In some appropriate units, time (t) and position (x) relation of a moving particle is given by t = x² + x. The acceleration of the particle is

We are given the relation between time *t* and position *x* as

$t=x^2+x$

To find the acceleration \(a=\dfrac{d^{2}x}{dt^{2}}\) we first differentiate *t* with respect to *x*:

$\frac{\mathrm{dt}}{\mathrm{dx}}=2x+1$

The velocity *v* is the reciprocal of this derivative because

$v=\backslash dfrac\{dx\}\{dt\}=\; \backslash dfrac\{1\}\{\backslash dfrac\{dt\}\{dx\}\}$

Thus

$v=\backslash dfrac\{1\}\{2x+1\}$

Now differentiate *v* with respect to *x*:

$\frac{\mathrm{dv}}{\mathrm{dx}}=\backslash dfrac\{-2\}\{\backslash left(2x+1\backslash right)^\{2\}\}$

Finally, use the chain rule \(a = \dfrac{dv}{dt} = \dfrac{dv}{dx}\cdot\dfrac{dx}{dt}\). Since \(\dfrac{dx}{dt}=v=\dfrac{1}{2x+1}\), we obtain

$a=\backslash dfrac\{-2\}\{\backslash left(2x+1\backslash right)^\{2\}\}\backslash cdot\backslash dfrac\{1\}\{2x+1\}=\; \backslash dfrac\{-2\}\{\backslash left(2x+1\backslash right)^\{3\}\}$

Therefore the acceleration of the particle is

$a=\backslash dfrac\{-2\}\{\backslash left(2x+1\backslash right)^\{3\}\}$