CBSE Class 11-science Answered

Pleas answer question number 2

Asked by arunavamitra50 | 06 Jul, 2018, 04:52: PM

Expert Answer

begin mathsize 12px style W e space a r e space g i v e n space t h a t comma space bold italic v bold left parenthesis bold italic t bold right parenthesis space equals space left parenthesis 12 minus 6 t right parenthesis i with rightwards arrow on top space plus space 4 j with rightwards arrow on top space........... left parenthesis 1 right parenthesis l e t space p o s i t i o n space v e c t o r space b e space bold italic r bold left parenthesis bold italic t bold right parenthesis bold space bold equals bold space x left parenthesis t right parenthesis i with rightwards arrow on top space plus space y left parenthesis t right parenthesis space j with rightwards arrow on top t h e n space V left parenthesis t right parenthesis space equals space fraction numerator d over denominator d t end fraction r left parenthesis t right parenthesis space equals space bold space fraction numerator d x left parenthesis t right parenthesis over denominator d t end fraction i with rightwards arrow on top space plus space fraction numerator d y left parenthesis t right parenthesis over denominator d t end fraction space j with rightwards arrow on top space........... left parenthesis 2 right parenthesis c o m p a r i n g space left parenthesis 1 right parenthesis space a n d space left parenthesis 2 right parenthesis comma space w e space g e t space fraction numerator d x left parenthesis t right parenthesis over denominator d t end fraction space equals space left parenthesis 12 minus 6 t right parenthesis space o r space x left parenthesis t right parenthesis space equals space 12 t space minus space 3 t squared space plus C subscript 1 space end style

It is given that particle starts from origin, i.e., when t=0, x=0 ;

If we apply this condition we get,

begin mathsize 12px style X left parenthesis t right parenthesis space equals space 12 t space minus space 3 t squared space equals space 3 t open parentheses 4 minus t close parentheses space.................. left parenthesis 3 right parenthesis end style

similarly by comparing (1) and (2) we get

begin mathsize 12px style fraction numerator d over denominator d t end fraction y left parenthesis t right parenthesis space equals space 4 space semicolon space h e n c e space y left parenthesis t right parenthesis space equals space 4 t space plus C subscript 2 end style

as earlier, since particle starts from origin, we have y=0, when t=0; applying this condition we get, y(t) = 4t ...................(4)

from (3), it is known that x(t) becomes zero again when t =4; at that instant y(t) = 4×4 = 16 m

Answered by Thiyagarajan K | 08 Jul, 2018, 10:11: PM

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