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

$\begin{array}{1 1} 2av^2 \\ -2av^2 \\ 2bv^2 \\-2abv^2 \end{array}$

Answer :$-2av^2$
$t=ax^2+bx$
$\large\frac{dt}{dx} $$=2ax +b \large\frac{dx}{dt}$$=(2ax +b)^{-}$
$v=(2ax +b)^{-}$
acceleration $=\large\frac{dv}{dt}$
$\qquad= \large\frac{dv}{dx} .\frac{dx}{dt}$
$\qquad= (-1) (2ax +b)^{-2}(2a)+(2ax +b)^{-1}$
$\qquad= -2a(2a+b)^{-3} =-2av^3$