Explanation

(a) Here, $\overrightarrow{v_{1}}$=30km $h^{- 1}$ due north $= \overrightarrow{OA}$
$\overrightarrow{v_{2}} = 40kmh^{- 1}$ due east $= \overrightarrow{OB}$
Change in velocity in 20 s
$\mathrm{\Delta}\overrightarrow{v} = \overrightarrow{v_{2}} - \overrightarrow{v_{1}} = \overrightarrow{v_{2}} + (\overrightarrow{- v_{1}})$
$= \overrightarrow{OB} + \overrightarrow{OC} = \overrightarrow{OD}$
$\left| \mathrm{\Delta}\overrightarrow{v} \right| = \sqrt{v_{2}^{2} + v_{1}^{2}} = \sqrt{40^{2} + 30^{2}}$
$= 50\ kmh^{- 1}$
Acceleration , $\overrightarrow{a} = \frac{\left| \mathrm{\Delta}\ \overrightarrow{v} \right|}{\mathrm{\Delta}t}$
$= \frac{50}{20} = 2.5\ kmh^{- 2}$
Tan $\beta = \frac{v_{1}}{v_{2}} = \frac{30}{40}$
$= 0.75 = \tan 37{^\circ}$
$\therefore\ \beta = 37{^\circ}$ north of east
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