Question

a body travels a distance of 3km towards east,then 4km towards north and finally 9km towards east the distance and displacement values are​

Answer 1

Answer:

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

Hope it helps!! :)