Question

6. A body of mass 2 kg has an initial velocity of 3 m/s along OE and it is subjected to a force of 4 N in a direction perpendicular to OE. The distance of the body from O after 4 sec will be: (a) 12 m (b) 28 m (c) 20 m (d) 48 m

JEE/Engineering Exams

Physics

Solution

118

4.0 (1 ratings)

1 ( sqrt{2 k g} rightarrow u=left.3 mright|_{s} ) ( S=u t+frac{1}{2} ) at ( (text { in } x- ) direction ( S_{x}=3(4) ) ( =12 mathrm{m} ) ( s_{y}=frac{1}{2}(c)(4)^{n} quad begin{array}{l}text { fin y-direction } u=0end{array} ) ( F=m a quad=16 mathrm{m} ) ( begin{array}{l}frac{4}{2}=a text { . } quad frac{S_{8}}{A} quad begin{array}{l}S_{text {resultan }}=sqrt{16^{2}+12^{2}} a=2^{m} mid Send{array} & begin{array}{l}=sqrt{400} =20 mathrm{m}end{array}end{array} ) ( Rightarrow S_{1}=20 mathrm{m} )