PhysicsMedium68×since 2002Q7586A 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 :A(100i^+400j^)m(100\widehat i + 400\widehat j)m(100i+400j)mB(100i^+100j^)m(100\widehat i + 100\widehat j)m(100i+100j)mC(400i^+100j^)m(400\widehat i + 100\widehat j)m(400i+100j)mD(400i^+400j^)m(400\widehat i + 400\widehat j)m(400i+400j)mCheck answerSkip