Answer:
22.9 miles
Step-by-step explanation:
We are given that
Acceleration=Deceleration=a=[tex]4ft/s^2[/tex]
Maximum cruising speed,v=[tex]90mi/h=90\times \frac{5280}{3600}=132ft/s[/tex]
1 hour=3600 s
1 mile=5280 feet
Time,t=15 minutes=[tex]15\times 60=900 s[/tex]
1 min=60 s
Initial speed,u=0
[tex]v=u+at[/tex]
Substitute the values
[tex]132=0+4t[/tex]
[tex]t=\frac{132}{4}=33 s[/tex]
[tex]s=u+\frac{1}{2}at^2=0+\frac{1}{2}(4)(33)^2=2178 ft[/tex]
Distance,d=[tex]speed\times time=vt=132\times 900=118800ft[/tex]
Total distance=s+d=2178+118800=120978ft
Total distance=[tex]\frac{120978}{5280}=22.9miles[/tex]
Hence, the maximum distance traveled by train =22.9 miles
