A car is driven east for a distance of 50 km, then north for 30 km, and then in a direction ${\mathbf{30}}^{\mathbf{°}}$east of north for 25 km . Sketch the vector diagram and determine (a) the magnitude and (b) the angle of the car’s total displacement from its starting point.

• The distance traveled in the east is,$\overrightarrow{\mathrm{a}}=(50\mathrm{km})\hat{\mathrm{i}}.$

• The distance traveled in the north is,$\overrightarrow{\mathrm{b}}=(30\mathrm{km})\hat{\mathrm{j}}$.
• The distance traveled in the east of north is,$\overrightarrow{\mathrm{c}}=(25\mathrm{km})\cos \left(60\right)\hat{\mathrm{i}}+(25\mathrm{km})\sin \left(60\right)\hat{\mathrm{j}}$

The vector is a quantity with magnitude and direction. The addition of the vectors is done by the laws of vector addition as normal mathematical operations cannot be used. The direction of the vector can be found in the components of the vectors along the unit vectors are known.

The problem involves the vector law of addition. Using the concept of the addition of two vectors the sum. Further using the general formula for the magnitude and the angle, the magnitude and the angle of the given vector can be calculated.

Formula:

The magnitude of the vector is given by,

$\mathrm{r}=\sqrt{{{\mathrm{r}}_{\mathrm{x}}}^2+{{\mathrm{r}}_{\mathrm{y}}}^2}$ (i)

Here, $r_x$and $r_y$are the components of the vector along x and y axis respectively.

The direction is given by,

$\mathrm{\theta }={\tan }^{-1}\left(\frac{{\mathrm{r}}_{\mathrm{y}}}{{\mathrm{r}}_{\mathrm{x}}}\right)\left(\mathrm{ii}\right)$

The total displacement from the figure is given by,

$$\begin{gathered} \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \\ =\left(50\mathrm{km}\right)\hat{\mathrm{i}}\left(30\mathrm{km}\right)\hat{\mathrm{j}}\left(12.5\mathrm{km}\right)\hat{\mathrm{i}}\left(21.7\mathrm{km}\right)\hat{\mathrm{j}} \\ =(62.5\mathrm{km})\hat{\mathrm{i}}+(51.7\mathrm{km})\hat{\mathrm{j}} \end{gathered}$$

Thus, the total displacement is$(62.5\mathrm{km})\hat{\mathrm{i}}+(51.7\mathrm{km})\hat{\mathrm{j}}$ .

To find the magnitude, use equation (i).

$r=\sqrt{{r_x}^2+{r_y}^2}$

Substitute the values of x and y components from part (a).

$$\begin{gathered} \mathrm{r}=\sqrt{{\left(62.5\mathrm{km}\right)}^2{\left(51.7\mathrm{km}\right)}^2} \\ =81\mathrm{km} \end{gathered}$$

Thus, the magnitude is 81 km

Using equation (ii), the magnitude is given by,

$\mathrm{\theta }={\tan }^{-1}\left(\frac{51.7\mathrm{km}}{62.5\mathrm{km}}\right)$

Thus, the magnitude is ${40}^{°}$.