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inkyvoyd
 3 years ago
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.
inkyvoyd
 3 years ago
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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this is diff eq'n? sounds like it can be solved via physics.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Well, it can be solved by physics. But it's in my sister's diff eq text book...

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1@lgbasallote help you took the course don't run D:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let accleration =a then, velocity(v)= at+c

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now, let total time taken By Alison to complete the race be x then dw:1347849846129:dw

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Saura, I have to go to bed, but rest assured I will read this tomorrow. Thank you so much for helping out :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1347850773445:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let y be the timetaken by KELVIN to complete the race.^

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1347851115531:dwdw:1347851153379:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Solving i and ii....... we get: dw:1347851218710:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thus, ALISON won the race by 1/2 seconds

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Unfortunately, that's not what my book tells me...

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1Alison 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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1To get speed: v(t) = s'(t) = Int(k,t) = kt + C

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Yessir. Then, we have \(\large s(t)=\frac{1}{2}at^2+v_0t+s(0)\)

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1You'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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1So s(t) = (V_i)t+(1/2)at^2

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Start velocity is 0 as well.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1but, start velocity for the last 1/4 and 1/3 of the race isn't 0.

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1It doesn't say that...

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Nor, of course is starting distance.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1It doesn't say that, but I paraphrased te problem badly. It should be start velocity=0

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1So, s(t) = (1/2)at^2

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1This 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.

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1Wait 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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1geezz that took a while to type out.

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1For 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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1did you follow? distance 1 is the first part of the distance, and d2 is the whole distance.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Why are t2's 3 and 4? I mean t is total time, and t1 is elapsed time before that last part?

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1t1 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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1now 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)

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1omg that's a lot of algebras.

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1For 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)

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1hmm 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.

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1man 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.

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1okay, I solved for the time it took Alison to complete the whole thing

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1actually I did something wrong.... YIKES!

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1Ok, I'm done working out this algebra... Just found this: http://www.chegg.com/homeworkhelp/questionsandanswers/driversalisonkevinparticipatingdragracebeginningstandingstartproceedconstantaccq303648 And no, you don't need a subscription to view that answer.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Loool. 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)

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1i miss ap cal bc, it was my fave class in hs :/

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1man, 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.

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1I'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 12

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1omg, I just copied the text, and all the subscripts and superscripts were there. :D

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Wait, what was the name of the linear book o.o

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Thanks. I'll get a copy somehow.

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1I think there's a free pdf of it online. Just google it

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Found it. Thanks so much man!
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