Question

If $S^2 = at^2 + 2bt + c$, then acceleration is

• A. Directly proportional to $S$
• B. Inversely proportional to $S$
• C. Directly proportional to $S^2$
• D. Inversely proportional to $S^3$

Answer 1

Answer: (D)

$S^2 = at^2 + 2bt + c$ .....(i)
Differentiating (i) w.r.t $'t'$, we get
$2S \frac{dS}{dt} = 2at +2b$
$ \Rightarrow \frac{dS}{dt} =\left(\frac{at +b}{S}\right)$ .....(ii)
Differentiating (ii) w.r.t. $'t'$, we get
$\frac{d^{2}S}{dt^{2}} = \frac{S \frac{d}{dt}\left(at+b\right)-\left(at+b\right) \frac{dS}{dt}}{S^{2}}$
$ \Rightarrow \:\: \:\: \frac{d^{2}S}{dt^{2}} = \frac{aS -\left(at+b\right) \frac{\left(at+b\right)}{S}}{S^{2}}$ (Using (ii))
$ \Rightarrow \:\: \:\: \frac{d^{2}S}{dt^{2}} = \frac{aS^{2} -\left(at+b\right)^{2} }{S^{3}}$
$ \Rightarrow \:\: \:\: \frac{d^{2}S}{dt^{2}} = \frac{a\left(at^{2}+2bt+c\right)-\left[\left(at\right)^{2}+b^{2}+2at b\right] }{S^{3}}$ [Using (i)]
$ \Rightarrow \:\: \:\: \frac{d^{2}S}{dt^{2}} = \frac{a^{2}t^{2} +2abt +ac - a^{2}t^{2} -b^{2} -2atb}{S^{3}}$
$ \Rightarrow \:\: \:\: \frac{d^{2}S}{dt^{2}} = \frac{ac-b^{2}}{S^{3}}$
$ \Rightarrow $ Acceleration = $ \left(\frac{d^{2}S}{dt^{2}} \right) \propto \frac{1}{S^3}$