Question

A car travels due east on a level road for 30 km It then turns due north at an intersection and travels 40 km before stopping. The resultant displacement of the car is

• A. $50km,53^{\circ }$ north of east
• B. $50km,53^{\circ }$ east of north
• C. $100km,37^{\circ }$ north of east
• D. $100km,37^{\circ }$ east of north

Answer 1

Answer: (A)

Vector $\vec{A}$ represents a displacement of $30km$ due east and vector $\vec{B}$ represents a displacement of $40km$ due north. Therefore resultant displacement of the car is $\vec{C}$.

Now its magnitude is
$\mid \vec{C}\models \sqrt{A^2+B^2}$
$=\sqrt{(30{)}^2+(40{)}^2}=50km$
The direction of the resultant displacement is given by
$\tan \theta =\frac{B}{A}=\frac{40}{30}=1.33$or
$\theta ={\tan }^{-1}(1.33)=53^{\circ }$
Thus the resultant displacement has a magnitude of $50km$ and makes an angle $53^{\circ }$ north of east.