Question

Challenging math problem.

Answer 1

Six Flags Theme Parks are considering the design of a roller coaster whose position function in seconds can be modeled by the following equation:
f(x)=40.17x^3-20.34x^2-5.74x+2.03

1. What is the velocity function for this hill on the coaster?

2. How fast (in MPH) is a rider traveling 0.743 second into this hill?

3. What's his acceleration at that point? How does this compare to the acceleration of gravity?

4. Given that a normal person can stand an acceleration about 5 times greater than that of gravity (5G) before experiencing a G-force Induced Loss of consciousness, would you support the roller coaster's construction?

Answer 2

oh ok. 120.51x^2-40.68x-5.74

Answer 3

@stamp

Answer 4

\[s(x)=40.17x^3-20.34x^2-5.74x+2.03\]\[v(x)=3(40.17)x^2-2(20.34)x-5.74\]Onward to part 2. How fast is a velocity, so \[Find\ v(0.743)\]

Answer 5

How would you go about that?

Answer 6

\[v(0.743)=3(40.17)(0.743)^2-2(20.34)(0.743)-5.74\]

Answer 7

ohh ok. =30.56

Answer 8

The acceleration? Yay! Can you show me that step too? This is not too bad. I used wolfram alpha to get the ans btw.

Answer 9

it is the 3rd deriv isn't it? of the velocity???

Answer 10

or is acc the 2nd deriv?

Answer 11

@stamp

Answer 12

\[s'(t)=v(t)\]\[v'(t)=a(t)\]

Answer 13

\[s''(t)=v'(t)=a(t)\]

Answer 14

awesome, thanks man!
That is 241.02x-40.68

@stamp

Answer 15

oh boy...

241.02(0.743) -40.68?

That is 138.4

Answer 16

Units?

Answer 17

ummm...well I guess Acc should be in m/s^2.

Answer 18

The velocity we found said in MPH, so I assume our acceleration is \[mi/h^2\]

Answer 19

True.

Answer 20

Compare this to the mi/h^2 acceleration of gravity
Perhaps take the difference?

Answer 21

Thats what I was thinking. I've got that. As far as part 4, the accl of gravity is 9.8 m/s, so does that need to be converted to mph? How would you say yes or no?

Answer 22

Hmm... I am a bit at odds with the units... I think we got a bit disorganized. Does the position function tell you the unit? Is it meters, miles?

Answer 23

The position function is in seconds.

Answer 24

I can't move on a roller coaster x seconds. Position is a distance, while seconds is a time.

Answer 25

The Position is a function of t seconds, but the output is s(t) meters? Miles?

Answer 26

Six Flags Theme Parks are considering the design of a roller coaster whose position function in seconds can be modeled by the following equation:
f(x)=40.17x^3-20.34x^2-5.74x+2.03

1. What is the velocity function for this hill on the coaster?

2. How fast (in MPH) is a rider traveling 0.743 second into this hill?

3. What's his acceleration at that point? How does this compare to the acceleration of gravity?

4. Given that a normal person can stand an acceleration about 5 times greater than that of gravity (5G) before experiencing a G-force Induced Loss of consciousness, would you support the roller coaster's construction?'

Answer 27

I see that from your original post. I find it hard to believe that they did not tell you the unit for your position function... that is troubling.

Answer 28

I'm assuming to constantly use MPH. I agree, thats why I'm assuming.

Answer 29

The function is in seconds though, that means if s(t seconds)=miles, s'(t)=v(t)=miles/second, and s''(t)=v(t)=(miles/second)/second

Answer 30

Yea, I think that is how I'm going to finish it.

Answer 31

And if our v(t) at point t = 0.743 seconds is v(0.743) miles/sec, we have to convert to give our answer in miles/hour

Answer 32

See how this is all going to hell if we do not have a unit for our position vector :(

Answer 33

Fortunately I think this is an easy enough fix. Convert the answer we obtained for v(0.743s) miles/second into v(0.743) miles/hour, then compare this value to \[a_{g}≈21.937 mph/s\]

Answer 34

Well, that is 110016 MPH

Answer 35

miles/s to mph should be a smaller number. \[v(0.743)\frac{mi}{sec}\]\[sec=(\frac{60\ sec}{1\ min}
\frac{60\ min}{1\ hour})^{-1}\]\[v(0.743)(36^{-1}10^{-2})\ mph\]110,016 mph is FAST. Even for a rollercoaster.

