Chapter 4.8, Problem 57E

a.

To determine

To find the maximum distance that train can travel.

Answer to Problem 57E

  $s=22.88\text{ miles}$

Explanation of Solution

Given:

A high speed bullet train travels at rate of $4$ ft/sec² and its maximum cruising speed is $90$ miles/hour.

Calculation:

  $s=2$ (Distance travelled when accelerating) $+$ Distance travelled at constant speed.

Acceleration and deceleration have same value so distance will be the same.

  $\begin{array}{l}s=a{t}^2+v\\ t=\frac{90\text{ mph}}{4(ft/{\sec }^2)}=22.5\sec \end{array}$

  $s=4{(22.5)}^2+90(1/4)=0.38\text{miles}+22.5\text{miles}=22.88\text{miles}$

b.

To determine

To find the maximum distance that can travel under the given conditions.

Answer to Problem 57E

  $21.75\text{ miles}$

Explanation of Solution

Given:

The train starts from rest and must complete to stop in $15$ minutes.

Calculation:

  $a=4$ Ft/sec2. Initial velocity and final velocity both are zero.

Therefore maximum distance covered in $15$ minutes is

  $\begin{array}{l}s=2\left(\frac{1}{2}a{t}_1{}^2\right)+90(15-2t_1)\\ t_1=\frac{dv}{a}=22.5\sec \\ t_2=15\min -2t_1\\ s=2025\text{feet}+21.375\text{miles}\\ s=21.75\text{ miles}\end{array}$

c.

To determine

To find the minimum time that the train takes to travel between two consecutive stations that are $45$ miles apart.

Answer to Problem 57E

  $t=30\min 48\sec$ .

Explanation of Solution

Given:

A high speed bullet train travels at rate of $4$ ft/sec² and its maximum cruising speed is $90$ miles/hour.

Calculation:

The minimum time is

  $\begin{array}{l}t=2t_1+\frac{(45-2s_1)}{90mph}\\ t=45+\frac{(45-0.38)}{90mph}\\ t=1859\text{ sec}\\ t=30\min 48\sec \end{array}$

d.

To determine

To find the distance between the stations.

Answer to Problem 57E

  $d=56.25\text{ miles}$ .

Explanation of Solution

Given:

The trip from one station to other station takes $37.5$ minutes.

Calculation:

A high speed bullet train travels at rate of $4$ ft/sec² and its maximum cruising speed is $90$ miles/hour.

The distance between the stations is

  $\begin{array}{l}d=\text{ speed }\times \text{time}\\ d=90\text{ mph}\times \text{37}\text{.5min}\\ d=90\times \frac{37.5}{60}\text{ miles}\\ d=56.25\text{ miles}\end{array}$