Question

A girl delivering newspapers covers her route by traveling $3.00$blocks west, $6.00$ blocks north, and then $6.00$blocks east.

(a) What is her resultant displacement?

(b) What is the total distance she travels?

Short answer

a) The resultant displacement of the girl is 5 blocks at $53.1°$(North of East).

b) The total distance travelled by the girl is 13 blocks.

Step-by-step solution

Define distance and displacement

Displacement of a body is defined as the body moving from its rest position to another one. Distance is defined as the length of space between two points.

(a) Find the resultant displacement

To distribute newspapers, the girl begins at point A and follows the course ABCD.

The resultant displacement

$\text{AO}=\sqrt{{\text{AD}}^2+{\text{OD}}^2}$

Now substitute $4.0$for AD and $3.0$for OD.

$$\begin{gathered} \text{AO}=\sqrt{{4.0}^2+{3.0}^2} \\ =5 \end{gathered}$$

Calculate the angle$\theta$ :

$\tan \theta =\frac{\text{AD}}{\text{OD}}$

Substitute 4.0 for AD and 3.0 for OD.

$$\begin{gathered} \tan \theta =\frac{4}{3} \\ \theta ={\tan }^{-1}\left(\frac{4}{3}\right) \\ \theta =53.1°\text{(North of East)} \end{gathered}$$

Here, $\theta$is the direction.

Hence, the resultant displacement of the girl is 5 blocks at $53.1°$(North of East).

(b) Find the total distance she travelled

The total distance travelled.

$\text{D}=\text{AB}+\text{BC}+\text{CO}$

Substitute 3.0 for AB, 4.0 for BC and 6.0 for CO.

$$\begin{gathered} \text{D}=3.0+4.0+6.0 \\ =13\text{blocks} \end{gathered}$$

Hence, the total distance travelled by the girl is 13 blocks.