Question

The relation between time $t$ and distance $x$ is $t=\alpha x^2+\beta x$ where $\alpha$ and $\beta$ are constants. The retardation is

• A. $2\alpha v^3$
• B. $2\beta v^3$
• C. $2\alpha \beta v^3$
• D. $2{\beta }^2{v}^3$

Answer 1

The correct option is A $2\alpha v^3$

Given $t=\alpha x^2+\beta x$
Differenting w.r.t time, we get $1=2\alpha \frac{dx}{dt}.x+\beta \frac{dx}{dt}$ $\therefore v=\frac{dx}{dt}=\frac{1}{\beta +2\alpha x}$
Differentiating one more time w.r.t $t$, we get $\frac{d^{2}x}{d{t}^2}=\frac{(\beta +2\alpha x).\frac{d}{dt}\left(1\right)-1.\frac{d}{dt}(\beta +2\alpha x)}{(\beta +2\alpha x{)}^2}$
$=\frac{-2\alpha v}{(\beta +2\alpha x{)}^2}=-2\alpha v^3$
Negative sign represents retardation.