Vectors in Mechanics

In Resolving Forces, we mentioned how the components of a force can be described in vector form. We can do the same for equations of motion.

Make sure you are happy with the following topics before continuing.

Using SUVAT

We can use SUVAT more or less in the usual way, but we need to be careful of which equations we choose to use. Adapting the equations for vector form, we have

$textcolor{red}{v} = u + textcolor{blue}{a}textcolor{purple}{t} to begin{pmatrix}textcolor{red}{v_1} textcolor{red}{v_2} end{pmatrix} = begin{pmatrix}u_1 u_2 end{pmatrix} + begin{pmatrix}textcolor{blue}{a_1} textcolor{blue}{a_2} end{pmatrix}textcolor{purple}{t}$

$textcolor{limegreen}{s} = utextcolor{purple}{t} + dfrac{1}{2}textcolor{blue}{a}textcolor{purple}{t}^2 to begin{pmatrix}textcolor{limegreen}{s_1} textcolor{limegreen}{s_2} end{pmatrix} = begin{pmatrix}u_1 u_2 end{pmatrix}textcolor{purple}{t} + dfrac{1}{2}begin{pmatrix}textcolor{blue}{a_1} textcolor{blue}{a_2} end{pmatrix}textcolor{purple}{t}^2$

$textcolor{limegreen}{s} = dfrac{1}{2}(u + textcolor{red}{v})textcolor{purple}{t} to begin{pmatrix}textcolor{limegreen}{s_1} textcolor{limegreen}{s_2} end{pmatrix} = dfrac{1}{2}left( begin{pmatrix}u_1 u_2 end{pmatrix} + begin{pmatrix}textcolor{red}{v_1} textcolor{red}{v_2} end{pmatrix}right) textcolor{purple}{t}$

$textcolor{limegreen}{s} = textcolor{red}{v}textcolor{purple}{t} – dfrac{1}{2}textcolor{blue}{a}textcolor{purple}{t}^2 to begin{pmatrix}textcolor{limegreen}{s_1} textcolor{limegreen}{s_2} end{pmatrix} = begin{pmatrix}textcolor{red}{v_1} textcolor{red}{v_2} end{pmatrix}textcolor{purple}{t} – dfrac{1}{2}begin{pmatrix}textcolor{blue}{a_1} textcolor{blue}{a_2} end{pmatrix}textcolor{purple}{t}^2$

Of course, we can extend that to $n$ dimensions, but this is just in 2D form, to demonstrate the idea. We might also demonstrate the vector in $textbf{i}$ , $textbf{j}$ and further components.

You might notice that $textcolor{red}{v}^2 = u^2 + 2textcolor{blue}{a}textcolor{limegreen}{s}$ has been removed…

Multiplying two vectors by each other, i.e. $(textcolor{red}{v}cdot textcolor{red}{v})$ , $(ucdot u)$ , $(textcolor{blue}{a}cdot textcolor{limegreen}{s})$ will not give us the solution we perhaps expect. But don’t worry about that, it’s not important for this section, anyway.

The point is, we can only use equations involving vector-by-scalar calculations. $textcolor{purple}{t}$ is our only scalar variable.

Using Non-Uniform Acceleration

Again, we just use integration and differentiation from the Non-Uniform Acceleration section in the same way. In vector form:

$begin{pmatrix}textcolor{red}{v_1} textcolor{red}{v_2} end{pmatrix} = dfrac{dbegin{pmatrix}textcolor{limegreen}{s_1} textcolor{limegreen}{s_2} end{pmatrix}}{dtextcolor{purple}{t}}$

$begin{pmatrix}textcolor{blue}{a_1} textcolor{blue}{a_2} end{pmatrix} = dfrac{dbegin{pmatrix}textcolor{red}{v_1} textcolor{red}{v_2} end{pmatrix}}{dtextcolor{purple}{t}}$

$begin{pmatrix}textcolor{limegreen}{s_1} textcolor{limegreen}{s_2} end{pmatrix} = {LARGE int} begin{pmatrix}textcolor{red}{v_1} textcolor{red}{v_2} end{pmatrix} dtextcolor{purple}{t}$

$begin{pmatrix}textcolor{red}{v_1} textcolor{red}{v_2} end{pmatrix} = {LARGE int} begin{pmatrix}textcolor{blue}{a_1} textcolor{blue}{a_2} end{pmatrix} dtextcolor{purple}{t}$

We can extend this to $n$ dimensions, and demonstrate our vectors in $textbf{i}$ , $textbf{j}$ and further components, also.

Here, we simply differentiate and integrate each component separately.

Example 1: SUVAT

A particle has initial displacement $underline{s}_0 = begin{pmatrix}3 5end{pmatrix}$ . Given that $underline{u} = begin{pmatrix}1 -1end{pmatrix}$ and $textcolor{blue}{underline{a}} = begin{pmatrix}textcolor{blue}{0} textcolor{blue}{-2}end{pmatrix}$ , find the displacement (in vector form) of the particle at time $textcolor{purple}{t} = textcolor{purple}{4}$ .

[3 marks]

$textcolor{limegreen}{underline{s}} = underline{u}textcolor{purple}{t} + dfrac{1}{2}textcolor{blue}{underline{a}}textcolor{purple}{t}^2 + underline{s}_0$

$= left( begin{pmatrix}1 -1end{pmatrix} times textcolor{purple}{4}right) + left( dfrac{1}{2}begin{pmatrix}textcolor{blue}{0} textcolor{blue}{-2}end{pmatrix} times textcolor{purple}{4}^{2}right) + begin{pmatrix}3 5end{pmatrix}$

$= begin{pmatrix}4 -4end{pmatrix} + begin{pmatrix}0 -16end{pmatrix} + begin{pmatrix}3 5end{pmatrix} = begin{pmatrix}7 -15end{pmatrix}$

Alternatively, we can work this out in $textbf{i}$ and $textbf{j}$ notation.

