Answer to Question #208483 in Mechanics | Relativity for Coleman

Question

A man start from a point A and walk a distance of 8.0 km due northeast to point B from B he then walks 4.0 km due South East to point x. Calculate the shortest distance between A and x.

Answer 1

$\vec{AX}=\vec{AB}+\vec{BX}$

$\angle(\vec{AB},\vec{BX})=\angle(\vec{NE},\vec{SE})=\frac{\pi}{2}$

$\vec{AX}^2=(\vec{AB}+\vec{BX})^2$

$\vec{AX}^2=\vec{AB}^2+\vec{BX}^2+2\vec{AB}*\vec{BX}$

$2\vec{AB}*\vec{BX}=2* AB*BX*\cos(\angle(\vec{AB},\vec{BX}))=0$

$AX^2 =AB^2+BX^2$

$AX =\sqrt{AB^2+BX^2}=\sqrt{8^2+4^2}=8.94$

$\text{Answer : }AX =8.94\ km$