Question

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

• A. $s^{-3}$
• B. $s^{3/2}$
• C. $s^{-2/3}$
• D. $s^{2}$

Answer 1

Answer: (A)

$s^{2} = at^{2} + 2bt + c \therefore 2s \frac{ds}{dt} = 2at + 2b $
or $\frac{ds}{dt} = \frac{at+b}{s}$ , again differentiating
$ \frac{d^{2}s}{dt^{2}} = \frac{a.s - \left(at+b\right)}{s^{2}} . \frac{ds}{dt} $
$= \frac{as -\left(at+b\right)\left(\frac{at+b}{s}\right)}{s^{2}}$
$ \therefore \frac{d^{2}s}{dt^{2}} = \frac{as^{2} - \left(at +b\right)^{2}}{s^{3}}$
$ \therefore a = \frac{d^{2}s}{dt^{2}} \infty s^{-3} $