JEE Main 202125 Jul, Evening shiftPhysicsMediumLaws of Motion70× since 2002A force F→=(40i^+10j^)N\overrightarrow F = (40\widehat i + 10\widehat j)NF=(40i+10j)N acts on a body of mass 5 kg. If the body starts from rest, its position vector r→\overrightarrow rr at time t = 10 s, will be :AOption A. (100i^+400j^)m(100\widehat i + 400\widehat j)m(100i+400j)mBOption B. (100i^+100j^)m(100\widehat i + 100\widehat j)m(100i+100j)mCOption C. (400i^+100j^)m(400\widehat i + 100\widehat j)m(400i+100j)mDOption D. (400i^+400j^)m(400\widehat i + 400\widehat j)m(400i+400j)mCheckSkip · show solution
