Question

a particle moves from a point 2i + 4j metre ,at t=0 sec, with initial velocity 5i+4j m/s .it is axted upon by a constant force which produces a constant acc 4i+4j m/s2 .what will its distance of the particle from origin at t = 2 sec
Ans-20√2

Answer 1

Answer:

20√2

Explanation:

a particle moves from a point 2i + 4j metre ,at t=0 sec, with initial velocity 5i+4j m/s .it is axted upon by a constant force which produces a constant acc 4i+4j m/s2

S = ut + (1/2)at²

U = 5i+4j  m/s

a = 4i+4j m/s²

t = 2 sec

S = (5i+4j)2 + (1/2) (4i+4j)2²

= 10i + 8j  + 8i + 8j

= 18i + 16j

Distance covered = 18i + 16j

Initial Position = 2i + 4j

Final Position = 18i + 16j + 2i + 4j

= 20i  + 20j

Distance from origin = √(20² + 20² )

= 20√2