Ahigh-speed bullet train accelerates and decelerates at the rate of 10 ft/s 2 . its maximum cruising speed is 105 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?
(b) suppose that the train starts from rest and must come to a complete stop in 15 minutes. what is the maximum distance it can travel under these conditions? (c) find the minimum time that the train takes to travel between two consecutive stations that are 37.5 miles apart.(d) the trip from one station to the next takes at minimum 37.5 minutes. how far apart are the stations?

a) 26.47 miles

b) 138600 ft

c) 2.74 secs

d) 1.33 miles

Step-by-step explanation:

Rate of acceleration and deceleration = 10ft/s^2

Maximum cruising speed= 105 mi/h

Convert the cruising speed to ft/s

105*5280 / 3600

= 154 ft/s

15 mins = 15*60

= 900 secs

a) acceleration from rest = a(t)

a(t) = 10ft/s

Integrate a(t) to find v(t)

v(t) = 10t + C

Since v(0) = 0 , C = 0

V(t) = 10ft/s

To find s(t), integrate v(t)

s(t) = 10t^2/2 + C

s(t) = 5t^2 + C

When t= 0 , C = 0

s(t) = 5t^2

Time to get to maximum cruising speed = t

10t = 154

t = 154/10

t = 15.4 secs

Distance traveled with the acceleration time

s(15.4) = 5(15.4)^2

= 1185.8 ft

Maximum distance traveled = 900*154

= 138600ft

Total distance traveled = 1185.8 + 138600 =139785.8 ft

Convert to miles

139785.8/5280

= 26.47 miles

When s = 37.5

Recall that s(t) = 5t^2

37.5 = 5t^2

t^2 = 37.5/5

t^2 = 7.5

t = √7.5

t = 2.74 secs

When t = 37.5

s(37.5) = 5(37.5)^2

= 7031.25ft

Convert to miles

S= 7031.25/5280

s. = 1.33 miles

v v cj van config k tbs eng r.c r.c fcb fb gb frfr