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inkyvoyd Group Title

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.

  • one year ago
  • one year ago

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

    • one year ago
  2. inkyvoyd Group Title
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    Well, it can be solved by physics. But it's in my sister's diff eq text book...

    • one year ago
  3. inkyvoyd Group Title
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    @lgbasallote help you took the course don't run D:

    • one year ago
  4. sauravshakya Group Title
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    Let accleration =a then, velocity(v)= at+c

    • one year ago
  5. sauravshakya Group Title
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    now, let total time taken By Alison to complete the race be x then |dw:1347849846129:dw|

    • one year ago
  6. inkyvoyd Group Title
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    Saura, I have to go to bed, but rest assured I will read this tomorrow. Thank you so much for helping out :D

    • one year ago
  7. sauravshakya Group Title
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    |dw:1347850773445:dw|

    • one year ago
  8. sauravshakya Group Title
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    Let y be the timetaken by KELVIN to complete the race.^

    • one year ago
  9. sauravshakya Group Title
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    |dw:1347851115531:dw||dw:1347851153379:dw|

    • one year ago
  10. sauravshakya Group Title
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    Solving i and ii....... we get: |dw:1347851218710:dw|

    • one year ago
  11. sauravshakya Group Title
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    Thus, ALISON won the race by 1/2 seconds

    • one year ago
  12. inkyvoyd Group Title
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    Unfortunately, that's not what my book tells me...

    • one year ago
  13. inkyvoyd Group Title
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    @zepp

    • one year ago
  14. bahrom7893 Group Title
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    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

    • one year ago
  15. bahrom7893 Group Title
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    To get speed: v(t) = s'(t) = Int(k,t) = kt + C

    • one year ago
  16. inkyvoyd Group Title
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    Yessir. Then, we have \(\large s(t)=\frac{1}{2}at^2+v_0t+s(0)\)

    • one year ago
  17. bahrom7893 Group Title
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    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

    • one year ago
  18. bahrom7893 Group Title
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    So s(t) = (V_i)t+(1/2)at^2

    • one year ago
  19. inkyvoyd Group Title
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    Start velocity is 0 as well.

    • one year ago
  20. inkyvoyd Group Title
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    but, start velocity for the last 1/4 and 1/3 of the race isn't 0.

    • one year ago
  21. bahrom7893 Group Title
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    It doesn't say that...

    • one year ago
  22. inkyvoyd Group Title
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    Nor, of course is starting distance.

    • one year ago
  23. bahrom7893 Group Title
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    oh drag racing

    • one year ago
  24. inkyvoyd Group Title
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    It doesn't say that, but I paraphrased te problem badly. It should be start velocity=0

    • one year ago
  25. bahrom7893 Group Title
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    So, s(t) = (1/2)at^2

    • one year ago
  26. inkyvoyd Group Title
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    yup

    • one year ago
  27. inkyvoyd Group Title
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    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.

    • one year ago
  28. bahrom7893 Group Title
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    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

    • one year ago
  29. bahrom7893 Group Title
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    geezz that took a while to type out.

    • one year ago
  30. bahrom7893 Group Title
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    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

    • one year ago
  31. inkyvoyd Group Title
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    Mmm

    • one year ago
  32. bahrom7893 Group Title
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    did you follow? distance 1 is the first part of the distance, and d2 is the whole distance.

    • one year ago
  33. inkyvoyd Group Title
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    Why are t2's 3 and 4? I mean t is total time, and t1 is elapsed time before that last part?

    • one year ago
  34. bahrom7893 Group Title
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    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

    • one year ago
  35. bahrom7893 Group Title
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    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)

    • one year ago
  36. inkyvoyd Group Title
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    omg that's a lot of algebras.

    • one year ago
  37. bahrom7893 Group Title
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    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)

    • one year ago
  38. bahrom7893 Group Title
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    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.

    • one year ago
  39. bahrom7893 Group Title
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    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.

    • one year ago
  40. inkyvoyd Group Title
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    kk.

    • one year ago
  41. bahrom7893 Group Title
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    okay, I solved for the time it took Alison to complete the whole thing

    • one year ago
  42. bahrom7893 Group Title
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    actually I did something wrong.... YIKES!

    • one year ago
  43. inkyvoyd Group Title
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    oh no~!

    • one year ago
  44. bahrom7893 Group Title
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    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.

    • one year ago
  45. inkyvoyd Group Title
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    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)

    • one year ago
  46. bahrom7893 Group Title
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    i miss ap cal bc, it was my fave class in hs :/

    • one year ago
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  47. inkyvoyd Group Title
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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.

    • one year ago
  48. bahrom7893 Group Title
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    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

    • one year ago
  49. inkyvoyd Group Title
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    omg, I just copied the text, and all the subscripts and superscripts were there. :D

    • one year ago
  50. bahrom7893 Group Title
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    coool!

    • one year ago
  51. inkyvoyd Group Title
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    Wait, what was the name of the linear book o.o

    • one year ago
  52. inkyvoyd Group Title
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    Thanks. I'll get a copy somehow.

    • one year ago
  53. bahrom7893 Group Title
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    I think there's a free pdf of it online. Just google it

    • one year ago
  54. inkyvoyd Group Title
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    Found it. Thanks so much man!

    • one year ago
  55. bahrom7893 Group Title
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    np

    • one year ago
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