Question

1.21 •• a postal employee drives a delivery truck along the route shown in fig. e1.21. determine the magnitude and direction of the resultant displacement by drawing a scale diagram. (see also exercise 1.28 for a different approach.) figure e1.21

Explanation:

Step1: Resolve the angled - displacement into components

The angled displacement of $d_3 = 3.1$ km at $45^{\circ}$ has $x$-component $d_{3x}=3.1\cos45^{\circ}$ km and $y$-component $d_{3y}=3.1\sin45^{\circ}$ km.
$d_{3x}=3.1\cos45^{\circ}\approx3.1\times\frac{\sqrt{2}}{2}\approx2.2$ km
$d_{3y}=3.1\sin45^{\circ}\approx3.1\times\frac{\sqrt{2}}{2}\approx2.2$ km
The first displacement $d_1 = 2.6$ km is in the $y$-direction and the second displacement $d_2 = 4.0$ km is in the $x$-direction.

Step2: Calculate the total $x$-component of the displacement

$D_x=d_2 + d_{3x}=4.0+2.2 = 6.2$ km

Step3: Calculate the total $y$-component of the displacement

$D_y=d_1 + d_{3y}=2.6+2.2 = 4.8$ km

Step4: Calculate the magnitude of the resultant displacement

Using the Pythagorean theorem $D=\sqrt{D_x^{2}+D_y^{2}}$.
$D=\sqrt{(6.2)^{2}+(4.8)^{2}}=\sqrt{38.44 + 23.04}=\sqrt{61.48}\approx7.8$ km

Step5: Calculate the direction of the resultant displacement

The direction $\theta$ is given by $\tan\theta=\frac{D_y}{D_x}$.
$\theta=\arctan(\frac{4.8}{6.2})\approx37.9^{\circ}$ north - of - east

Answer:

Magnitude: approximately $7.8$ km, Direction: approximately $37.9^{\circ}$ north - of - east