(a) A(-1, 2), B(3, 5) and C(4, 8) are the vertices of triangle ABC. Forces whose magnitudes are 5N and \(3\sqrt{10}\)N act along \(\overrightarrow{AB}\) and...

(a) Direction of the resultant of the two forces.

\(\overrightarrow{AB}=B-A=(3-(-1),\,5-2)=(4,3),\) with \(|\overrightarrow{AB}|=\sqrt{16+9}=5.\) A force of 5 N along \(\overrightarrow{AB}\) is \(\mathbf{F_1}=5\cdot\dfrac{(4,3)}{5}=(4,3).\)

\(\overrightarrow{CB}=B-C=(3-4,\,5-8)=(-1,-3),\) with \(|\overrightarrow{CB}|=\sqrt{1+9}=\sqrt{10}.\) A force of \(3\sqrt{10}\) N along \(\overrightarrow{CB}\) is \(\mathbf{F_2}=3\sqrt{10}\cdot\dfrac{(-1,-3)}{\sqrt{10}}=(-3,-9).\)

Resultant \(\mathbf{R}=\mathbf{F_1}+\mathbf{F_2}=(4-3,\,3-9)=(1,-6).\)

The resultant lies in the fourth quadrant. Its direction with the positive x-axis:

\[\theta=\tan^{-1}\!\left(\frac{-6}{1}\right),\qquad \tan^{-1}6=80.5^\circ.\]

So the resultant acts at \(80.5^\circ\) below the positive x-axis (equivalently a direction of \(-80.5^\circ,\) or \(279.5^\circ\) measured anticlockwise). Its magnitude, for reference, is \(\sqrt{1^2+6^2}=\sqrt{37}\approx6.08\text{ N}.\)

(b) Straight-line motion from rest.

(i) Acceleration. \(u=0,\ v=20\text{ m/s},\ s=8\text{ m}.\) Using \(v^2=u^2+2as:\)

\[400=0+2a(8)\Rightarrow a=\frac{400}{16}=25\text{ m/s}^2.\]

(ii) Time to cover 40 m. Using \(s=ut+\tfrac12 at^2\) with \(u=0:\)

\[40=\tfrac12(25)t^2\Rightarrow t^2=\frac{40}{12.5}=3.2\Rightarrow t=\sqrt{3.2}\approx1.79\text{ s}.\]