Hey guys. I am taking AP Calc in my Senior year and we had to write a problem for the class to solve. I am pretty proud of mine and i want to see if anyone here can solve it. Take it as a fun challenge if you will. If there is anything little I can do for you like a new signature or somthing I will try it if you get the right answer. Here's the problem:

Two cars are at a Test N’ Tune at the local dragstrip. The first car, a Honda Civic, decides he thinks he can take on the track record holder who is driving a Chevrolet Camaro. They line up about 6 feet apart at the dragstrip and get ready to race. When the light turns green, the Civic takes off, but the Camaro, knowing that he has the race won, gives the Civic a head start. After 4.5 seconds, the Camaro takes off in pursuit. The car launches so hard that it instantly reaches it’s top speed, and is 5 feet ahead of the Civic when the civic is 660 feet down the dragstrip. At this point, t = 7 seconds., and the civic is only traveling 80 MPH, but has reached his top speed. How quickly is the distance between the two cars changing when the Camaro crosses the finish line at the end of the quarter mile dragstrip?

Good luck!

Note: this is purely theoretical. My teacher wouldn't let me include acceleration in the problem. Hense the "launch" bit.

 

How about nooooo!! lol

 

Awww cmon. At least try it

heres a tip, the answer is not horizontal distance, it is a diagonal between the two cars because, after all, you don't drag race on top of the other car!

 

I'll bet your answer is wrong.

 

no faith!

It's not wrong. My teacher got the same answer after yelling at me for making it too hard.

 

I'll bet your teacher got it wrong, too... you didn't consider the length of the two vehicles.

 

consider both to be particles in motion, and also assume that "about" 6 feet is actually exactly 6 feet and that the car is traveling in an exactly straight line.


I guess i took for granted that someone would actually think it through that deeply. My math class is still considering everything as a particles in motion. My AP Physics class would consider the length though

/ponders.

 

you plan on going into the engineering field in college? people dont take those AP classes for nothing.

 

Yes. If all goes as I have set it up to go, next year i'll be heading to CMU for mechanical Engineering. My real prep class for that is AP Physics. It's so incredibly difficult but i lvoe it. The book we use for APP is a 2nd year university book! It even says "University physics" on the front


edit: i suppose i should mention i am free to do whatever i please basically though. Along with AP math and Science classes, I have taken (and got all A's) in AP history, AP english, and AP german classes as well.

 

Rounded to 10 decimal places: The distance between the two is changing at 148.7876936369 fps when the Camaro finishes the race.

 

The Camaro crosses the finish line in exactly 9.46s and is traveling 181.88mph (or 266ft/s). The Civic crosses the finish line in 12.627s and is traveling 80mph (or 117ft/s). The Civic is 3.131s behind the Camaro and the distance is 366.327ft.

This is an algebra problem, not calculus, btw.

Physics is easy - wait until you take Statics as a Sophmore.... Washout central

 

It looks like he was asking a different question than the one you answered.

 

This is close but not correct. The first answer you edited out was much much closer (only 1 one hundredth off when rounded to 2 decimals.) I think i could say you got it right though, because thats just internal rounding.

just 1 question though, what prompted you to edit out the correct answer for something alittle higher?? I'm not sure where that logic came in.


if i was looking for what you gave, this then becomes a 5th grade problem (I was in advanced math and began learning algebra in 5th grade). pyro is right in his suggestion that i was looking for a different answer. One which, for all the ways i've solved it, involves very basic calculus. Although im sure there are other ways than the 2 i've seen.


and physics begins easy. but in my APP class we're getting into some pretty deep stuff. Just the sheer algebra after you have what you're looking for is incredible. So far all we've put into the equations is friction, air resistance, and pully friction (on problems that it applies too) though... Still, it's pretty hard to do i think. At least as a highschool class.

 

College level engineering classes are about 5x as hard as the toughest ones in HS. Believe me - I learned the hard way when I went in 1989.

I'm of the mindset of using simple math (algebra in this case) if possible instead of calculus or anything more challenging. But of course, do what your instructor asks when you are a student...

 

You might call that close but it's also above the upper bound of the solution.
----------
Here's the proof it's too high:
> restart;
> #So, I guess I got to this a day late but:
> #Using the spherical chicken within a vacuum assumptions:
> #Lets call X the distance between the lanes (which we will model as parallel lines).
> #Lets call y the distance down the 1/4 track (1320 feet)
> #The Camaro goes 665 ft in 7-4.5 seconds such that it's speed is:
> speed_camaro:= 665/(7-4.5);#feet per second

speed_camaro := 266.0000000

> #The civic is going 80 mph, but this is more useful in feet per second (ft/s)
> speed_civic:= 80*5280/3600.;

speed_civic := 117.3333333

> #Now, what time do we have when the Camaro wins?
> t_camaro:= 1320/speed_camaro+4.5;

t_camaro := 9.462406015

> #Just for fun, how long did it take the Camaro to run its quarter (ignoring reaction time)?
> t_camaro-4.5;

4.962406015

> #Yeah, a 4(barely)-second Camaro, please remember that unlike architects and physicists, engineers should live in the real world.
> #Where is the Camaro at the end, 1320 ft. Where is the Civic at this point?
> p_civic:= (t_camaro-7)*speed_civic+660;

p_civic := 948.9223057

> #What is the distance between them?
> d:= sqrt((1320-p_civic)^2+6^2);

d := 371.1261985

> #What is the upper limit that they could be moving apart at (infinite distance between them or x spacing of 0?
> speed_camaro-speed_civic;

148.6666667

>

 

Here is the solution to the speed between the two vehicles:

> #What is the velocity between the two objects at this point?
> #I'm going to do this with Trig rather than Vectors to prove it's not needed. A better Calc question would be to ask for the velocity between the two as a function of time.
> #Lets draw a triangle, and find the angle between the line the Camaro is on and the line going through the Camaro and Civic.
> theta:= arctan(6/d);#This is in radians

theta := 0.01616559909

> #What is the magnitude of the y-direction speed difference in the direction of the line between the two cars?
> (speed_camaro-speed_civic)*cos(theta);

148.6472418

Also, if you wanted to take vehicle length into account you would just put an offset on the y-axis distance between them (subtract length_civic+length_camaro) assuming that they are rectangles and the 6 feet is an inner to inner distance. Curving the corners wouldn't serve to make this harder given that it's only an instantaneous solution, it would just make it take more time

 

There we go. That is the answer i was looking for. Approximately 101.35MPH.


