Mass m=100 g=0.1 kg
Force F=5i^+10j^ N
Acceleration a=mF=0.15i^+10j^=50i^+100j^ m/s2
Since the body starts from rest, initial velocity u=0.
Position after t=2 s is given by r=ut+21at2
r=21(50i^+100j^)(2)2=100i^+200j^ m
Comparing with the given position r=2xi^+5yj^:
2x=100⇒x=50
5y=200⇒y=40
The ratio x:y=50:40=5:4
Answer: 5:4