Question

An object moves 3m north then 4m east and 6m south. Calculate the distance and airship displacement

Answer 1

The distance covered by the airship

= (3 + 4 + 6) m = 13 m

And, magnitude of diplacement

$= \sqrt{ {4}^{2} + {3}^{2} } \\ = \sqrt{16 + 9 \: } \\ = \sqrt{25}$

= 5 m towards South-East

HOPE THIS COULD HELP!!!

Answer 2

Solution :

Distance travelled by the airship

= (3 + 4 + 6)m

= 13m

By using the Pythagoras theorem,

AC^2 = AB^2 + BC^2

=> AC^2 = AB^2 + BC^2

=> AC = √AB^2 + BC^2

=> √4^2 + 3^2

=> √25 = 5

.°. AC = 5m

.°. Displacement = 5m