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inkyvoyd

  • 2 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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  1. lgbasallote
    • 2 years ago
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    this is diff eq'n? sounds like it can be solved via physics.

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

  3. inkyvoyd
    • 2 years ago
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    @lgbasallote help you took the course don't run D:

  4. sauravshakya
    • 2 years ago
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    Let accleration =a then, velocity(v)= at+c

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

  6. inkyvoyd
    • 2 years ago
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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

  7. sauravshakya
    • 2 years ago
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    |dw:1347850773445:dw|

  8. sauravshakya
    • 2 years ago
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    Let y be the timetaken by KELVIN to complete the race.^

  9. sauravshakya
    • 2 years ago
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    |dw:1347851115531:dw||dw:1347851153379:dw|

  10. sauravshakya
    • 2 years ago
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    Solving i and ii....... we get: |dw:1347851218710:dw|

  11. sauravshakya
    • 2 years ago
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    Thus, ALISON won the race by 1/2 seconds

  12. inkyvoyd
    • 2 years ago
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    Unfortunately, that's not what my book tells me...

  13. inkyvoyd
    • 2 years ago
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    @zepp

  14. bahrom7893
    • 2 years ago
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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

  15. bahrom7893
    • 2 years ago
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    To get speed: v(t) = s'(t) = Int(k,t) = kt + C

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

  17. bahrom7893
    • 2 years ago
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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

  18. bahrom7893
    • 2 years ago
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    So s(t) = (V_i)t+(1/2)at^2

  19. inkyvoyd
    • 2 years ago
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    Start velocity is 0 as well.

  20. inkyvoyd
    • 2 years ago
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    but, start velocity for the last 1/4 and 1/3 of the race isn't 0.

  21. bahrom7893
    • 2 years ago
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    It doesn't say that...

  22. inkyvoyd
    • 2 years ago
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    Nor, of course is starting distance.

  23. bahrom7893
    • 2 years ago
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    oh drag racing

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

  25. bahrom7893
    • 2 years ago
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    So, s(t) = (1/2)at^2

  26. inkyvoyd
    • 2 years ago
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    yup

  27. inkyvoyd
    • 2 years ago
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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.

  28. bahrom7893
    • 2 years ago
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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

  29. bahrom7893
    • 2 years ago
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    geezz that took a while to type out.

  30. bahrom7893
    • 2 years ago
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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

  31. inkyvoyd
    • 2 years ago
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    Mmm

  32. bahrom7893
    • 2 years ago
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    did you follow? distance 1 is the first part of the distance, and d2 is the whole distance.

  33. inkyvoyd
    • 2 years ago
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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?

  34. bahrom7893
    • 2 years ago
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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

  35. bahrom7893
    • 2 years ago
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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)

  36. inkyvoyd
    • 2 years ago
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    omg that's a lot of algebras.

  37. bahrom7893
    • 2 years ago
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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)

  38. bahrom7893
    • 2 years ago
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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.

  39. bahrom7893
    • 2 years ago
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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.

  40. inkyvoyd
    • 2 years ago
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    kk.

  41. bahrom7893
    • 2 years ago
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    okay, I solved for the time it took Alison to complete the whole thing

  42. bahrom7893
    • 2 years ago
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    actually I did something wrong.... YIKES!

  43. inkyvoyd
    • 2 years ago
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    oh no~!

  44. bahrom7893
    • 2 years ago
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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.

  45. inkyvoyd
    • 2 years ago
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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)

  46. bahrom7893
    • 2 years ago
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    i miss ap cal bc, it was my fave class in hs :/

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  47. inkyvoyd
    • 2 years ago
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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.

  48. bahrom7893
    • 2 years ago
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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

  49. inkyvoyd
    • 2 years ago
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    omg, I just copied the text, and all the subscripts and superscripts were there. :D

  50. bahrom7893
    • 2 years ago
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    coool!

  51. inkyvoyd
    • 2 years ago
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    Wait, what was the name of the linear book o.o

  52. inkyvoyd
    • 2 years ago
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    Thanks. I'll get a copy somehow.

  53. bahrom7893
    • 2 years ago
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    I think there's a free pdf of it online. Just google it

  54. inkyvoyd
    • 2 years ago
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    Found it. Thanks so much man!

  55. bahrom7893
    • 2 years ago
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    np

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