Solution:

To find the acceleration, we need to compute the second derivatives of $x$ and $y$ with respect to $t$.

1. $x = 7t + 4t^2$
   
   • First derivative (velocity in x-direction): $$\frac{dx}{dt}=7+8t$$
   • Second derivative (acceleration in x-direction): $$\frac{d^2x}{dt^2} = 8 \, \text{m/s}^2$$
     
2. $y = 5t$
   • First derivative (velocity in y-direction): $$\frac{dy}{dt}=5$$
   • Second derivative (acceleration in y-direction): $$\frac{d^{2}y}{d{t}^2}=0{\text{m/s}}^2$$

Now, the total acceleration is given by:

$$a=\sqrt{(a_x{)}^2+(a_y{)}^2}=\sqrt{(8{)}^2+(0{)}^2}=8{\text{m/s}}^2$$

Thus, the acceleration of the particle at $t = 5 \, \text{s}$ is $8 \, \text{m/s}^2$