Question

A hiker makes four straight-line walks A 15 km at 59 ◦ B 23 km at 340 ◦ C 22 km at 125 ◦ D 30 km at 163 ◦ in random directions and lengths starting at position (41 km, 41 km), A B C D How far from the starting point is the hiker after these four legs of the hike? All angles are measured in a counter-clockwise direction from the positive x-axis.

Answer 1

The distance from the starting point is (-11.97 Km, 31.78 Km)

Step-by-step explanation:

The easiest method for this problem is the analytical calculation of the distance traveled by the hiker, we will decompose each vector and find the result in each direction

Va= 15 Km 59º

Vax= 15 cos 59= 7.73 Km

Vay= 15 sin 59= 12.86 Km

Vbx= 23 cos 340= 21.61 Km

Vby= 23 sin 340= -7.87 Km

Vcx= 22 cos 125= -12.62 Km

Vcy= 22 sin 125 = 18.02 Km

Vdx= 30 cos 163= -28.69 Km

Vdy = 30 sin163 = 8.77 Km

We find each component of the total movement

Vtx= Vax+ Vbx+ Vcx+ Vdx

Vty = Vay+ Vby+ Vcy+Vdy

Vtx= 7.73+ 21.61-12.62-28.69

Vtx = -11.97 Km

Vty= Vay+ Vby+ Vcy+ Vdy

Vty= 12.86-7.87+ 18.02+ 8.77

Vty= 31.78 Km

The distance from the starting point is (-11.97 Km, 31.78 Km)

To know what distance to advanced we add the total distances to the value of the initial distance

(41Km,41Km) = (Vox,Voy)

Dx= Vtx +Vox

Dy = Vty + Voy

Dx= -11.97 + 41= 29.03 Km

Dy = 31.78 + 41= 72.78 Km

This is the total distance advanced by the hiker