Monday, September 28, 2009

Period1: IROC

Please contribute your calculations for the instananteous rate of change of the instance of Usain Bolt's 2008 Olympic run that you were responsible for. Please remember to demonstrate good mathematical form including proper units.

(Complete today's homework assignment from the homework log.)
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26 comments:

  1. magzSeptember 28, 2009 at 5:10 PM

    Assigned Point of Race: 2.5 seconds

    D(2.5)=0.006163(2.5)^4-0.191217(2.5)^3+2.09964(2.5)^2+2.34451(2.5)-0.65173
    =15.58527156

    t: delta t:
    2.5 15.58527156
    2.6 16.55836672
    2.51 15.68178688
    2.501 15.59491504

    AROC(2.5,2.6)
    =(16.55836672-15.58527156) / 2.6-2.5
    =9.7 m/s

    AROC(2.5,2.51)
    =(15.68178688-15.58527156) / 2.51-2.5
    =9.65 m/s

    AROC(2.5, 2.501)
    =(15.59491504-15.58527156) / 2.501-2.5
    =9.64 m/s

    Therefore, I would approximate the slope of the tangent at 2.5 seconds to be about 9.64 m/s.

    ReplyDelete
  2. alex.pSeptember 28, 2009 at 6:51 PM

    slope of tangent at 1 second

    t: delta t:
    1.0 3.607366
    1.1 4.2223088
    1.01 3.66747
    1.001 3.6133624
    1.0001 3.6079655

    AROC (1.0,1.1)
    (4.2223088-3.607366)/(1.1-1.0)
    = 6.14928

    AROC (1.0,1.01)
    (3.66747-3.607366)/(1.01-1.0)
    = 6.0104

    AROC (1.0,1.001)
    (3.6133624-3.607366)/(1.001-1.0)
    = 5.9964

    AROC (1.0,1.0001)
    (3.6079655-3.607366)/(1.0001-1.0)
    = 5.995

    Therefore the slope of the tangent at 1 second is approximately 5.995 m/s

    ReplyDelete
  3. VickyK.September 28, 2009 at 7:00 PM

    Assigned Point of Race: 5 sec.

    D(5)=0.006163(5)^4-0.191217(5)^3+2.09964(5)^2+2.34451(5)-0.65173
    =43.51157

    tv chart ...

    5 43.511757
    5.1 44.721175
    5.01 43.632397
    5.001 43.523652

    AROC[5,5.1]
    = (44.721175 - 43.511757)/(5.1 - 5)
    = 12.09418 m/s

    AROC[5,5.01]
    = (43.632397 - 43.511757)/(5.01 - 5)
    = 12.064 m/s

    AROC[5,5.001]
    = (43.523652 - 43.511757)/(5.001 - 5)
    = 11.894 m/s

    Therefore, I would approximate the slope of the tangent at 5 sec. to be about 11.894 m/s.

    ReplyDelete
  4. rennSeptember 28, 2009 at 7:25 PM

    Assigned Point of Race: 5s

    D(5) = 0.006163(5)^4 - 0.191217(5)^3 + 2.09964(5)^2 + 2.34451(5) - 0.65173
    = 43.51157 m

    t: delta t:
    5 43.51157
    5.1 44.721175
    5.01 43.632397
    5.001 43.523651
    5.0001 43.5127781


    AROC [5, 5.1]
    = (44.721175 - 43.51157) / (5.1 - 5)
    = 12.09605 m/s

    AROC [5, 5.01]
    = (43.632397 - 43.51157) / (5.01 - 5)
    = 12.0827 m/s

    AROC [5, 5.001]
    = (43.526351 - 43.51157) / (5.001 - 5)
    = 12.081 m/s

    AROC [5, 5.0001]
    = (43.5127781 - 43.51157) / (5.0001 - 5)
    = 12.081 m/s

    Therefore, I would approximate the slope of the tangent at 5s to be about 12.081 m/s.

