Asked by Anonymous
A train normally travels at a uniform speed of 74km/h on a long stretch of straight, level track. On a particular day, the train must make a 3.0min stop at a station along this track.
If the train decelerates at a uniform rate of 1.3m/s2 and, after the stop, accelerates at a rate of 0.70m/s2 , how much time is lost because of stopping at the station?
If the train decelerates at a uniform rate of 1.3m/s2 and, after the stop, accelerates at a rate of 0.70m/s2 , how much time is lost because of stopping at the station?
Answers
Answered by
Henry
Vo = 74km/h = 74000m/3600s = 20.56 m/s.
V = Vo + a*t
t1 = (V-Vo)/a = (0-20.56)/-1.3 = 15.8 s.
t2 = 3 min = 180 s.
t3 = 20.56-0)/0.7 = 29.37 s
T=t1 + t2 + t3=15.8 + 180 + 29.37=225.2s
= 3.75 min. = Time lost.
V = Vo + a*t
t1 = (V-Vo)/a = (0-20.56)/-1.3 = 15.8 s.
t2 = 3 min = 180 s.
t3 = 20.56-0)/0.7 = 29.37 s
T=t1 + t2 + t3=15.8 + 180 + 29.37=225.2s
= 3.75 min. = Time lost.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.