Answer to Question #252951 in Mechanics | Relativity for Ahmed

Question

A car travels 7 km to east and then 4 km in a direction 60° north of east. Find the magnitude and direction of the car’s displacement vector relative to positive x-axis.

Answer 1

Let's first find $x$- and $y$-components of the car's resultant displacement:

$$d_x=7\ km\cdot cos0^{\circ}+4\ km\cdot cos60^{\circ}=9\ km,$$ $$d_y=7\ km\cdot sin0^{\circ}+4\ km\cdot sin60^{\circ}=3.46\ km.$$

We can find the magnitude of the car's resultant displacement from the Pythagorean theorem:

$$d=\sqrt{d_x^2+d_y^2}=\sqrt{(9\ km)^2+(3.46\ km)^2}=9.64\ km.$$

We can find the direction of the resultant displacement from the geometry:

$$\theta=tan^{-1}({\dfrac{d_y}{d_x}}),$$ $$\theta=tan^{-1}({\dfrac{3.46\ km}{9\ km}})=21^{\circ}\ N\ of\ E.$$