Question

If $S^2=a{t}^2+2bt+c$, 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)

$S^2=a{t}^2+2bt+c$ .....(i)
Differentiating (i) w.r.t ${}^{\prime }{t}^{\prime }$, we get
$2S\frac{dS}{dt}=2at+2b$
$\Rightarrow \frac{dS}{dt}=\left(\frac{at+b}{S}\right)$ .....(ii)
Differentiating (ii) w.r.t. ${}^{\prime }{t}^{\prime }$, we get
$\frac{d^{2}S}{d{t}^2}=\frac{S\frac{d}{dt}\left(at+b\right)-\left(at+b\right)\frac{dS}{dt}}{S^2}$
$\Rightarrow \frac{d^{2}S}{d{t}^2}=\frac{aS-\left(at+b\right)\frac{\left(at+b\right)}{S}}{S^2}$ (Using (ii))
$\Rightarrow \frac{d^{2}S}{d{t}^2}=\frac{a{S}^2-{\left(at+b\right)}^2}{S^3}$
$\Rightarrow \frac{d^{2}S}{d{t}^2}=\frac{a\left(a{t}^2+2bt+c\right)-\left[{\left(at\right)}^2+b^2+2atb\right]}{S^3}$ [Using (i)]
$\Rightarrow \frac{d^{2}S}{d{t}^2}=\frac{a^2{t}^2+2abt+ac-a^2{t}^2-b^2-2atb}{S^3}$
$\Rightarrow \frac{d^{2}S}{d{t}^2}=\frac{ac-b^2}{S^3}$
$\Rightarrow$ Acceleration = $\left(\frac{d^{2}S}{d{t}^2}\right)\propto \frac{1}{S^3}$