Assume a maximum stopping acceleration of g/2 where g is acceleration due to gravity.
Answer:
2.99 m/s
Step-by-step explanation:
Stopping distance, s = 3 ft = 0.914 m
final velocity, v = 0
a = g/2 = 4.9 m/s²
Use third equation of motion:
substitute the values to find the speed of train: