WAEC 2014 · Paper 2 · Q15✱

(a)
$A(-1, 2)$ , $B(3, 5)$ and $C(4, 8)$ are the vertices of triangle $ABC$ . Forces whose magnitudes are $5\text{ N}$ and $3\sqrt{10}\text{ N}$ act along $\overrightarrow{AB}$ and $\overrightarrow{CB}$ respectively. Find the direction of the resultant of the forces. Give the direction as a bearing.
(b)(i)
A particle starts from rest and moves in a straight line. It attains a velocity of $20\text{ m s}^{-1}$ after covering a distance of 8 metres. Calculate its acceleration.
(b)(ii)
Calculate the time it will take to cover a distance of 40 metres.

Worked solution (try it first)

$\overrightarrow{AB} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$ , of length 5, so the 5 N force is $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$ N.
$\overrightarrow{CB} = \begin{pmatrix} -1 \\ -3 \end{pmatrix}$ , of length $\sqrt{10}$ , so the $3\sqrt{10}$ N force is $3\begin{pmatrix} -1 \\ -3 \end{pmatrix} = \begin{pmatrix} -3 \\ -9 \end{pmatrix}$ N.
Add: the resultant is $\begin{pmatrix} 1 \\ -6 \end{pmatrix}$ N.
It points right and down, at $\tan^{-1}6 \approx 80.5^\circ$ below the positive $x$ -axis.
As a bearing, this is $90^\circ + 80.5^\circ \approx 170.5^\circ$ , about $171^\circ$ (S $9.5^\circ$ E).