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IA Type II -- Stopping Distances

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I'm confused about the second question asking you to "develop functions."

Does that mean to find an equation for the function?

And if yes, then how do you find the equation for the curve? Do you just use excel to graph a treadline or something like that?

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I'm confused about the second question asking you to "develop functions."

Does that mean to find an equation for the function?

And if yes, then how do you find the equation for the curve? Do you just use excel to graph a treadline or something like that?

Hey. OMG i did that portfolio in Grade 10 pre-ib math. And yes u have to develop a function algebraically. Many people did it using the calculator but do u really think ib is going to make a portfolio like that. Well anyways, u gotta work around with systems of equations. To find 2 variables u need 2 equations. to find a b c u need 3 equations. thats all i can give u.

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Hey. OMG i did that portfolio in Grade 10 pre-ib math. And yes u have to develop a function algebraically. Many people did it using the calculator but do u really think ib is going to make a portfolio like that. Well anyways, u gotta work around with systems of equations. To find 2 variables u need 2 equations. to find a b c u need 3 equations. thats all i can give u.

You can develop a function using SinReg

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Hi, I'm a junior in the IB Programme. In my Math SL class, we are doing a practice portfolio, which is the Stopping Distances. I've gotten the first equation, and I know that for the second its a 3-variable 3-substitution type of problem, but I've worked it several times and I'm still not getting the right answer. I would just use QuadReg on the calculator, but we have to show out work on how we derived the equation. Is there anyone out there who could maybe tell me what I'm doing wrong, or point me in the right direction? Any help is GREATLY appreciated because this is really stressing me out. Thanks so much!

-Kelsey :D

Stopping Distances Assignment

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Hey people,

I know it is not a hard topic for IA, however I'm stuck at one point and I'm not sure how to continue. I'm doing the 2nd step which is developing model functions.

"Speed versus thinking distance" graph is a linear, and I did that part. "Speed versus braking distance" graph is a quadratic, however I am not sure how to continue. By using GDC, I found the values, but my teacher would want to see how I solved it. I also tried derivatives, but I don't think my teacher would assign me that without studying derivatives and calculus. Anyone can give me a hint how to continue?

Thanks in advance!!!

Link for the portfolio: www.mrvignolini.com/CCDS/IB%20Maths/Stopping%20Distance.rtf

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sorry I swear I typed my reply already just now but my browser crashed before I posted it so now I have to rewrite my reply :D

I have read the task and yes it is very easy compared to the 2011-2012 tasks :)

You are asked to develop the functions using your knowledge of functions and not using technology so you cannot just use your GDC to show the function and be done :)

Let v=speed

Let st=thinking distance

Let sb=breaking distance

For v and st,

Firstly state the general expression of a linear function.

v=mst+c

There is a formula you can use to find m using two points in the graph. Then substitute another point into the equation to find c. And you are done! :S

For v and sb,

Firstly state the general expression of a quadratic function.

v=asb²+bsb+c

To find c, substitute the value of y when x=0 into the general equation.

Then substitute 2 sets of values of x and y into the equation after you find c to solve for a and b :)

After that, if you want to go the extra mile, remember the 3 conditions of discriminant related to the number of roots of the quadratic function? :)

From >0, =0 and <0, which one do you think is the case in this task? Then substitute the values of a, b and c that you have found to kind of prove it :)

Any further doubt or question just ask!

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Thanks so much!

Again, I have a problem. I'm comparing my results with the results I've got in the calculator and they do not quite match.

Let me tell you what I've done using your explanations.

I found c = 0: because breaking distance is 0, the car is at rest so its speed is 0 as well.

0 = a . (0)2 + b.0 + c

0 = 0 + c

c = 0

Things get complicated when I try to substitute 2 sets of values into the equation.

As v is speed which is at the same time equal to y, and sb ('x') therefore,

32 = a. (0.006)2 + b. (0.006)

32 = 0.000036a + 0.006b

64 = a. (0.024)2 + b. (0.024)

64 = 0.000576a + 0.024b

After these steps, I tried to solve them simultaneously (I couldn't think of any other method, but simultaneous equations).

However, in the end I got some bizarre numbers which don't match my calculator's results.

What have I done wrong? I guess I'm having a whole day brainfart! :) :)

Thank God, I think I found it!

Is your answer 5.859375 x 10-6x2?

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that is for the 2nd graph, right?

SORRY!! I forgot to point this out, to find c, you look at the graph, where the curve cuts the y-axis. If you use TI, I think you can trace the graph and see the value of y when x=0. Let's say y=20, then yeah c=20 :)

So not 0 :)

And your next step is correct :D to solve them you can use matrices (remember inverse? :S), elimination or substitution :)

And no, my answer is different. I think the equation of the curve can vary, depending on which points you choose to make the graph.

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vivo per lei, you substitute the wrong values when trying to find the equation for the quadratic. You must equal the Y values with braking distance, because remember the graph is speed versus braking distance.

for instance

0.006 = a(32)square + b(32)+0.0006

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