IB Mathematics AI AHL Vector applications to kinematics MAI Study Notes- New Syllabus
IB Mathematics AI AHL Vector applications to kinematics MAI Study Notes
LEARNING OBJECTIVE
- Vector applications to kinematics.
Key Concepts:
- Kinematics with Vectors
- Constant & Variable Velocity
- IBDP Maths AI SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IBDP Maths AI SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Maths AI HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Maths AI HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Maths AI HL- IB Style Practice Questions with Answer-Topic Wise-Paper 3
KINEMATICS
A nice application of the vector equation of a line is the following:
◆ VELOCITY AND SPEED
Suppose that a body is moving along a straight line with a constant velocity, and its position at time \( t \) is given by:
$\vec{r} = \vec{a} + t\vec{b}$
Then:
\(\vec{a}\) is the position of the body at time \( t = 0 \).
\(\vec{b}\) is the velocity vector of the body (usually denoted as \(\vec{v}\)).
\(|\vec{b}|\) is the speed of the body (usually denoted as \(|\vec{v}|\)).
The vectors (and thus the motion) can be in either 2D or 3D space.
Example Suppose that a body is moving according to the equation: $\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ The initial position (at \( t = 0 \)) ▶️Answer/ExplanationSolution: The initial position (at \( t = 0 \)) of the body is \((1, 2)\). It is \(\sqrt{1^2 + 2^2} = \sqrt{5} = 2.23 \, \text{m}\) far from the origin. |
NOTE
If \(\vec{r} = \vec{a} + \lambda \vec{b}\) is an equation of a line, the direction vector \(\vec{b}\) can be substituted by any multiple of \(\vec{b}\).
However, if \(\vec{r} = \vec{a} + t\vec{b}\) is an equation of motion, the velocity vector \(\vec{b}\) cannot be substituted by a multiple of \(\vec{b}\).
Explanation:
Suppose a body is initially at position \( A(1, 2) \) and after 1 second at position \( B(5, 8) \). Then:
Velocity vector: \(\vec{v} = \overrightarrow{AB} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}\).
Equation of motion: \(\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 4 \\ 6 \end{pmatrix}\).
Suppose the body is initially at \( A(1, 2) \) and after 2 seconds at \( B(5, 8) \). Then:
The direction vector \(\vec{b} = \overrightarrow{AB} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}\) corresponds to 2 seconds.
Velocity vector: \(\vec{v} = \frac{1}{2} \vec{b} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\).
Equation of motion: \(\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 2 \\ 3 \end{pmatrix}\).
Example Suppose that a body is moving on a straight line (in 3D space) in the direction of the vector \(\vec{b} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}\) with speed \( 15 \, \text{ms}^{-1} \). Its initial position is \( A(1, 1, 1) \). Find the equation of the motion of the body. ▶️Answer/ExplanationSolution: The magnitude of \(\vec{b}\) is \(|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = 3\). |
MOTION WITH VARIABLE VELOCITY IN 2D
Until now, the direction vector \(\vec{b}\) (or the velocity vector \(\vec{v}\)) was a constant vector. For example, if:
$\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 3 \\ 4 \end{pmatrix},$
the velocity vector is \(\vec{v} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\) (constant).
An alternative way to find the velocity vector is by using derivatives. If the motion is given by:
$\vec{r} = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix},$
then the velocity vector is:
$\vec{v} = \frac{d\vec{r}}{dt} = \begin{pmatrix} \dot{x}(t) \\ \dot{y}(t) \end{pmatrix}.$
Similarly, the acceleration is:
$\vec{a} = \frac{d\vec{v}}{dt} = \frac{d^2\vec{r}}{dt^2} = \begin{pmatrix} \ddot{x}(t) \\ \ddot{y}(t) \end{pmatrix}.$
In the case of constant velocity, \(\vec{a} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\).
Example The equation of motion for a body is given by: $\vec{r} = \begin{pmatrix} 1 + 3t \\ 2 + 4t + 5t^2 \end{pmatrix}.$ Find Velocity: \(\vec{v} = \). At \( t = 0 \): ▶️Answer/ExplanationSolution: Velocity: \(\vec{v} = \frac{d\vec{r}}{dt} = \begin{pmatrix} 3 \\ 4 + 10t \end{pmatrix}\). At \( t = 0 \): After 1 second: |
Example If: $\vec{r} = \begin{pmatrix} 1 + e^{2t} \\ 2 + t^3 \end{pmatrix},$ Find Velocity: ▶️Answer/ExplanationSolution: Velocity: \(\vec{v} = \frac{d\vec{r}}{dt} = \begin{pmatrix} 2e^{2t} \\ 3t^2 \end{pmatrix}\). |
◆ CIRCULAR MOTION
Suppose that:
$\vec{r} = \begin{pmatrix} r \cos \omega t \\ r \sin \omega t \end{pmatrix}.$
Then:
$x^2 + y^2 = (r \cos \omega t)^2 + (r \sin \omega t)^2 = r^2 (\cos^2 \omega t + \sin^2 \omega t) = r^2.$
The equation \( x^2 + y^2 = r^2 \) represents a circle with center \( O(0, 0) \) and radius \( r \).
Thus, the body is moving on a circular path.
Example If: $\vec{r} = \begin{pmatrix} 2 \cos 0.1t \\ 2 \sin 0.1t \end{pmatrix},$ Find Path: ▶️Answer/ExplanationSolution: Path: \( x^2 + y^2 = 4 \) (circle of radius 2). |
