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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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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} $

Can you answer this question?
 
 

1 Answer

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Answer :$-2av^2$
$t=ax^2+bx$
$\large\frac{dt}{dx} $$=2ax +b$
$\large\frac{dx}{dt}$$=(2ax +b)^{-}$
$v=(2ax +b)^{-}$
acceleration $=\large\frac{dv}{dt}$
$\qquad= \large\frac{dv}{dx} .\frac{dx}{dt}$
$\qquad= (-1) (2ax +b)^{-2}(2a)+(2ax +b)^{-1}$
$\qquad= -2a(2a+b)^{-3} =-2av^3$
answered Aug 13, 2014 by meena.p
 

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