Answer 36

@AmTran_Bus Can you simplify that to mph?

Answer 37

I'm trying...

Answer 38

In other words\[\frac{v(0.743)}{3600}mph\]

Answer 39

http://www.wolframalpha.com/input/?i=%280.743%29%2F3600

Answer 40

wait.....

Answer 41

That is http://www.wolframalpha.com/input/?i=30.56%2F3600

Answer 42

Here is v(0.743)
http://www.wolframalpha.com/input/?i=-5.74-40.68+x%2B120.51+x^2+for+x%3D0.743
I am not doing anything with this, just leaving it here for when we need to refer back to v(0.743).

Answer 43

I am more confused than a dead fish on a hot summers day.

Ok...now that makes sense.

Answer 44

Graph of s(t), miles: http://www.wolframalpha.com/input/?i=40.17x^3-20.34x^2-5.74x%2B2.03
Graph of v(t), miles/s: http://www.wolframalpha.com/input/?i=-5.74-40.68+x%2B120.51+x^2&lk=1&a=ClashPrefs_*Math-
Graph of a(t), mps/s: http://www.wolframalpha.com/input/?i=-40.68%2B241.02+x&lk=1&a=ClashPrefs_*Math-

Answer 45

Let's think back to Part 2, how fast in mph is the coaster at t = 0.743 s? Let's refer to our graph of v(t), and find v(0.743).

Answer 46

Visually, it looks to be ~ 30.
v(0.743), mi/s = http://www.wolframalpha.com/input/?i=-5.74-40.68+x%2B120.51+x^2+for+x%3D0.743

Answer 47

And I got 30.56

Answer 48

so we need MPH.

Answer 49

Yes so v(0.743) is 30.56 miles / second. What do we get converting to mph, per Part 2's request? Miles per hour should be a smaller number.\[\frac{30.56\ mi}{1\ second}\]\[1\ second=\frac{1}{3600}hour\]

Answer 50

Ok haha I think I am confused on the unit conversion... the number is bigger. Perhaps your original of 110,000 was correct?

Answer 51

@stamp considering my teacher and he made this test and part 4's question, I think it is right. Lets go on from here.

Answer 52

Yes, let us think of it like this... miles per second, you have less time to go a distance, so mps will be less than miles per hour. In one hour, you can cover more distance, so mph will be greater than mps.
\[v(0.743s)=3600(30.56)mph\]\[a(0.743s)=138.398\ mps/s\]\[a_{g}≈21.937 mph/s\]

Answer 53

How would you answer Part IV?

Answer 54

well, I think I would say NO, this is a stupid idea.

Answer 55

Amen.

Answer 56

@AmTran_Bus Thank you for the problem, and thank you for sticking it out with me. I would ask your teacher next class what was going on with the position function, and ask why it did not have units. It is pretty ugly to go through a problem having to "assume" things, your teacher should understand the importance of units, especially if he is having you convert and consider them in a problem.

Answer 57

The most important thing to take from this problem going into a future quiz or test is \[s'(t)=v(t)\]\[s''(t)=v'(t)=a(t)\]

Answer 58

THANK YOU SOOOO MUCH.. You will never know how much I appreciate you!

Will remember!

BTW, for a formal answer for part 4, should I change 9.8 m/s^2 to MPH...and then say NO?

Answer 59

It depends on the context of this problem. Is this a take-home quiz, where no outside sources are to be used? In that case, I really do not know if you are supposed to have the conversion from m/s and mph on deck. I had to use the internet to look up my acceleration for gravity to compare it; normally an instructor will provide any constants or conversions needed for a problem. Answer it however you deem appropriate, but I will tell you that I had to look up my acceleration of gravity in mph/s, that is not a standard constant to have memorized like 9.8 is.

Answer 60

Ok. Yes, this was take home, but we could use sources.

Thanks and God Bless again!