Antiderivatives

Open that Hampster! · #1 $Old$ November 2nd, 2009, 06:40 PM

A high-speed bullet train accelerates and decelerates at the rate of 3 ft/s2. Its maximum cruising speed is 90 mi/h. (Give your answers correct to one decimal place.)

$a(t) = 3$
$v(t) = 3x+C$ x < $\frac{90-C}{3}$
$s(t) = \frac{3x^2}{2}+Cx+D$

(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.

I was trying to solve for C originally, at t=0, v = 0, which gave me C = 0, but doing the same thing for distance gives me that D = 0 too. This seems...wrong.

I'm too tempted to figure out how to solve these using equations of motion, so I'm drawing a blank on what to do for these above. Any hints would be appreciated.

Scott H · #2 $Old$ November 3rd, 2009, 05:04 AM

You are correct that the initial velocity, $C$ , and the initial distance, $D$ , are both $0$ . This gives us

$\begin{aligned}a(t)&=3\\v(t)&=3t\\s(t)&=\frac{3}{2}t^2.\end{aligned}$

We must also remember to convert miles per hour into feet per second:

$90\frac{\mbox{mi}}{\mbox{h}}=90\cdot\frac{5280\,\mbox{ft}}{3600\,\mbox{s}}=132\frac{\mbox{ft}}{\mbox{s}}.$

The maximum speed of the train is therefore attained in $\frac{132}{3}=44$ seconds. Now we may solve the problems, remembering that

$\begin{aligned}15\,\mbox{min}&=15\cdot 60\,\mbox{s}=900\,\mbox{s}\\45\,\mbox{min}&=45\cdot 60\,\mbox{s}=2700\,\mbox{s}.\end{aligned}$

Open that Hampster! · #3 $Old$ November 3rd, 2009, 05:01 PM

Still not conceptualizing this one properly.

For (a) I did that since it's maximum cruising speed is $132\frac{ft}{s}$ , I multiplied it by 856 to get the distance it will travel after reaching max speed, and then to find the distance traveled while accelerating I plugged t=44 into s(t). Came up with 21 miles (or something around there) and was told that the answer was incorrect. Where did I go wrong? I know (b) is a similar situation, just with the 44s on the beginning and the end.

Scott H · #4 $Old$ November 3rd, 2009, 07:56 PM

Sounds like you put part of the statement of (a) together with part of (b) in your calculation. As the car accelerates and then runs for $15$ minutes, we obtain

$s(44)+132\cdot 900$

for the total distance the car travels in (a).

Open that Hampster! · #5 $Old$ November 4th, 2009, 04:26 PM

I'm starting to lose my ability to reason things.

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

45 miles = 237600 feet

Time to accelerate to max velocity is 44s.

s(44) = 2904

$237600-2904 = 132t_2$
Total time = 1778 + 44 = 1822s

They want it in minutes:

1822s = 30.4 min

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

$s(t) = \frac{3}{2}t^2 => s(2250) = \frac{3}{2}2250^2$
$s(2250) = 7593750$

Which leads to like 1400 miles or something, which doesn't make any sense.

Where am I going wrong on these two? ._.

Scott H · #6 $Old$ November 4th, 2009, 06:44 PM

The train must slow down before arrival at the next station. Therefore, in (b),

$132t_2=237600-2s(44).$