Let $underline{s}_0 = (3textbf{i} + 5textbf{j})$ , $underline{u} = (textbf{i} – textbf{j})$ and $textcolor{blue}{underline{a}} = textcolor{blue}{-2textbf{j}}$

Then $textcolor{limegreen}{underline{s}} = textcolor{purple}{4}(textbf{i} – textbf{j}) + dfrac{textcolor{purple}{4}^2}{2}(textcolor{blue}{-2textbf{j}}) + (3textbf{i} + 5textbf{j}) = (textcolor{limegreen}{7textbf{i} – 15textbf{j}})$

Example 2: Non-Uniform Acceleration

A particle has acceleration $textcolor{blue}{underline{a}} = (textcolor{blue}{2ttextbf{i} + t^{2}textbf{j}})$ . It is initially stationary at the point $underline{s}_0 = (3textbf{i} + 3textbf{j})$ . Find an expression for its displacement vector $textcolor{limegreen}{s}$ , at time $textcolor{purple}{t}$ .

[3 marks]

$begin{aligned}textcolor{red}{underline{v}} = int textcolor{blue}{underline{a}} dtextcolor{purple}{t} = int begin{pmatrix}textcolor{blue}{2t} textcolor{blue}{t^{2}}end{pmatrix} dtextcolor{purple}{t} = begin{pmatrix}textcolor{red}{t^{2}} textcolor{red}{dfrac{1}{3}t^{3}}end{pmatrix} + begin{pmatrix}0 0 end{pmatrix} = begin{pmatrix}textcolor{red}{t^{2}} textcolor{red}{dfrac{1}{3}t^{3}}end{pmatrix}end{aligned}$

$begin{aligned}textcolor{limegreen}{underline{s}} &= int textcolor{red}{underline{v}} textcolor{purple}{t} = int begin{pmatrix}textcolor{red}{t^{2}} textcolor{red}{dfrac{1}{3}t^{3}}end{pmatrix} dtextcolor{purple}{t} = begin{pmatrix}textcolor{limegreen}{dfrac{1}{3}t^{3}}[6pt] textcolor{limegreen}{dfrac{1}{12}t^{4}}end{pmatrix} + underline{s}_0[1.2em]&=begin{pmatrix}textcolor{limegreen}{dfrac{1}{3}t^{3}}[6pt] textcolor{limegreen}{dfrac{1}{12}t^{4}}end{pmatrix} + begin{pmatrix}3 3end{pmatrix}[1.2em]&=begin{pmatrix}textcolor{limegreen}{dfrac{1}{3}t^{3} + 3}[6pt] textcolor{limegreen}{dfrac{1}{12}t^{4} + 3}end{pmatrix}end{aligned}$

In component form:

$begin{aligned}textcolor{red}{underline{v}} = int textcolor{blue}{underline{a}} dtextcolor{purple}{t} = int (textcolor{blue}{2ttextbf{i} + t^{2}textbf{j}}) dtextcolor{purple}{t} = textcolor{red}{t^{2}textbf{i} + dfrac{1}{3}t^{3}textbf{j}}end{aligned}$

$begin{aligned}textcolor{limegreen}{underline{s}} &= int textcolor{red}{underline{v}} dtextcolor{purple}{t} = int left(textcolor{red}{t^{2}textbf{i} + dfrac{1}{3}t^{3}textbf{j}}right) dtextcolor{purple}{t} = textcolor{limegreen}{dfrac{1}{3}t^{3}textbf{i} + dfrac{1}{12}t^{4}textbf{j}} + underline{s}_0[1.2em]&=textcolor{limegreen}{dfrac{1}{3}t^{3}textbf{i} + dfrac{1}{12}t^{4}textbf{j}} + (3textbf{i} + 3textbf{j})[1.2em]&=textcolor{limegreen}{left(dfrac{1}{3}t^{3} + 3right) textbf{i} + left( dfrac{1}{12}t^{4} + 3right) textbf{j}}end{aligned}$

Vectors in Mechanics Example Questions

Question 1: A particle has an initial velocity of $(5textbf{i} + textbf{j})$ and acceleration $(0.3textbf{i} + 0.4textbf{j})$ . Find the particle’s velocity after $8textbf{ s}$ .

[2 marks]

Use $underline{v} = underline{u} + underline{a}t$ to give

$underline{v} = (5textbf{i} + textbf{j}) + 8(0.3textbf{i} + 0.4textbf{j})$

$=left( (5 + 2.4)textbf{i} + (1 + 3.2)textbf{j}right)$

$= (7.4textbf{i} + 4.2textbf{j})$

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Question 2: A particle sets off from the origin with a velocity of $begin{pmatrix}t^3 6tend{pmatrix}$ . Find an expression for the displacement vector $underline{s}$ and the acceleration vector $underline{a}$ .

[4 marks]

$underline{a} = dfrac{dunderline{v}}{dt} = begin{pmatrix}3t^2 6end{pmatrix}$

$underline{s} = int underline{v} dt = {LARGE int} begin{pmatrix}t^3 6tend{pmatrix} = begin{pmatrix}dfrac{1}{4}t^4[6pt] 3t^2 end{pmatrix} + begin{pmatrix}0 0 end{pmatrix} = begin{pmatrix}dfrac{1}{4}t^4[6pt] 3t^2 end{pmatrix}$