Question

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

Answer 1

The question is incomplete. The complete question is :

A high-speed bullet train accelerates and decelerates at the rate of 4 ft/s^2. Its maximum cruising speed is 90 mi/h. 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?

Solution :

Given :

Speed of the bullet train, v = 90 mi/h

                                            = $90 \times \frac{5280}{3600}$

                                            = 132 ft/s

Time = 15 minutes

        = 15 x 60

        = 900 s

Acceleration from rest,

$a(t) = 4 \ ft/s^2$

$v(t) = 4t + C$

Since, v(0) = 0, then C = 0, so velocity is

v(t) = 4t ft/s

Then find the position function,

$s(t) = \frac{4}{2}t^2 + C$

      $=2t^2+C$

It is at position 0 when t = 0, so C = 0, and the final position function for only the time it is accelerating is :

$s(t) = 2t^2$

Time to get maximum cruising speed is :

4t = 132

t = 33 s

Distance travelled (at cruising speed) by speed to get the remaining distance travelled.

$900 \ s \times 132 \ \frac{ft}{s} = 118800 \ ft$

Total distance travelled, converting back to miles,

$2178 + 118800 = 120978\ ft . \ \frac{mi}{5280 \ ft}$

                      = 22.9125 mi

Therefore, the distance travelled is 22.9125 miles