JEE Main202125 Jul 2021Evening ShiftPhysicsLaws of MotionActual
A force F → = ( 40 i ^ + 10 j ^ ) N acts on a body of mass 5 kg . If the body starts from rest, its position vector r → at time t = 10 s will be
Options
- A( 100 i ^ + 400 j ^ )   m
- B( 100 i ^ + 100 j ^ )   m
- C( 400 i ^ + 100 j ^ )   m
- D( 400 i ^ + 400 j ^ )   m
Correct answer
C. ( 400 i ^ + 100 j ^ )   m
Step-by-step solution
d v → d t = a → = F → m = ( 8 i ^ + 2 j ^ )   m   s - 2 d r → dt = v → = ( 8 t i ^ + 2 t j ^ )   m   s - 1 r → = ( i ^ + 2 j ^ ) t 2 2   m At t = 10   sec r → = [ ( 8 i ^ + 2 j ^ ) 50 ]   m ⇒ r → = ( 400 i ^ + 100 j ^ )   m