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Question: Answered & Verified by Expert
A force F=(40i^+10j^) 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
PhysicsLaws of MotionJEE MainJEE Main 2021 (25 Jul Shift 2)
Options:
  • A (100i^+400j^) m
  • B (100i^+100j^) m
  • C (400i^+100j^) m
  • D (400i^+400j^) m
Solution:
1095 Upvotes Verified Answer
The correct answer is: (400i^+100j^) m

dvdt=a=Fm=(8i^+2j^) m s-2

drdt=v=(8ti^+2tj^) m s-1

r=(i^+2j^)t22 m

At t=10 sec

r=[(8i^+2j^)50] m

r=(400i^+100j^) m

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