A high-speed bullet train accelerates and decelerates at the rate of 10 ft/s2. Its maximum cruising speed is 90 mi/h.

Question

A high-speed bullet train accelerates and decelerates at the rate of 10 ft/s2. Its maximum cruising speed is 90 mi/h. 

(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 45 miles apart.

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

Answer 1

a) Split up this situation into two kinematic equations:

     1) Accelerating: d = 1/2*10*t^2 (No initial velocity) - UNITS: ft, s

         To find t, we know the final velocity: v = 90mi/h = 132 ft/s = 10*t => t = 13.2s

         d = 5*(13.2^2) = 871.2 ft

     2) Constant speed for 15 minutes = 900s

         d = 132ft/s * 900s = 118800 ft

     Total Distance: 119671.2 ft = 22.665 mi

 

b) Split this situation up into 3 parts, Maximum distance would be maximum cruising time:

     1) Accelerating to cruising speed: d = 871.2 ft, t = 13.2s

     2) Decelerating from cruising speed:

         d = -5*t^2 + 132*t, v = 0 = -10*t + 132, t = 13.2s, d = 871.2ft

     3) Cruising:

        We know t = 900s - 13.2s - 13.2s  = 873.6s

        d = 132*873.6 = 115315.2ft

     Total Distance: 117057.6 ft = 22.17 mi

 

c) Similar to part b, only now we know the cruising distance and need to solve for time (minimum time is maximum speed, so maximum cruising time).

      1) Accelerating: d = 871.2 ft, t = 13.2s

      2) Decelerating: d = 871.2 ft, t = 13.2s

      3) Cruising:

          d = 45 mi - 871.2ft - 871.2ft = 237600ft - 1742.4ft = 235857.6ft

          t = 235857.6/132 = 1786.8s

      Total Time: 1786.8+13.2+13.2 = 1813.2s = 30.22 minutes

 

d) This part is the same as part c, only the reverse situation:

      1) Accelerating: d = 871.2ft, t = 13.2s

      2) Decelerating: d = 871.2ft, t = 13.2s

      3) Cruising:

           t = 2250 - 26.4 = 2223.6s

           d = 132*2223.6 = 293515.2ft

      Total Distance: 293515.2 + 1742.4 = 295257.6 ft = 55.92 miles