Question
In a panic stop, a 109 ton train can slow down with an acceleration of 0.08 g's. What is the time required to stop a train from a speed of 53 mph?
Answers
drwls
53 mph = 77.7 ft/s
Deceleration rate = 0.08g
0.08*32.2 ft/s^2 = = 2.58 ft/s^2
(deceleration)*(Time) = (initial velocity) = Vo
2.58 ft/s^2 * T = 77.7 ft/s
T = 30.2 s
You don't need to use the 109 ton mass.
Deceleration rate = 0.08g
0.08*32.2 ft/s^2 = = 2.58 ft/s^2
(deceleration)*(Time) = (initial velocity) = Vo
2.58 ft/s^2 * T = 77.7 ft/s
T = 30.2 s
You don't need to use the 109 ton mass.