Answer:
AFter 3.5 s, the wagon is moving at: [tex]4.62\,\,\frac{m}{s}[/tex]
Explanation:
Let's start by finding first the net force on the wagon, and from there the wagon's acceleration (using Newton's 2nd Law):
Net force = 250 N + 178 N = 428 N
Therefore, the acceleration from Newton's 2nd Law is:
[tex]F=m\,*\,a\\a = \frac{F}{m} \\a= \frac{428}{325}\, \frac{m}{s^2} \\a\approx 1.32 \,\,\frac{m}{s^2}[/tex]
So now we apply this acceleration to the kinematic expression for velocity in an object moving under constant acceleration:
[tex]v_f=v_i+a\,*\,t\\v_f=0+1.32\,*\,3.5\,\,\frac{m}{s} \\v_f=4.62\,\,\frac{m}{s}[/tex]
