Question

The relation between time $t$ and distance $x$ is $t=a{x}^2+bx$ where $a$ and $b$ are constants. The acceleration is

• A. $-2a{v}^3$
• B. $2a{v}^2$
• C. $-2ab{v}^2$
• D. $2b{v}^3$

Answer 1

The correct option is A $-2a{v}^3$

$\begin{array}{c}t=a{x}^2+bx\\ \Rightarrow \frac{dt}{dx}=2ax+b\\ \Rightarrow v=\frac{dx}{dt}=\frac{1}{2ax+b}\\ \Rightarrow \frac{dv}{dx}=\frac{-2a}{{\left(2ax+b\right)}^2}\\ a=\frac{vdv}{dx}=\frac{-2a}{{\left(2ax+b\right)}^2}+\frac{1}{\left(2ax+b\right)}\\ =\frac{-2a}{{\left(2ax+b\right)}^3}-2a{V}^3\\ Ans.\left(A\right)\end{array}$