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Physics Question on Motion in a straight line

A particle is moving along a straight line path according to the relation s2=at2+2bt+cs^2 = at^2 + 2bt + c s represents the distance travelled in t seconds and a, b, c are constants. Then the acceleration of the particle varies as.

A

s−3s^{-3}

B

s3/2s^{3/2}

C

s−2/3s^{-2/3}

D

s2s^{2}

Answer

s−3s^{-3}

Explanation

Solution

s2=at2+2bt+c∴2sdsdt=2at+2bs^{2} = at^{2} + 2bt + c \therefore 2s \frac{ds}{dt} = 2at + 2b or dsdt=at+bs\frac{ds}{dt} = \frac{at+b}{s} , again differentiating d2sdt2=a.s−(at+b)s2.dsdt\frac{d^{2}s}{dt^{2}} = \frac{a.s - \left(at+b\right)}{s^{2}} . \frac{ds}{dt} =as−(at+b)(at+bs)s2= \frac{as -\left(at+b\right)\left(\frac{at+b}{s}\right)}{s^{2}} ∴d2sdt2=as2−(at+b)2s3 \therefore \frac{d^{2}s}{dt^{2}} = \frac{as^{2} - \left(at +b\right)^{2}}{s^{3}} ∴a=d2sdt2∞s−3\therefore a = \frac{d^{2}s}{dt^{2}} \infty s^{-3}