Let d = the required stopping distance of the air bag.
The initial velocity is
u = 76 km/h
= (76000/3600 m/s)
= 21.111 m/s
The maximum acceleration (actually deceleration) is -250 m/s².
The final velocity is zero, therefore
0² = (21.111 m/s)² + 2*(-250 m/s²)*(d m)
Obtain
d = 0.8914 m
Because the acceleration decreases when the stopping distance increases, a stopping distance of 1 m would be a good design choice.
Answer:
The stopping distance is 0.9 m (nearest tenth)
