KinematicsMedium
Question
In some appropriate units, time $(t)$ and position $(x)$ relation of a moving particle is given by $t = x^{2} + x$. The acceleration of the particle is
Options
A.$- \frac{2}{(x + 2)^{3}}$
B.$- \frac{2}{(2x + 1)^{3}}$
C.$+ \frac{2}{(x + 1)^{3}}$
D.$+ \frac{2}{2x + 1}$
Solution
$t = x^{2} + x$
$$\begin{matrix} & \frac{dt}{dx} = 2x + 1 \\ & v = \frac{dx}{dt} = \frac{1}{(2x + 1)} \\ & \frac{dv}{dx} = \frac{- 2}{(2x + 1)^{2}} \\ & a = v\frac{dv}{dx} = \frac{1}{(2x + 1)}\left\lbrack \frac{- 2}{(2x + 1)^{2}} \right\rbrack \\ & \ = - \frac{2}{(2x + 1)^{3}} \end{matrix}$$
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