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

To find the acceleration, we need to differentiate the velocity function with respect to time. We can use the chain rule here:

a = rac{dv}{dt} = rac{dv}{dx} imes rac{dx}{dt}

Given rac{dx}{dt} = v, we can substitute to get:

a = rac{dv}{dx} imes v

Differentiating $v = x^2 + 2$:

rac{dv}{dx} = 2x

Substituting $x = 1$ into the acceleration formula:

1. Find $v$ at $x = 1$: $v = (1)^2 + 2 = 3 ext{ ms}^{-1}$
2. Find rac{dv}{dx} at $x = 1$: rac{dv}{dx} = 2(1) = 2
3. Now calculate acceleration: $a = 2 imes 3 = 6 ext{ ms}^{-2}$

The acceleration of the particle at $x = 1$ is $6 ext{ ms}^{-2}$. Thus, the correct answer is C.