   Chapter 3.9, Problem 73E

Chapter
Section
Textbook Problem

# A high-speed bullet train accelerates and decelerates at the rate of 4 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 37.5 minutes. How far apart are the stations?

To determine

a)

To find:

The maximum distance the train can travel.

Explanation

1) Concept:

If F is an antiderivative of  f on an interval I, then the most general antiderivative of f on I is, Fx+c, where c is an arbitrary constant.

2) Given:

Acceleration=4 ft/s2

Maximum cruising speed is 90 mi/h

3) Calculations:

First to convert acceleration unit of feet per second square to meter per minute square

Acceleration=4 ft/s2

αt=4ft/s2

Taking the general antiderivative of αt,

α't=4t+v0, where v0 is the initial velocity.

As α't=vt,

vt=4t+v0,

But initial velocity is v0=0

So, substitute t=0 in vt we get,

v0=40+v0

v0=0

Substitute v0=0 in vt,

vt=4tft/s

Maximum cruising speed is 90 mi/h

vt=90mih=90·52803600=132ft/s

So, maximum time is

4t=132

t=1324

t=33 sec

Now, taking the general antiderivative of vt we get,

v't=4t22+D , where D is constant

To determine

b)

To find:

The maximum distance when the train starts from rest and must come to a complete stop in 15 minutes.

To determine

c)

To Find:

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

To determine

d)

To find:

The distance between stations.

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