A commuter train has stations spaced 1 mi apart. The top speed of the train is 8
ID: 1712569 • Letter: A
Question
A commuter train has stations spaced 1 mi apart. The top speed of the train is 80 mph. The train accelerates at 5.5f/sec and decelerates at 4.5 ft/sec. It dwells at the platform for 20 sec to pick up passengers. (Part c is a real PE exam problem. Parts a and b are given herein to give you some hints of solving part c) Estimate the time it takes for the train to reach the top speed, assuming it starts from 0 mph. What is the distance it travels during the time period? (Hint: use Eqs. 5-7 through 5-9 in the textbook, remember you have to covert mph to f/sec first for you to use those equations). Estimate the time it takes for the train to brake to stop from the top speed. What is the distance it travels during the time period? (Hint: use Eqs. 5-7 through 5-9 in the textbook, remember you have to covert mph to ft/sec first for you to use those equations). a. b. What are the running speed and overall speed of the train? (Hint: you need to know the difference between running speed and overall speed. In other word, you need to know the different between running time and total time). c.Explanation / Answer
a)top speed of train=80 mph=117.33 ft/s
Acceleration=5.5 ft/s²
Time taken to reach top speed=117.33/5.5=21.3 seconds
Distance travelled=0.5*5.5*21.3²=1251.48 ft
b)deceleration=4.5 ft/s²
Time taken to stop from top speed=117.33/4.5=26.07 seconds
Distantance travelled while stopping=117.33²/(2*4.5)=1529.6 ft
c)distance between stations= 1 mile=5280 ft
Out of 5280 ft, 1251.48+1529.6=2781.08 ft is travelled in acceleration and deceleration=
Remaining distance=5280-2781.08=2498.92 ft is travelled at speed of 117.33 ft /s
Time taken to travel 2498.92 ft=2498.92/117.33=21.3 seconds
Total running time of train=21.3+26.07+21.3=68.67 seconds
Total time between two stations including dwell=88.67 second
Running speed=5280/68.67=76.9 ft/s
Overall speed=5280/88.67=59.54 ft/s
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