Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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.

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 2 years ago
  2. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  3. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  4. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  5. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  6. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  7. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1347850773445:dw|

    • 2 years ago
  8. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  9. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  10. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  11. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Thus, ALISON won the race by 1/2 seconds

    • 2 years ago
  12. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  13. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @zepp

    • 2 years ago
  14. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • 2 years ago
  15. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  16. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  17. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • 2 years ago
  18. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  19. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Start velocity is 0 as well.

    • 2 years ago
  20. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  21. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    It doesn't say that...

    • 2 years ago
  22. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Nor, of course is starting distance.

    • 2 years ago
  23. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    oh drag racing

    • 2 years ago
  24. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  25. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  26. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yup

    • 2 years ago
  27. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 2 years ago
  28. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • 2 years ago
  29. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    geezz that took a while to type out.

    • 2 years ago
  30. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • 2 years ago
  31. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Mmm

    • 2 years ago
  32. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  33. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  34. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • 2 years ago
  35. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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)

    • 2 years ago
  36. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    omg that's a lot of algebras.

    • 2 years ago
  37. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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)

    • 2 years ago
  38. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 2 years ago
  39. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 2 years ago
  40. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    kk.

    • 2 years ago
  41. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  42. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    actually I did something wrong.... YIKES!

    • 2 years ago
  43. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    oh no~!

    • 2 years ago
  44. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 2 years ago
  45. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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)

    • 2 years ago
  46. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
    1 Attachment
  47. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 2 years ago
  48. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • 2 years ago
  49. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  50. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    coool!

    • 2 years ago
  51. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  52. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Thanks. I'll get a copy somehow.

    • 2 years ago
  53. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  54. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Found it. Thanks so much man!

    • 2 years ago
  55. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    np

    • 2 years ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.