Question

A body a travels a distance of 3km towards East, then 4km towards North and finally 9 km towards East. a) What is the total distance travelled b) What is the resultant displacement?

Answer 1

Given A body travels travels a distance of 3km towards East, then 4km towards North and finally 9 km towards East.
So total distance travelled by body is length of the path traversed =AB+BC+CA=(3+4+9)km=16km

Displacement is the shortest path between source and destination.
So in this case it is Path OC.

If we see the diagram, as ABCD is a rectangle.
AB=CD=4KM
BC=AD=9KM
OD=OA+AD=3+9=12KM
Consider right angle triangle ODC
By Pythagoras theorem,
OC²=OD²+CD²=12²+4²
=144+16
OC²=160KM
OC=√160=12.65Km

Hence displacement is 12.65Km

Answer 2

Answer:

distance travelled = 3km + 4km + 9km = 16 km

displacement = by using pythagoreos theorem 12^2 +4^2 =144+ 16

=160 km

taking 160 underroot and by solving we get 12.65km that is the displacement