Question

A particle is moving along a straight line path according to the relation $s^2=a{t}^2+2bt+c$ 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)

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