Question

A body travels a distance of 3 km towards East, then 4 km towards North and finally 9 km towards East.

(i) What is the total distance traveled?

(ii) What is the resultant displacement?

Answer 1

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

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

(b) If we see the diagram, as ABCD is a rectangle.
AB = CD =4 km
BC = AD = 9 km
OD = OA + AD = 3 + 9 = 12 km
Consider right angle triangle ODC,
By Pythagoras theorem,
${OC}^2$ = ${OD}^2$ + ${CD}^2$ = ${12}^2$ + $4^2$ = 144 + 16
${OC}^2$ = 160 km
OC== 12.65 Km

Hence, displacement is 12.65 Km