Can you please explain me how have we resolved v (rain,road) for the second diagram as
v(rain,road)sin(30°+a)=12km/h cos30° in detail ?
 

Asked by Varsneya Srinivas | 12th May, 2017, 04:31: PM

Expert Answer:

  • Taking the components along OA indicates that OA has to be taken as a base reference to resolve the forces around it which are mentioned in equation (i).
  • When OA is taken as a base reference, it is found that by the geometry of the velocity v man,road=2kmph, OA happens to be the horizontal component of this velocity.
  • Therefore, OA=12kmph.Cos30.
  • Let the quadrilateral in the above diagram be parallelogram OPQR.
  • In triangle OQA, angle Q=(30+a) as angle Q and angle (30+a) are the alternate angles formed between two parallel sides OP and RQ.
  • Now by applying trigonometry in triangle OQA
             begin mathsize 12px style Sin left parenthesis 30 plus straight a right parenthesis equals OA over OQ
Sin left parenthesis 30 plus straight a right parenthesis equals fraction numerator 12 kmph. Cos 30 over denominator straight v subscript rain comma road end subscript end fraction

straight v subscript rain comma road end subscript cross times Sin left parenthesis 30 plus straight a right parenthesis equals 12 kmph. Cos 30
end style
  • This is how we get equation (iii) of the solution mentioned in your query.

Answered by Abhijeet Mishra | 14th May, 2017, 12:44: PM