Question

The relation between time and distance is t=$a{x}^2+\beta x$, 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. $2{\beta }^3{v}^3$

Answer 1

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

$\begin{array}{c}So,ift=a{x}^2+bx\\ then,differentiatebothsidewithrespecttotime\left(t\right)\\ \frac{dt}{dt}=2ax.\frac{dx}{dt}+b.\frac{dx}{dt}\\ 1=2axv+bv\\ v.\left(2ax+b\right)=\frac{1}{v}\\ Againdifferentiatebothsidewithrespecttotime\left(t\right)\\ 2a.\frac{dx}{dt}=-v^{-2}.\frac{dv}{dt}\\ 2av=v^{-2}.acceleration\\ a=-2a{v}^3\\ Hence,retardation=2a{v}^3\end{array}$