Question

A man has to go 50m due north, 40m due east and 20m due south to reach a cafe from home.
(A) What distance he has to walk to reach the cafe ? (B) What is his displacement from his home to the cafe.

Answer 1

(A) Total distance travelled = 50 m + 40 m + 20 m = 110 m
(B) Let east direction is `hati` and north direction is `hatj` then
`vec(AB) = 50 hatj, vec(BC) = 40 hati, vec(CD) = -20 hatj`

According to law of polygon `vec(AD) = vec(AB) + vec(BC) + vec(CD) or vec(AD) = 50 hatj + 40 hati - 20 hatj = 40 hati + 30 hatj`
`" "|vec(AD)| = sqrt(40^(2) + 30^(2)) = 50`m
Now, if angle of `barAD` with east towards north is `theta`, then
`tan theta = (30)/(40) = (3)/(4)`, so` theta = tan^(-1)((3)/(4)) = 37^(@)`
So direction of displacement is `E 37^(@) N ( 37^(@) N ` of `E)`