Answer
a) 1.35 km, 21.0$^{\circ}$ north of west
b) 0.540 km/h
Work Step by Step
AE
= BD – 0.50
= 2.15cos(35$^{\circ}$) - 0.50
$\approx$ 1.26
CE
= CD – 0.75
= 2.15sin(35$^{\circ}$) - 0.75
$\approx$ 0.483
Displacement
= AC
= $\sqrt ((AE)^{2}+(CE)^{2})$
= $\sqrt (1.26^{2}+0.483^{2})$
$\approx$ 1.35 km
$\theta$
= $tan^{-1}(\frac{CE}{AE})$
= $tan^{-1}(\frac{0.483}{1.26})$
$\approx$ 21.0$^{\circ}$ north of west
Average velocity
= $\frac{displacement}{time}$
= $\frac{1.35}{2.50}$
$\approx$ 0.540 km/h
