Question:
The relation between time $t$ and distance $x$ is $t = ax^2 + bx$, where $a$ and $b$ are constants. The acceleration is:
The relation between time $t$ and distance $x$ is $t = ax^2 + bx$, where $a$ and $b$ are constants. The acceleration is:
Updated On: Mar 26, 2024
- $-2abv^2$
- $2bv^3$
- $-2av^3$
- $2av^2$
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The Correct Option is C
Solution and Explanation
Given $t = ax^2 + bx$
Differentiating $w.r.t. \,t$
$\frac{dt}{dt}=2\,ax \frac{dt}{dt}+b \frac{dx}{dt}$
$v=\frac{dx}{dt}=\frac{1}{\left(2\,ax+b\right)}$
Again differentiating, $w.r.t. \,t$
$\frac{d^{2}x}{dt^{2}}=\frac{-2a}{\left(2\,ax+b\right)^{2}}.\frac{dx}{dt}$
$\therefore f=\frac{d^{2}x}{dt^{2}}$
$=-\frac{-1}{\left(2\,ax+b\right)^{2}}. \frac{2a}{\left(2\,ax+b\right)}$
or $f=\frac{-2a}{\left(2\,ax+b\right)^{3}}$
$\therefore f=-2\,av^{3}$
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