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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
Option: 1
$-2av^{3}$

Option: 2
$2av^{2}$

Option: 3
$-2abv^{2}$

Option: 4
$2bv^{3}$

Answers (1)

$t=ax^{2}+bx$

Differentiate the equation with respect to $t$

$\therefore$ $1=2ax\frac{dx}{dt}+b\frac{dx}{dt}$

or $1=2axv+bv$ as $\frac{dx}{dt}=v$

or $v=\frac{1}{2ax+b}$

or $\frac{dv}{dt}=\frac{-2a(dx/dt)}{(2ax+b)^{2}}=-2av\times v^{2}$

or Acceleration = $-2av^{3}$ .

Posted by

Sumit Saini

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