <font color="#B0DEA9">║</font>
<font color="#E8B0D3">╒══════╩════════════════════╕</font> <font color="#B0DEA9">║</font>
<font color="#E8B0D3">│</font>Start by <font color="#2B83C3">halving</font> <font color="#BD398C">the number</font><font color="#E8B0D3">│</font> (x )² <font color="#B0DEA9">║</font>
<font color="#E8B0D3">│</font><font color="#BD398C">in-front of the x term</font> and <font color="#E8B0D3">├—————————————┐</font> <font color="#B0DEA9">║</font>
<font color="#E8B0D3">│</font>put into the brackets: <font color="#E8B0D3">│ </font> (x <font color="#BD398C">+ 5</font>)² <font color="#B0DEA9">║</font>
<font color="#E8B0D3">╘═══════════════════════════╛</font> <font color="#B0DEA9">║</font>
Then subtract: (<font color="#BD398C">5</font>)² (x <font color="#BD398C">+ 5</font>)² - (<font color="#BD398C">5</font>)² <font color="#B0DEA9">║
║</font>
Simplify: (x <font color="#BD398C">+ 5</font>)² - <font color="#BD398C">...</font> <font color="#62BE53"> </font><font color="#B0DEA9">║</font>
<font color="#B0DEA9">▼</font>
Now bring down the units term: <font color="#62BE53">+</font><font color="#62BE53">12</font> (x <font color="#BD398C">+ 5</font>)² - <font color="#BD398C">...</font> <font color="#62BE53">+</font><font color="#62BE53"> ...</font>
<font color="#9499B1">╘════╤══════╛
┌────┴─────┐
│ simplify │
└────┬─────┘
╒══╧═══╕</font>
The answer is: <font color="#2B83C3">(x + 5)² - ...</font>
To show this GEOMETRICALLY, click on the GEOGEBRA applet (above) and then edit the numbers
in the BLUE box and the RED box.
Then keep clicking NEXT <font color="#E54239">(if it says PAUSE but nothing is happening, CLICK-AGAIN)</font>
Then you'll be able to produce a drawing that explains your completed square</pre>
Question 3
Question 3: Write \(y\;=\;2{x}^{2}\,\color{#bd398c}{-\,12}x\,\color{#3C7850}{+\,10}\) as a completed square
We cannot complete the square if the coefficient of x² ≠ 1 ┌ ┌─────────────┐ ┐ So FIRST, we need to factorise the ‘2’ out: 2│ │ x² - …x + … │ │ └ └──────┬──────┘ ┘ COMPLETE-SQUARE FOR BIT IN BRACKETS ┌──────────────────┘ │ Halve the coefficient of ‘x’ ┌─────┴─────────────┐ and put into the brackets: │(x - 3)² │ │ │ Then subtract: (3)² │(x - 3)² - (3)² │ │ │ Simplify: │(x - 3)² - …… │ │ │ Then add the units term: +10 │(x - 3)² - …… + …… │ │ │ Simplify: │(x - 3)² + …… │ └─────┬─────────────┘ └──────────────────┐ ┌ ┌──────┴──────┐ ┐ RE-INSERT INTO THE SQUARE-BRACKETS 2│ │ (x-3)² + …… │ │ └ └─────────────┘ ┘ Finally, multiply out: 2(x - 3)² + ……
Question 3
Question 3: Write \(y\;=\;2{x}^{2}\,\color{#bd398c}{-\,12}x\,\color{#3C7850}{+\,10}\) as a completed square
We cannot complete the square if the coefficient of x² ≠ 1 ┌ ┌─────────────┐ ┐ So FIRST, we need to factorise the ‘2’ out: 2│ │ x² - …x + … │ │ └ └──────┬──────┘ ┘ COMPLETE-SQUARE FOR BIT IN BRACKETS ┌──────────────────┘ │ Halve the coefficient of ‘x’ ┌─────┴─────────────┐ and put into the brackets: │(x - 3)² │ │ │ Then subtract: (3)² │(x - 3)² - (3)² │ │ │ Simplify: │(x - 3)² - …… │ │ │ Then add the units term: +10 │(x - 3)² - …… + …… │ │ │ Simplify: │(x - 3)² + …… │ └─────┬─────────────┘ └──────────────────┐ ┌ ┌──────┴──────┐ ┐ RE-INSERT INTO THE SQUARE-BRACKETS 2│ │ (x-3)² + …… │ │ └ └─────────────┘ ┘ Finally, multiply out: 2(x - 3)² + ……