a body travels a distance of 3 km towards east then 4 kilometre towards north and finally 9 kilometre towards east what is his resultant displacement
Answers
Explanation:
Let the distance of 3km will be a to b and 4 km is B to c 9 km is C to D
displacement is 9+3=12
Answer:
4√10 km along tan^-1(1/3) with east.
Explanation:
Sign convention used is given in above diagram.
According to sign convention, body first travels 3 km towards east, which makes a displacement of 3i km
Then it travels 4 km in North direction which means a further displacement of 4j km
Then finally it moves another 9 km towards east making further displacement of 9i km
Therefore, total displacement is. (12i + 4j )km
Magnitude of displacement comes to be 4√10 km by formula
|c| = √{(a)^2 +(b)^2 + 2abcosx}
where
c = magnitude of resultant vector (12i + 4j)
a = 12, b = 4 and x = 90° is the angle between 12i and 4j
For direction we use
tan z = (b sin x)/(a + b cos x)
where z is the angle made by resultant vector with positive x axis.
z comes out to be tan^-1(1/3)
