We are told that the velocity increases at the rate of $5\ \text{m s}^{-1}$ per meter.

This means $\frac{dv}{dx} = 5\ \text{s}^{-1}$

We know the relation between acceleration, velocity, and position: $a = \frac{dv}{dt}$ Using the chain rule, $\frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt}$ Since $\frac{dx}{dt}=v$, $a = v\frac{dv}{dx}$

At the given point, the velocity is $v = 20\ \text{m s}^{-1}$ and $\frac{dv}{dx} = 5$

Therefore, $a = 20 \times 5 = 100\ \text{m s}^{-2}$

Hence, the acceleration of the particle is $\boxed{100\ \text{m s}^{-2}}$