Question

Calculate the resultant displacement and the direction of a motorbike traveling 40 km to south then 70 km to east and finally 50 km to northwest.

Answer 1

Let's add up the components of the three vectors and find the components of the resultant displacement:

$$R_x=70\ km+50\ km\cdot cos(90^{\circ}+45^{\circ})=34.64\ km,$$ $$R_y=-40\ km+50\ km\cdot sin(90^{\circ}+45^{\circ})=-4.64\ km.$$

Then, we can find the magnitude of the resultant displacement from the Pythagorean theorem:

$$R=\sqrt{R_x^2+R_y^2}=\sqrt{(34.64\ km)^2+(-4.64\ km)^2}=34.95\ km.$$

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

$$\theta=cos^{-1}(\dfrac{R_x}{R})=cos^{-1}(\dfrac{34.64\ km}{34.95\ km})=7.64^{\circ}.$$

In order to find correct angle we must subtruct the obtained angle from $360^{\circ}$:

$$\theta=360^{\circ}-7.64^{\circ}=346.1^{\circ}.$$

The resultant displacement has the magnitude 32.65 km and direction $352.36^{\circ}$(counted conterclockwise from the $x$-axis).