A high-speed bullet train accelerates and decelerates at the rate of 10 ft/s 2 .

Question

A high-speed bullet train accelerates and decelerates at the rate of 10 ft/s2. 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?
mi

(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?
mi

(c) Find the minimum time that the train takes to travel between two consecutive stations that are 52.5 miles apart.
min

(d) The trip from one station to the next takes at minimum 37.5 minutes. How far apart are the stations?
mi

Explanation / Answer

Solution:

1)a.

It helps to put the measurements in the same units. Most of the questions are in terms of miles and minutes, so it is probably convenient to convert the acceleration to mi/min^2 and the speed to mi/min.

=> 10 ft/s^2 = (10 ft/s^2)*(1 mi)/(5280 ft)*(60 s/min)^2 = (10*60^2/5280) mi/min^2 = 6.81 mi/min^2
=> 105 mi/h = (90 mi/h)*(1 h)/(60 min) = 1.75 mi/min

The distance (d) required to accelerate to speed (v) is given in terms of acceleration (a) by
d1 = v2 /(2a) = (1.75 mi/min)2 /(2*6.81 mi/min^2) = 0.225 mi

The distance traveled in 15 minutes is   d2 = (1.75 mi/min)*(15 min) = 26.25 mi

The total distance traveled in this scenario is d1 + d2 = (0.225 mi) + (26.25 mi) = 26.475 mi

1)b.

The acceleration time is (1.75 mi/min)/(6.81 mi/min^2) = 0.257 min.
The train takes as long to decelerate as it does to accelerate, so it is running at speed for
15 min - 2(0.257 min) = 14.486 min

In that time, it will travel;

d3 = (14.486 min)(1.75 mi/min) = 25.35 mi

We know the total distance traveled during the acceleration and deceleration periods is
d4 = 2*(0.225 mi) = 0.45 mi
The total travel distance is given by d3+d4 = 23.35 mi + 0.45 mi = 23.8 mi

1)c.

We can subtract the acceleration and deceleration distances from the total and find the cruising time. Then we need to add the acceleration and deceleration times.
cruise time = (52.5 mi - 2(0.225 mi))/(1.75 mi/min) = 29.743 min
total time = (cruise time) + (accelerate time) + (decelerate time)
total time = (29.743 min) + (0.257 min) + (0.257 min) = 30.257 min

1)d.

The stations are farther apart than the distance in part 1b by the distance that can be traveled at cruising speed in 22.5 minutes.
d5 = (22.5 min)(1.75 mi/min) = 39.375 mi
d3+d4+d5 = 23.8 mi + 39.375 mi = 63.175 mi

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