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

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- lgbasallote

this is diff eq'n? sounds like it can be solved via physics.

- inkyvoyd

Well, it can be solved by physics. But it's in my sister's diff eq text book...

- inkyvoyd

@lgbasallote help you took the course don't run D:

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## More answers

- anonymous

Let accleration =a
then, velocity(v)= at+c

- anonymous

now, let total time taken By Alison to complete the race be x then
|dw:1347849846129:dw|

- inkyvoyd

Saura, I have to go to bed, but rest assured I will read this tomorrow. Thank you so much for helping out :D

- anonymous

|dw:1347850773445:dw|

- anonymous

Let y be the timetaken by KELVIN to complete the race.^

- anonymous

|dw:1347851115531:dw||dw:1347851153379:dw|

- anonymous

Solving i and ii....... we get:
|dw:1347851218710:dw|

- anonymous

Thus, ALISON won the race by 1/2 seconds

- inkyvoyd

Unfortunately, that's not what my book tells me...

- inkyvoyd

@zepp

- bahrom7893

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

- bahrom7893

To get speed:
v(t) = s'(t) = Int(k,t) = kt + C

- inkyvoyd

Yessir. Then, we have \(\large s(t)=\frac{1}{2}at^2+v_0t+s(0)\)

- bahrom7893

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

- bahrom7893

So s(t) = (V_i)t+(1/2)at^2

- inkyvoyd

Start velocity is 0 as well.

- inkyvoyd

but, start velocity for the last 1/4 and 1/3 of the race isn't 0.

- bahrom7893

It doesn't say that...

- inkyvoyd

Nor, of course is starting distance.

- bahrom7893

oh drag racing

- inkyvoyd

It doesn't say that, but I paraphrased te problem badly. It should be start velocity=0

- bahrom7893

So, s(t) = (1/2)at^2

- inkyvoyd

yup

- inkyvoyd

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.

- bahrom7893

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

- bahrom7893

geezz that took a while to type out.

- bahrom7893

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

- inkyvoyd

Mmm

- bahrom7893

did you follow?
distance 1 is the first part of the distance, and d2 is the whole distance.

- inkyvoyd

Why are t2's 3 and 4? I mean t is total time, and t1 is elapsed time before that last part?

- bahrom7893

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

- bahrom7893

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)

- inkyvoyd

omg that's a lot of algebras.

- bahrom7893

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)

- bahrom7893

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.

- bahrom7893

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.

- inkyvoyd

kk.

- bahrom7893

okay, I solved for the time it took Alison to complete the whole thing

- bahrom7893

actually I did something wrong.... YIKES!

- inkyvoyd

oh no~!

- bahrom7893

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.

- inkyvoyd

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)

- bahrom7893

i miss ap cal bc, it was my fave class in hs :/

##### 1 Attachment

- inkyvoyd

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.

- bahrom7893

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

- inkyvoyd

omg, I just copied the text, and all the subscripts and superscripts were there. :D

- bahrom7893

coool!

- inkyvoyd

Wait, what was the name of the linear book o.o

- bahrom7893

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.

- inkyvoyd

Thanks. I'll get a copy somehow.

- bahrom7893

I think there's a free pdf of it online. Just google it

- inkyvoyd

Found it. Thanks so much man!

- bahrom7893

np

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