Question

A mail carrier leaves the post office and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00 km. After dropping off a package, he drives 9.50 km 35.0° north of east to Starbucks. What is his distance from the starting point? (Hint: draw a picture) (3 sig figs) *

Answer 1

Answer:

12.97 km

Explanation:

In order to find the resultant displacement, we have to resolve each of the 3 displacements along the x and y direction.

Taking north as positive y direction and east as positive x-direction, we have:

- Displacement 1: 2.00 km to the north

So

$A_x = 0\\A_y = +2.00 km$

- Displacement 2: 60.0° south of east for 7.00 km

So

$B_x=(7.00)(cos (-45^{\circ}))=4.95 km\\B_y = (7.00)(sin (-45^{\circ}))=-4.95 km$

- Displacement 3: 9.50 km 35.0° north of east

So

$C_x=(9.50)(cos 35^{\circ})=7.78 km\\C_y=(9.50)(sin 35^{\circ})=5.45 km$

So the net displacement along the two directions is:

$R_x=A_x+B_x+C_x=0+4.95+7.78=12.73 km\\R_y=A_y+B_y+C_y=2.00+(-4.95)+5.45=2.50 km$

So, the  distance between the initial and final position is equal to the magnitude of the net displacement:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{12.73^2+2.50^2}=12.97 km$