    ReplyDelete
  5. celine123September 28, 2009 at 7:36 PM

    assigned point of race: 2seconds

    D(5)=0.006163(2)^4-0.191217(2)^3+2.09964(2)^2+2.34451(2)-0.65173= 11.004722

    Table of Values

    2= 11.004722
    2.1= 11.880151
    2.01= 11.091289
    2.001= 11.013369

    AROC [2,2.1]
    = (11.880151-11.004722)/(2.1-2)
    = 8.75429

    AROC [2,2.01]
    = (11.091289-11.004722)/(2.01-2)
    = 8.6567

    AROC [2, 2.001]
    = (11.013369-11.004722)/(2.001-2)
    =8.647

    Therefore, the slope of the tangent at 2 seconds, is approximately 8.65m.

    ReplyDelete
  6. AnsonHSeptember 28, 2009 at 7:51 PM

    Find the Instantaneous velocity at 9.5 seconds:

    D(9.5)=0.006163(9.5)^4-0.191217(9.5)^3+2.09964(9.5)^2+2.34451(9.5)-0.65173
    D(9.5)=97.36697

    D(9.6)=0.006163(9.6)^4-0.191217(9.6)^3+2.09964(9.6)^2+2.34451(9.6)-0.65173
    D(9.6)=98.527053

    D(9.51)=0.006163(9.51)^4-0.191217(9.51)^3+2.09964(9.51)^2+2.34451(9.51)-0.65173
    D(9.51)=97.482985

    D(9.501)=0.006163(9.501)^4-0.191217(9.501)^3+2.09964(9.501)^2+2.34451(9.501)-0.65173
    D(9.501)=97.378571

    D(9.5001)=0.006163(9.5001)^4-0.191217(9.5001)^3+2.09964(9.5001)^2+2.34451(9.5001)-0.65173
    D(9.5001)=97.36813

    AROC[9.5,9.6] = (Y2-Y1)/(X2-X1)
    AROC[9.5,9.6] = (98.527053m- 97.36697m) / (9.6s-9.5s)
    AROC[9.5,9.6] = 1.160083m / 0.1s
    AROC[9.5,9.6] = 11.60083m/s

    AROC[9.5,9.51] = (Y2-Y1)/(X2-X1)
    AROC[9.5,9.51] = (97.482985m- 97.36697m) / (9.51s-9.5s)
    AROC[9.5,9.51] = (0.116015m)/ (0.01s)
    AROC[9.5,9.51] = 11.6015m/s

    AROC[9.5,9.501] = (Y2-Y1)/(X2-X1)
    AROC[9.5,9.501] = (97.378571m- 97.36697m) / (9.501s-9.5s)
    AROC[9.5,9.501] = (0.011601m) / (0.001s)
    AROC[9.5,9.501] = 11.601m/s

    AROC[9.5,9.5001] = (Y2-Y1)/(X2-X1)
    AROC[9.5,9.5001] = (97.36813m- 97.36697m) / (9.5001s-9.5s)
    AROC[9.5,9.5001] = (0.00116m) / (0.0001s)
    AROC[9.5,9.5001] = 11.6m/s

    The tangent is tending towards 11.6m/s.

    Therefore, I would approximate the slope of the tangent at 9.5seconds will be about 11.6m/s.

    ReplyDelete
  7. GloriaaSeptember 28, 2009 at 8:04 PM

    This comment has been removed by the author.

    ReplyDelete
  8. AnonymousSeptember 28, 2009 at 8:07 PM

    Assigned point of race: 4.5 seconds

    D(4.5):37.5191406
    D(4.6):38.70885662
    D(4.51):37.63788019
    D(4.501):37.53101258

    AROC(4.5,4.6):11.8971602
    AROC(4.5,4.51): 11.873959
    AROC(4.5,4.501):11.87198

    Therefore I would approximate the slope of the tangent at 4.5 second to be about 11.87 m/s

