Question

A plane flies $50$ degrees east of south for $100km$ then $400km$ north and then $20$ degrees north of west for $250km$. What is the magnitude of the net displacement? (in km)

• A. $335$
  
• B. $365$
  
• C. $395$
  
• D. $345$
  

Answer 1

The correct option is A $335$

The net displacement is the vector sum of the three displacements.

The first displacement $\left(\overrightarrow{S_1}\right)$, the second displacement $\left(\overrightarrow{S_2}\right)$, the third displacement $\left(\overrightarrow{S_3}\right)$

$\overrightarrow{S_{net}}=\overrightarrow{S_1}+\overrightarrow{S_2}+\overrightarrow{S_3}$

$S_{netx}=S_{1x}+S_{2x}+S_{3x}$

$=100\times \cos \left(50°\right)+0-25\times \cos \left(20°\right)$

$=40.8㎞$

$S_{nety}=S_{1y}+S_{2y}+S_{3y}$

$=-100\times \sin \left(50°\right)+400+25\times \sin \left(20°\right)$

$=332㎞$

The magnitude of net displacement is

$S_{net}=\sqrt{S_{netx}^2+S_{nety}^2}=335㎞$