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 of the particle, we need to first determine the relationship between velocity and acceleration. The acceleration $a$ can be found using the formula:

a = rac{dv}{dt}

Using the chain rule, this can also be expressed as:

a = rac{dv}{dx} \cdot \frac{dx}{dt}

Since we know $v = x^2 + 2$, we begin by differentiating this with respect to $x$:

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

Next, we substitute $x = 1$ into this derivative:

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

Now, we also need to find $\frac{dx}{dt}$, which is simply the velocity $v$ at $x = 1$:

$v = 1^2 + 2 = 3 \text{ m/s}$

Now we can calculate acceleration:

$a = \frac{dv}{dx} \cdot \frac{dx}{dt} = 2 \cdot 3 = 6 \text{ m/s}^2$

Since this is a multiple-choice question, the correct answer is B. 3 m/s².