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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

  1. A( 100 i ^ + 400 j ^ )   m
  2. B( 100 i ^ + 100 j ^ )   m
  3. C( 400 i ^ + 100 j ^ )   m
  4. 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

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