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A force F=(10i+5j)N acts on a body of mass 10gm.If the body is at origine at time t=0,then what would be its position at time t=5s?
F = 10i + 5j as F=ma where m=0.01kg accl. of body(a) = 1000i+500j equation of motion is : S = ut + 0.5at2 with starting at rest , u =0 position = (0.5*1000*25 i + 0.5*500*25 j ) = (12500 i + 6250 j) -- Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.Regards,Ramesh IIT Kgp - 05 batch
F = 10i + 5j
as F=ma where m=0.01kg
accl. of body(a) = 1000i+500j
equation of motion is : S = ut + 0.5at2 with starting at rest , u =0
position = (0.5*1000*25 i + 0.5*500*25 j )
= (12500 i + 6250 j)
--
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.Regards,Ramesh IIT Kgp - 05 batch
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