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Q.
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
Motion in a Plane
Solution:
Vector $\vec{A}$ represents a displacement of $30 \,km$ due east and vector $\vec{B}$ represents a displacement of $40 \,km$ due north. Therefore resultant displacement of the car is $\vec{C}$.
Now its magnitude is
$\mid \vec{C} \models \sqrt{A^{2}+B^{2}}$
$=\sqrt{(30)^{2}+(40)^{2}}=50 km$
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 $50 \,km$ and makes an angle $53^{\circ}$ north of east.
