CBSE Class 11-science Answered

12?

Asked by smanishkumar2002 | 09 Jun, 2018, 07:55: AM

Expert Answer

Let T be the time to ascend a vertical distance y.

T = y/v₀............................(1)

Horizontal drift x(y) at y is given by

begin mathsize 12px style x left parenthesis y right parenthesis space equals space integral subscript 0 superscript T v subscript x d t space space equals space integral subscript 0 superscript T a cross times y cross times d t space equals space space integral subscript 0 superscript T a cross times v subscript 0 t cross times d t space equals space a cross times v subscript 0 cross times T squared over 2 space............ left parenthesis 2 right parenthesis B y space s u b s t i t u t i n g space T space f r o m space e q n left parenthesis 1 right parenthesis comma space w e space r e w r i t e space e q n. left parenthesis 2 right parenthesis space a s x left parenthesis y right parenthesis space equals fraction numerator a cross times y squared over denominator 2 v subscript 0 end fraction space....................... left parenthesis 3 right parenthesis end style

Resultant velocity v(y) is given by

begin mathsize 12px style v left parenthesis y right parenthesis space equals space square root of v subscript 0 superscript 2 plus a squared y squared end root end style

...............................(4)

By integrating eqn.(4) we get tangential acceleration

begin mathsize 12px style a subscript t space equals space fraction numerator d v over denominator d t end fraction equals space fraction numerator a squared v subscript 0 y over denominator square root of v subscript 0 superscript 2 plus a squared y squared end root end fraction end style

.......................(5)

Normal acceleration : a_r = v²/R ................................(6)

where R is radius of curvature. R is given by

begin mathsize 12px style R space equals space fraction numerator open square brackets 1 plus open parentheses begin display style fraction numerator d x over denominator d y end fraction end style close parentheses squared close square brackets to the power of begin display style bevelled 3 over 2 end style end exponent over denominator open parentheses fraction numerator d squared x over denominator d y squared end fraction close parentheses end fraction end style

..........................(7)

from eqn.(3), we get

begin mathsize 12px style fraction numerator d x over denominator d t end fraction equals space fraction numerator a y over denominator v subscript 0 end fraction space semicolon space space fraction numerator d squared x over denominator d t squared end fraction equals a over v subscript 0 end style

by substituting above derivatives in eqn.(7) , we get

begin mathsize 12px style R space equals space fraction numerator open parentheses v subscript 0 superscript 2 plus a squared y squared close parentheses to the power of begin display style bevelled 3 over 2 end style end exponent over denominator a cross times v subscript 0 superscript 2 end fraction space equals space fraction numerator v cubed over denominator a cross times v subscript 0 superscript 2 end fraction end style

....................(8)

using eqns.(4), (6) and (8), we get normal acceleration

begin mathsize 12px style a subscript r space equals space fraction numerator a cross times v subscript 0 superscript 2 over denominator square root of v subscript 0 superscript 2 plus a squared y squared end root end fraction end style

Total acceleration =

begin mathsize 12px style square root of a subscript r superscript 2 plus a subscript t superscript 2 end root space equals space square root of fraction numerator a squared v subscript 0 superscript 4 over denominator v squared end fraction plus fraction numerator a to the power of 4 v subscript 0 superscript 2 y squared over denominator v squared end fraction end root equals fraction numerator a v subscript 0 superscript 2 over denominator v end fraction square root of 1 plus a squared y squared end root end style

Answered by Thiyagarajan K | 09 Jun, 2018, 09:00: PM

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