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Differential equations: Alison and Kevin are drag racing. They both travel with a constant acceleration. It takes Alison 3 seconds to complete the last 1/4 of the race. It takes Kevin 4 seconds to complete the last 1/3 of the race. Who won and by how much? I have the answers, but I want to know how to get them. Thanks.

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this is diff eq'n? sounds like it can be solved via physics.
Well, it can be solved by physics. But it's in my sister's diff eq text book...
@lgbasallote help you took the course don't run D:

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Other answers:

Let accleration =a then, velocity(v)= at+c
now, let total time taken By Alison to complete the race be x then |dw:1347849846129:dw|
Saura, I have to go to bed, but rest assured I will read this tomorrow. Thank you so much for helping out :D
|dw:1347850773445:dw|
Let y be the timetaken by KELVIN to complete the race.^
|dw:1347851115531:dw||dw:1347851153379:dw|
Solving i and ii....... we get: |dw:1347851218710:dw|
Thus, ALISON won the race by 1/2 seconds
Unfortunately, that's not what my book tells me...
Alison and Kevin are drag racing. They both travel with a constant acceleration. It takes Alison 3 seconds to complete the last 1/4 of the race. It takes Kevin 4 seconds to complete the last 1/3 of the race. Who won and by how much? Okee, soo... If s(t) - distance travelled, then s''(t) = k
To get speed: v(t) = s'(t) = Int(k,t) = kt + C
Yessir. Then, we have \(\large s(t)=\frac{1}{2}at^2+v_0t+s(0)\)
You're right. I think we can just say s(0) as 0 because they're both travelling the same amount of distance. So we can just assume they both start at s=0 and end at s=s1
So s(t) = (V_i)t+(1/2)at^2
Start velocity is 0 as well.
but, start velocity for the last 1/4 and 1/3 of the race isn't 0.
It doesn't say that...
Nor, of course is starting distance.
oh drag racing
It doesn't say that, but I paraphrased te problem badly. It should be start velocity=0
So, s(t) = (1/2)at^2
yup
This is all I've gotten. I've drawn a picture and proved to myself that the problem is solvable, but i do want to solve it with diff eq and without a picture.
Wait I'm dumb, we're taking the somewhat wrong approach.. For Alison: d1 = (1/2)a*(t1)^2, v1 = a*(t1), d2 = (1/2)a*(t2)^2 + (v1)*(t2) + (d1), d1 = d2 - (d2/4), t = t1+t2, and t2=3
geezz that took a while to type out.
For Kevin: d1 = (1/2)a*(t1)^2, v1 = a*(t1), d2 = (1/2)a*(t2)^2 + (v1)*(t2) + (d1), d1 = d2 - (d2/3), t = t1+t2, and t2=4
Mmm
did you follow? distance 1 is the first part of the distance, and d2 is the whole distance.
Why are t2's 3 and 4? I mean t is total time, and t1 is elapsed time before that last part?
t1 is time before the last part of the distance for both, and t2 is the amount of time it took them for the last part. By the way for both sets t1s are not the same, and neither are t2s, I just didn't feel like writing out t11, t12 and t21, t22
now we just have ts to plugin and a few equations to solve, and this seems to be a pita, but, here goes: For Alison: d1 = (1/2)a*(t1)^2, v1 = a*(t1), d2 = (1/2)a*(t2)^2 + (v1)*(t2) + (d1), d1 = d2 - (d2/4), t = t1+t2, and t2=3 **** t = t1+3 d2 = (1/2)a*(3)^2 + (v1)*3 + (d1) = (9/2)a + 3(v1) + (d1) d2 = (9/2)a+3(v1)+(d2 - (d2/4)) 0 = (9/2)a + 3(v1) - (d2/4) (d2)/4 = (9/2)a + 3(v1) d2 = 18a + 12(v1) Now substitute: v1=a*(t1) d2 = 18a + a*(t1)
omg that's a lot of algebras.
For Kevin: d1 = (1/2)a*(t1)^2, v1 = a*(t1), d2 = (1/2)a*(t2)^2 + (v1)*(t2) + (d1), d1 = d2 - (d2/3), t = t1+t2, and t2=4 **** t = t1+4 d2 = (1/2)a*(4)^2 + (v1)*4 + (d1) d2 = 8a + 4(v1) + (d1) d2 = 8a + 4(v1) + (d2 - (d2/3)) 0 = 8a + 4(v1) - (d2/3) (d2)/3 = 8a + 4(v1) d2 = 24a + 12(v1) and plugging in v1=a*(t1) d2 = 24a + a*(t1)
hmm is there anything we didn't use yet? I'm assuming their accelerations are constant, but different, so t1s would also be different.. but d2s are the same.
man random messages are annoying me. New users are surprised to see a random level 99 pop up.. Hang on let me try this on paper and see what I get.
kk.
okay, I solved for the time it took Alison to complete the whole thing
actually I did something wrong.... YIKES!
oh no~!
Ok, I'm done working out this algebra... Just found this: http://www.chegg.com/homework-help/questions-and-answers/drivers-alison-kevin-participating-drag-race-beginning-standing-start-proceed-constant-acc-q303648 And no, you don't need a subscription to view that answer.
Loool. That was wonderful man. Don't worry, I'm not even enrolled in diff eq, so it doesn't really matter if I get full answers. (I'm in AP calc BC right now)
i miss ap cal bc, it was my fave class in hs :/
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man, I actually don't really get it, cause this guy won't use a carat. I'm sure I will tomorow though, when I'm less of a sleepy zombie. Thanks for the help man. I can finally move on to chap 2 of my sis' textbook. Btw, would you happen to have any good suggestions for a linear algebra textbook? I need to read up on linear if I hope to understand diff eq.
I'll pm you. The one I used was okay, but I suck at linear algebra.. it kinda seems that all I know is up to calculus 1-2
omg, I just copied the text, and all the subscripts and superscripts were there. :D
coool!
Wait, what was the name of the linear book o.o
http://www.amazon.com/Matrix-Analysis-Applied-Algebra-Solutions/dp/0898714540/ref=sr_1_15?ie=UTF8&qid=1348016409&sr=8-15&keywords=linear+algebra+and+its+applications I used this.
Thanks. I'll get a copy somehow.
I think there's a free pdf of it online. Just google it
Found it. Thanks so much man!
np

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