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A high-speed bullet train accelerates and decelerates at the rate of 4 ft/s 2 . Its maximum cruising speed is 90 mi/h . (Round your answers to three decimal places.) (a) What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes

User Kaanmijo
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1 Answer

2 votes

Answer:

22.9 miles

Explanation:

We are given that

Acceleration=Deceleration=a=
4ft/s^2

Maximum cruising speed,v=
90mi/h=90* (5280)/(3600)=132ft/s

1 hour=3600 s

1 mile=5280 feet

Time,t=15 minutes=
15* 60=900 s

1 min=60 s

Initial speed,u=0


v=u+at

Substitute the values


132=0+4t


t=(132)/(4)=33 s


s=u+(1)/(2)at^2=0+(1)/(2)(4)(33)^2=2178 ft

Distance,d=
speed* time=vt=132* 900=118800ft

Total distance=s+d=2178+118800=120978ft

Total distance=
(120978)/(5280)=22.9miles

Hence, the maximum distance traveled by train =22.9 miles

User Jean Vitor
by
8.3k points