Answer to Question #165022 in Mechanics | Relativity for Kian

Question

You found a treasure map and it said: start at the well, go 100 m straight South, then 30 m West, then 25 straight East, and finally 45 m, North. Using the component method of vector addition, how far from the well and in what direction is the hidden treasure?

Answer 1

$\text{let }\vec{X}(x,y) \text{ and }(x,y)\text{ vector component:}$

$\text{where }x=x*\vec{n};y=x*\vec{e}$

$\vec{n}-\text{unit vector pointing North}$

$\vec{e}-\text{unit vector pointing East}$

$\text{negative x component values South direction}$

$\text{negative н component values West direction}$

$\text{go 100 m straight South}-(-100,0)$

$\text{then 30 m West}-(0,-30)$

$\text{then 25 straight East}-(0,25)$

$\text{and finally 45 m, North} -(45,0)$

$\text{point of treasure}-(-100+0+45,0-30+25+0);$

$(-55,-5)$

$S= \sqrt{(-55)^2+(-5)^2}\approx52.23$

$cotg(\alpha)=\frac{55}{5}=11;\alpha\approx5.2\degree$

$\text{Answer: }\text{treasure is located 52.23 m }5.2\degree\text{southwestward}$