Question

A man has to go 50m due north , 40m due east and 20m due south to reach a field . A) what distance he has to walk to reah field? B) What is his displacement from his house to the field ? Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Solution:

Distance :

Distance is the total path covered by the man.
Distance travelled by man=50m+40m+20m=110m

Displacement : 
Displacement is the shortest path between source [house] and destination [ field]= AB

See the figure :
Draw Perpendicular BE to AC.
BE=CD=40m
AE=AC-CE=50-20=30

NOW, DISPLACEMENT AB=√AE²+EB²
AB=√30²+40²=50m
Angle with horizontal is given by tanθ=30/40=3/4
θ=tan⁻¹[3/4] direction north  to east.

∴His displacement from his house to field is 50m , tan⁻¹[3/4] direction north  to east.

Answer 2

Answer:

Explanation:

a) Distance travelled = 50 + 40 + 20 = 110 m b) AF = AB – BF = AB – DC = 50 – 20 = 30 M His displacement is AD AD = √  (AF2 -DF2) = √  (302 - 402) = 50m In ΔAED tanθ = DE/AE = 30/40 = 3/4 θ = tan–1 (3/4) His displacement from his house to the field is 50 m, tan–1 (3/4) north to east.