Question

A car traveling at 56 mph decelerates at 22 feet per second per second. how long does it take to stop?

Answer 1

Okay. Using the formula
v = v(knot) + a * t

you plug in the numbers (remember UNITS are IMPORTANT)
v=0
vknot=56 mph (82.133 fps)
a= -22 fps^2

0=82.133-22(t)
Solve for t by subtracting 82.133 to both sides and flipping the negatives
82.133=22t
divide by 22 on both sides

3.733 seconds

Hope this helped!

Answer 2

           (56 mile/hour) x (5,280 ft/mile) x (hour / 3,600 sec)

       =      82.133... ft/sec

If you lose 22 ft/sec of that speed every second,
then it'll take

                  (82.133... ft/sec) / (22 ft/sec²)

               =    3.733... seconds

to lose it all, and stop.