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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:
A
$-\frac{2}{(x + 2)^3}$
B
$-\frac{2}{(2x + 1)^3}$
C
$+\frac{2}{(x + 1)^3}$
D
$+\frac{2}{2x + 1}$
Explanation
Differentiating gives $v = \frac{1}{2x+1}$. Then $a = v \frac{dv}{dx} = \frac{1}{2x+1} \left(-\frac{2}{(2x+1)^2}\right) = -\frac{2}{(2x+1)^3}$.
Detailed Solution
Given $t = x^2 + x$. Differentiating with respect to $x$: $\frac{dt}{dx} = 2x + 1$. Velocity $v = \frac{dx}{dt} = \frac{1}{2x + 1}$. Acceleration is $a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$. Computing $\frac{dv}{dx}$: $\frac{d}{dx}(2x + 1)^{-1} = -2(2x + 1)^{-2}$. Therefore, $a = \left(\frac{1}{2x + 1}\right) \left(-\frac{2}{(2x + 1)^2}\right) = -\frac{2}{(2x + 1)^3}$.