The velocity of a particle, in metres per second, is given by $v = x^2 + 2$, where $x$ is its displacement in metres from the origin - HSC - SSCE Mathematics Extension 1 - Question 10 - 2018 - Paper 1

Given the velocity function $v = x^2 + 2$, we differentiate with respect to $x$:

$\frac{dv}{dx} = 2x$

Now substituting $x = 1$ into the derivative:

$\frac{dv}{dx} = 2(1) = 2$

Next, find the velocity at $x = 1$:

$v(1) = 1^2 + 2 = 3$

Finally, we can find the acceleration at $x = 1$:

$a = v \frac{dv}{dx} = 3 \cdot 2 = 6 \, \text{ms}^{-2}$

Thus, the acceleration of the particle at $x = 1$ is given by:

$6 \, \text{ms}^{-2}$

Thus, the correct answer is B. 6 ms$^{-2}$.