Also, i said he was close because he edited his post. The original post (which i got in my e-mail since it was, after all, the original) had a much closer answer.


And i like vectors


edit: also, I am aware of 99% of what you said. The problem is, the calc class which i had to write this problem for is about 2 years of calculus behind the APP class which i am taking. In fact it is suggested that you have taken this calc class before even considering to take my APP class, but i took ti anyways. Its the only math class that ever really challenged me.

 

> restart;
> #You'll have to pardon my using numbers like 9/2 instead of 4.5, it is required to keep my data type as integer rather than floating point thereby assuring no round off error in the intermediate calculations.
> speed_camaro:= 665/(7-9/2);#feet per second

speed_camaro := 266

> speed_civic:= 80*5280/3600;

speed_civic := 352/3

> t_camaro:= 1320/speed_camaro+9/2;

2517
t_camaro := ----
266

> #Just for fun, how long did it take the Camaro to run its quarter (ignoring reaction time)?
> t_camaro-4.5;

4.962406015

> p_civic:= (t_camaro-7)*speed_civic+660;

378620
p_civic := ------
399

> d:= sqrt((1320-p_civic)^2+6^2);

2 sqrt(5481873709)
d := ---------------
399

> speed_camaro-speed_civic;

446/3

> theta:= arctan(6/d);#This is in radians

1197 sqrt(5481873709)
theta := arctan(------------------)
5481873709

> (speed_camaro-speed_civic)*cos(theta);

223 sqrt(30058793839412535262)
---------------------------
8224959777

> evalf(%);

148.6472418

>

 

Just wait until you get into some proper college classes. I might be biased as the lowest math class offered where I got my undergrad (Mechanical) was calc 1. As for your year two physics book, physics from mechanics to electronics and optics is a sequence that should be done by the end of your freshman year (for mechanical majors). After that if you're lucky they'll start you on a proper conservation principles course and teach you that 1) You can't do anything useful with just AP level physics and 2) The assumptions made (like F=ma being remotely close to complete) are blinding. Not to say that physics didn't feel hard back when I was in high school (a mere 5 years ago) but, there's a couple orders of magnitude more out there if you get yourself into a challenging program.
Seeing as you like vectors, here's a more useful one (once you get into fluids in a couple years)

Given the following values (x,y,z) find z(P):

A(1,2,0)
B(4,5,24)
C(5,-2,4)
P(4,1,z(p))

Give me your estimate of z(P) and I'll tell you 1) The correct answer and 2) The closest you can come to the correct answer given the provided information.

 

We started fluids last year, though i missed the unit due to pneumonia (and didnt have to make it up oddly enough).

We are already almost finished with mechanics for now. we start electronics and optics later on in the year.


And dont worry. I am already way past thinking F=ma is 3 simple variables.. waaaaaay past. Although i often have a hard time remembering what i already put in the equation and what i didn't. and the sins and cos's. Man does it confuse me some times. But it's all good fun! Integrals are kinda interesting too. I hadnt done work with them until this year.

 

You don't need any fluids for that, it's just that the most common need for a set of linear interpolations is a fluids table. Due to the availability of three points rather than four, the vector approach to the above problem is however easier, or at least more through and generic when solved via a vector solution in comparison to a basic algebraic one.

 

I think you should rework the wording. To say that the car launches and
"it instantly reaches it’s top speed" implys instantaneous acceleration which is implausible. Unless this camaro has an LSX motor with infinite horsepower!!

Good thinking on the AP classes, they're good prep for college.

 

the impossibly fast quarter mile falls into that catagory as well

 

Ack. I had just come home from a bar and couldn't sleep... came across your challenge and decided to try to (attempt to) tackle it while still under the influence of some excellent American craft beer.

The first time I worked it out, I only took it out to 3 decimal places, which made the 6' distance between the cars paths of travel negligible. I redid the problem, taking it out to 10 decimal places and reposted my answer.

As I visualized the problem at the time, I missed the effect the distance between the paths of travel had on the over difference in speed. It is as obvious as the nose on my face, today, that as the distance between them approaches infinity, the difference between their speeds approachees zero. I had it as the opposite in my head when I originally posted.

I also oversimplified the problem. My calculation was:
x = sqrt((camaro_distance_travels_in_1_sec - honda_distance_travels_in_1_sec)^2 + (distance_between_the_paths_of_travel)^2))

which clearly put my answer over the upper bound of possible answers.

Thanks for the challenge, and thanks to all for the follow-up discussion. It felt pretty good to get some rust off of my math skills... even if my answer was ridiculous!

EDIT: Ok, even sober it's taking me a couple tries to explain my logic.

 

You aren't quite sober yet
As the distance between them approaches infinity the 6 feet becomes negligable, and the speed between them approaches speed_camaro-speed_civic (hence forth why this was the upper bound).
Neither car is accelerating, and therefore since their speeds are different, one car is still moving relative to the other.