Question

A high-speed bullet train accelerates and decelerates at the rate of \({\rm{4 }}{\bf{ft/}}{{\bf{s}}^{\bf{2}}}\)its maximum cruising speed \({\bf{90 mi/h}}\).

1. 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?
2. 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 this condition?
3. Find the minimum time that the train takes to travel between two consecutive stations that are 45 miles apart.
4. The trip from one station to the next takes 37.5 minutes. How far apart are the stations?

Short answer

1. Thus, the maximum distance of the train is \(22.0875{\rm{ }}mi\).
2. The maximum distance it can travel under this condition is about 21 miles.
3. The minimum time that is \(30.55{\rm{ }}\min \).
4. The distance between the stations is \(55.013{\rm{ mi}}\).

Step-by-step solution

Given information

It is given that, \(v = 90{\rm{ mi/h}}\).

It is known that, \(1{\rm{ mi/h}} = 1.467{\rm{ ft/s}}\).

Now we convert \({\rm{mi/s}}\) to \({\rm{ft/s}}\).

\(\begin{aligned}{c}v &= 90{\rm{ mi/h}}\\ &= 90 \times 1.467{\rm{ ft/s}}\\ &= 132{\rm{ ft/s}}\end{aligned}\)

Part (a)

It is given that, \(a = 4{\rm{ ft/}}{{\rm{s}}^2}\).

Differentiate \(a = 4{\rm{ ft/}}{{\rm{s}}^2}\) with respect to x,

\(\frac{{da}}{{dx}} = 4t{\rm{ ft/s}}\)

Let \(\frac{{{d^2}a}}{{d{x^2}}} = {d_1}\). Then,

\(\begin{aligned}{c}\frac{{{d^2}a}}{{d{x^2}}} &= \frac{1}{2} \times 4{t^2}{\rm{ ft/s}}\\ &= 2{t^2}{\rm{ ft/s}}\end{aligned}\)

\({d_1} = 2{t^2}{\rm{ ft/s }}......(i)\)

Here we find the Declaration of the rate

It is known that, \(4t = v\), where the value of the v is known.

Then we get,

\(\begin{aligned}{c}4t &= 132{\rm{ ft/s}}\\t &= \frac{{132}}{4}{\rm{ ft/s}}\\ &= 33{\rm{ }}s\end{aligned}\)

Put the value t in equation (1) then we get,

\(\begin{aligned}{c}{d_1} &= 2{\left( {33} \right)^2}\\ &= 2 \times 1089\\ &= 2178{\rm{ ft}}\end{aligned}\)

As the time is 15 minutes.

Here we change minutes in second,

\(\begin{aligned}{c}15{\rm{ minuts}} &= 15 \times 60{\rm{ s}}\\ &= 900{\rm{ s}}\end{aligned}\)

Then, second distance is,

\(\begin{aligned}{c}{d_2} &= 132{\rm{ ft/s}} \times 900{\rm{ s}}\\ &= {\rm{118800 ft}}\end{aligned}\).

Now here we find the maximum distance \(d = {d_2} - {d_1}\).

Put the value here \({d_1}\) and \({d_2}\) then we get.

\(\begin{aligned}{c}d &= 118800{\rm{ ft}} - 2178{\rm{ ft}}\\ &= 116622{\rm{ ft}}\\ &= 22.0875{\rm{ mi}}\end{aligned}\)

Hence, the maximum distance of the train is \(22.0875{\rm{ mi}}\).

Part (c)

Let \({d_1}\) is the minimum distance, \(t\) is declaration of the time.

From part (a), \({d_1} = 2178ft\), \(t = 33s\).

The distance between two stations is \(45{\rm{ mi}}\).

It is known that, \({\rm{1 mile}} = {\rm{5280 ft}}\). Then,

\(\begin{aligned}{c}45{\rm{ mi}} &= 45 \times 5280{\rm{ ft}}\\ &= 237600{\rm{ ft}}\end{aligned}\)

The, formula to find the new distance is, \(d = {\rm{ Distance between two stations}} - {\rm{2}}{{\rm{d}}_1}\).

\(\begin{aligned}{c}d &= 237600{\rm{ ft}} - 4356{\rm{ ft}}\\ &= 233244{\rm{ ft}}\end{aligned}\)

Next, the time for the obtained distance.

\(\begin{aligned}{c}{t_3} &= \frac{d}{v}\\ &= \frac{{233244}}{{132}}\\ &= 1767\end{aligned}\)

Now, find the minimum time as,

\(\begin{aligned}{c}{\rm{Mi}}n.{\rm{ time = }}{{\rm{t}}_3} + 2t\\{\rm{ = }}1767 + 2 \times 33\\ = 1767 + 66s\\ = 1833s\\ = 30.55\min utes\end{aligned}\)

Hence, the minimum time that is \(30.55{\rm{ }}\min \).

Part (d)

Let \({d_1}\) is the minimum distance, \(t\) is declaration of the time.

From part (a), \({d_1} = 2178ft\), \(t = 33s\).

The given time is\(37.5\)minutes.

We need to change the minutes in the second,

\(37.5{\rm{ }}\min = 2250{\rm{ }}\sec \)

Now, find the updated time as,

\(\begin{aligned}{c}{t_4} &= 37.5{\rm{ }}\min - 2t\\ &= 2250 - 2 \times 33\\ &= 2250 - 66\\ &= 2184{\rm{ s}}\end{aligned}\)

Now we find the distance,

\(d* = v \times {t_4}\).

Here put the value v and \({t_4}\)

\(\begin{aligned}{c}d &= 132 \times 2184\\ &= 288288{\rm{ ft}}\end{aligned}\)

The, the total distance will be,

\(\begin{aligned}{c}d &= 288288 + 2184\\ &= 290472{\rm{ ft}}\\ &= 55.013{\rm{ mi}}\end{aligned}\).

Thus, the distance between the stations is \(55.013{\rm{ mi}}\).