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

$8 \, \text{m/s}^2$