Question

The relation between $t$ and distance $x$ is $t=a{x}^2+bx$ where $a$ and $b$ are constants. Express the instantaneous acceleration in terms of instantaneous velocity.

Answer 1

$t=a{x}^2+bx1=2ax\frac{dx}{dt}+b\frac{dx}{dt}⟶1v=\frac{dx}{dt}=\frac{1}{(2ax+b)}(2ax+b)=\frac{1}{v}⟶2$

differentiating$1$ w.r.t $t$

$0=2ax\frac{d^{2}x}{d{t}^2}+2a\frac{dx}{dt}+b\frac{d^{2}x}{d{t}^2}2av=(2ax=b)\frac{dx}{d{t}^2}$

using equation $2$

$\frac{d^{2}x}{d{t}^2}=\stackrel{⟶}{\underset{ins}{a}}=2a{v}^2$