Question

A car travels 6.0 km due North and then 12.0 km ${60}^o$ west of North. Find the magnitude of displacement.

• A. 18 km
• B. 14 km
• C. 16 km
• D. 12 km

Answer 1

The correct option is C 16 km

D = resultant of two vectors
$=\sqrt{A^2+B^2+2AB\cos \theta }$
$=\sqrt{6^2+{12}^2+\mathrm{2.6.12}\cos {60}^{\circ }}$
$=\sqrt{36+144+72}$
$=\sqrt{252}\simeq 16$