Vectors in kinematics

In a nutshell

Position vectors are quantities that determine the position of one point in space, always relative to another point. Vectors are useful tools used to describe motion in a plane, as they have a direction and a magnitude.

Equations

DESCRIPTION

EQUATION

Movement with constant velocity.

$\bold r = \bold r_{\bold 0} + \bold v t$

Movement with constant acceleration.

$\bold r = \bold u t + \cfrac{1}{2} \bold a t^2$

$\bold v = \bold u + \bold a t$

Variable definitions

Quantity name	symbol	UNIT NAME	UNIT
$Displacement \ vector$	$\bold r$	$Metres$	$m$
$Initial \ displacement \ vector$	$\bold r_0$	$Metres$	$m$
$Velocity \ vector$	$\bold v$	$Metres \ per\ second$	$ms^{-1}$
$Initial \ velocity \ vector$	$\bold u$	$Metres \ per\ second$	$ms^{-1}$
$Acceleration \ vector$	$\bold a$	$Metres\ per \ second\ squared$	$ms^{-2}$
$Time$	$t$	$Seconds$	$s$

Vectors in kinematics

PROCEDURE

1.	Identify the variables and data in the question.
2.	Recall the relevant equations of motion.
3.	Group the terms according to the unit vectors $\bold i$ and $\bold j$ , which are unit vectors often defined to be due east and north respectively.

Example 1

A particle starts from the point with position vector $(5\bold{i} + 10 \bold{j}) \ m$ and moves with constant velocity $(2\bold{i} - 2 \bold{j})\ ms^{-1}$ .

i) Find the position vector of this particle after $10$ seconds.

ii) Show that the particle is never at the point $(0,0).$

i) Find the position vector of this particle after $10$ seconds.

Using the equation $\bold r = \bold r_0 + \bold vt$ , substitute $\bold r_0 =(5\bold{i} + 10 \bold{j})\, m$ and $\bold v=(2\bold{i} - 2 \bold{j})\ ms^{-1}$ :

$\bold r = \bold r_0 + \bold v t = \left[(5\bold i + 10 \bold{j}) +(2\bold{i} - 2 \bold{j})t\right] \ m$ 

Group the terms according to the unit vectors:

 $\bold r = \left[(5+2t)\bold{i} +(10-2t)\bold j\right]\ m$ 

Now substitute the value of the time $t=10 \ s$ :

$\begin{aligned}\bold r &= (5+2\times 10)\bold{i} +(10-2\times 10)\bold j \\&=(5+20)\bold{i} +(10-20)\bold j \\&= (25\bold{i} -10\bold j) \ m\end{aligned}$

In conclusion, $\underline{\bold r = (25 \bold i - 10 \bold j) \ m}$ .

ii) Show that the particle is never at the point $(0,0).$

Using $\bold r = \left[(5+2t)\bold{i} +(10-2t)\bold j\right]m$ (from the first part), you only have to substitute $\bold r = (0 \bold i + 0 \bold j) \ m$ . Therefore:

$\bold r = 0\bold i + 0 \bold j = (5+2t)\bold{i} +(10-2t)\bold j\implies\begin{cases}0 \bold i = (5 +2t)\bold i \\0 \bold j = (10 -2 t )\bold j\end{cases} \implies \begin{cases}t=-0.4 \ s\\t = 5 \ s\end{cases}$

The particle is never at the origin because the two times are not consistent.

Example 2

A particle $P$ initially moves with velocity $(3\bold i - 8\bold j)\ ms^{-1}$ . It has an acceleration of $(-2 \bold i + 3 \bold j) \ ms^{-2}$ .

i) Find the velocity of $P$ after $t=3 \ s.$ 

ii) Find the speed of $P$ at $t = 3 \ s$ to $2\space d.p.$ 

i) Find the velocity of $P$ after $t=3 \ s.$ 

Using the equation $\bold v=\bold u +\bold a t$ you have to substitute $\bold u =(3\bold i - 8\bold j)\ ms^{-1}$ and $\bold a=(-2 \bold i + 3 \bold j) \ ms^{-2}$ . Therefore:

$\bold v = \left[(3\bold i - 8 \bold j) + (-2 \bold i + 3 \bold j)t\right]\ ms^{-1}$

Group the terms according to the unit vectors.

$\bold v = \left[(3 - 2t)\bold i + (-8 + 3t)\bold j\right] \ ms^{-1}$ 

Now substitute the value of the time $t = 3 \ s:$ 

$\begin{aligned}\bold v &= (3 - 2\times 3)\bold i + (-8 + 3\times 3)\bold j \\&=(3 -6)\bold i + (-8 + 9)\bold j \\&= (-3\bold i + \bold j) \ ms^{-1}\end{aligned}$ 

In conclusion $\underline{\bold v = (-3 \bold i +\bold j) \ ms^{-1}}$ 

ii) Find the speed of $P$ at $t = 3 \ s.$

The formula for speed is $\sqrt{v_x^2 + v_y^2}$ , therefore the speed of $P$ is:

$Speed \ of \ P = \sqrt{(-3)^2 +1^2 }=\sqrt{9+ 1}=\sqrt{10} = \underline{3.16 \ ms^{-1} \ (2 \ d.p.)}$ 