The distance d traveled by an object under uniformly accelerated motion while it changes its speed from v_o to v_f with an acceleration a is given by the formula:
[tex]d=\frac{v^2_f-v^2_0}{2a}[/tex]Since the car travels initially at 16m/s and it stops, then v_0=16m/s and v_f=0. Replace those values into the formula, as well as a=-4m/s^2 to find the stopping distance for the car:
[tex]d=\frac{(0)^2-(16\frac{m}{s})^2}{2(-4\frac{m}{s^2})}=\frac{-256\frac{m^2}{s^2}}{-8\frac{m}{s^2}}=32m[/tex]Therefore, the stopping distance for the car is 32 meters.
