Acar traveling at 47 ft/sec decelerates at a constant 3 feet per second per second. how many feet does the car travel before coming to a complete stop?

What you do is use equations of motions. Since the vehicle is decelerating so the acceleration is negative. 

Using the first equation of motion
v=u+a*t
u=47 (initial velocity)
a=-3  (final velocity)
v=0 (Complete stop)
therefore, 
→t=( v - u ) / a
   t= (0 - (-47) ) / (-3)
   t=15.67 second

Now,
Using second equation of motion,

→s=u*t +1/2*a*t²

    s=47 * (15.67) + 1/2 * (-3) * (15.67²)
    s=736.49 -368.32

Answer= 368.16 feet