Question

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

Answer 1

Answer:

22.9 miles

Step-by-step explanation:

We are given that

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

Maximum cruising speed,v=$90mi/h=90\times \frac{5280}{3600}=132ft/s$

1 hour=3600 s

1 mile=5280 feet

Time,t=15 minutes=$15\times 60=900 s$

1 min=60 s

Initial speed,u=0

$v=u+at$

Substitute the values

$132=0+4t$

$t=\frac{132}{4}=33 s$

$s=u+\frac{1}{2}at^2=0+\frac{1}{2}(4)(33)^2=2178 ft$

Distance,d=$speed\times time=vt=132\times 900=118800ft$

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

Total distance=$\frac{120978}{5280}=22.9miles$

Hence, the maximum distance traveled by train =22.9 miles