Answer to Question #136365 in Mechanics | Relativity for Alfred John Aguilar

Question

A hiker walks 2.5 km, 53.1°N of E. Then he walks another 2.0 km eastward. How far is the hiker from his starting point?

Answer 1

Solution

According to question diagram can be shown as below

Components of Vector A are

$A_x=2.5\times cos53.1^0\\ \space\space=1.5km$

$A_y=2.5\times sin53.1^0=2 km$

Components of Vector B

$B_x=2\times cos0^0=2km$

$B_y=0 km$

Resultant vector C can be written as

$C_x=A_x+B_x=1.5+2\\=3.5 km$

$C_y=A_y+B_y=2+0\\$ =2 km

Therefore distance from starting point

$C= \sqrt{C_x^2+C_y^2} \\= \sqrt{(3.5)^2+2^2}\\=4.03 km$

Therefore hiker is 4.03 Km from starting point.