Question

A pedestrian steps out into the road without looking both ways beforehand, 67.5 m in front of a car that is traveling 14.3 m/s. What must the car’s braking acceleration be to avoid hitting the pedestrian (in m/s2)?

Answer 1

$$v^2=2as\\14.3^2=2(67.5)a\\a=1.52\frac{m}{s^2}$$