Assume a maximum stopping acceleration of g/2 where g is acceleration due to gravity.
Answer:
2.99 m/s
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:
[tex]v^2-u^2 = 2as[/tex]
substitute the values to find the speed of train:
[tex]0 -u^2 = 2\times -4.9 \times 0.914 \\u^2=8.96 \\u=2.99 m/s[/tex]