A force F → = ( 40 i ^ + 10 j ^ )   N acts on a body of mass 5   kg . If the body starts…

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

Solution

dv→dt=a→=F→m=(8i^+2j^) m s-2

dr→dt=v→=(8ti^+2tj^) m s-1

r→=(i^+2j^)t22 m

At t=10 sec

r→=[(8i^+2j^)50] m

⇒r→=(400i^+100j^) m

Asked in: JEE Main 2021 (25 Jul Shift 2)

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