1 mile = 5280 feet

We can first convert the acceleration value to mi/h2 and that would be more convenient.

10 ft/s2 = `(10/5280) * 3600 *3600` ` ` mi/h2

= 24545.454 mi/h2

The time to accelerate to maximum speed and decelerate from maximum speed to zero are same = t1

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1 mile = 5280 feet

We can first convert the acceleration value to mi/h2 and that would be more convenient.

10 ft/s2 = `(10/5280) * 3600 *3600` ` ` mi/h2

= 24545.454 mi/h2

The time to accelerate to maximum speed and decelerate from maximum speed to zero are same = t1

t1 = `(120 / 24545.454)` h = 0.005 h (rounded to three decimal places, the actual answe is 0.00488888 h)

t1 = 0.005 * 60 minutes = 0.3 minutes.

a) The distance travelled when accelerating = d1

`d1 = (1/2) * 0.005 h * 120 (mi)/h = 0.3 mi`

Distance travelled during the rest of time = d2

`d2 = 120 (mi/h) * (0.25 - 0.005) h = 29.4 mi`

Therefore total disdance travelled is = 29.4 mi + 0.3 mi = 29.7 mi

b)

The distance travelled during acceleration = 0.3 mi

The distnace travelled during deceleration is also same because both are same values.

Time needed to accelerate = time needed to decelerate = 0.005 h

The time travelled in constant speed = (0.25 - 0.005 - 0.005) h

= 0.24 h

The distance travelled in constant speed = d4

`d4 = 120 (mi/h) * 0.24 h = 28.8 mi`

The total distance travelled = 0.3 mi + 28.8 mi + 0.3 mi = 29.4 mi

c)

The minimum time is achieved when train is travelling at maximum speed

T min = `(60 (mi))/(120 (mi)/h) = 0.5 h = 30 minutes`

d)

The distance = the maximum speed * minimum time

Distance = `120 (mi)/h * (37.5/60) h = 75 mi`

The distance between two stations are 75 miles