    ReplyDelete
  9. GloriaaSeptember 28, 2009 at 8:15 PM

    D(4.5) = 37.5191406
    D(4.6) = 38.70885662
    D(4.51) = 37.63788019
    D(4.501) = 37.53101258

    AROC(4.5, 4.6)
    = 38.70885662-37.5191406/4.6-4.5
    = 11.8971602

    AROC(4.5, 4.51)
    = 37.63788019-37.5191406/4.51-4.5
    = 11.873959

    AROC(4.5, 4.501)
    = 37.53101258-37.5191406/4.501-4.5
    = 11.87198

    Therefore the approximate instantaneous speed of Usain at 4.5 s is 11.87198m/s

    ReplyDelete
  10. RachelleSeptember 28, 2009 at 8:30 PM

    Assigned Time: 3 seconds

    D(3) = 20.614904
    D(3.1) = 21.666412
    D(3.01) = 20.719426
    D(3.001) = 20.62535

    AROC (3,3.1)
    = 21.666412 - 20.614904/3.1-3
    = 10.51508m/s

    AROC (3,3.01)
    = 20.719426-20.614904/3.01-3
    = 10.4522 m/s

    AROC (3,3.001)
    = 20.62535-20.614904/3.001-3
    = 10.446 m/s

    Therefore the approximate instantaneous speed at 3 seconds is 10.466 m/s

    ReplyDelete
  11. John. ISeptember 28, 2009 at 8:43 PM

    D(4)= 31.66039
    D(4.1)= 32.818359
    D(4.01)= 31.775839
    D(4.001)= 31.671931

    AROC (4,4.1)
    = 32.818359 - 31.66039/4.1-4
    = 11.57969m/s

    AROC (4,4.01)
    = 31.775839-31.66039/4.01-4
    = 11.5449 m/s

    AROC (4,4.001)
    = 31.671931-31.66039/4.001-4
    = 11.541 m/s

    the instantaneous speed at 4 seconds is 11.541 m/s

    ReplyDelete
  12. corradocSeptember 28, 2009 at 8:50 PM

    D(4.5)=0.006163(4.5)^4-0.191217(4.5)^3+2.09964(4.5)^2+2.34451(4.5)-0.65173
    =37.51884106m

    t: delta t:
    4.5 37.51884106s
    4.49 37.40015534s
    4.6 37.53071258s
    4.501 37.52002819s

    AROC[4.49,4.5]
    =(37.51884106-37.40015534) / (4.5-4.49)
    =11.868572 m/s

    AROC[4.5, 4.501])
    =(37.53071258-37.51884106) / (4.501-4.5)
    =11.87142 m/s

    Therefore, I would approximate the slope of the tangent at seconds to be about 11.871 m/s.

    ReplyDelete
  13. rebecca CSeptember 28, 2009 at 8:57 PM

    Assigned time of race: 8.5 s

    Table of Values:
    8.5=85.71570006
    8.51=85.83300476
    8.501=85.72743147
    8.5001=85.71687321

    AROC[8.51-8.5]
    =(85.83300476-85.71570006)/(8.51-8.5)
    =11.73047

    AROC[8.501-8.51]
    =(85.72743147-85.83300476)/(8.501-8.51)
    =11.73036556

    AROC[8.5001-8.501]
    =(85.71687321-85.72743147)/(8.5001-8.501)
    =11.7314

    Therefore the approximate IROC at 8.5s is 11.7314m/s

    :)

    ReplyDelete
  14. BradleySeptember 28, 2009 at 9:01 PM

    To calculate the instantaneous rate of change at 1.5 seconds:

    AROC [1.5, 1.51] = 7.44913

    AROC [1.5, 1.501] = 7.4373

    AROC [1.5, 1.5001] = 7.436

    Therefore, the approximate average rate of change at 1.5 seconds is 7.436. The number is slowly approaching 7.4

    ReplyDelete
  15. Cath W.September 28, 2009 at 9:07 PM

    5.5 seconds

    D(5.5) = 0.006163(5.5)^4 - 0.191217(5.5)^3 + 2.09964(5.5) + 2.34451(5.5) - 0.65173
    ≈49.5829681

    AROC [5.5, 5.6] = 50.802472m-49.582987m / 5.6s - 5.5s
    ≈12.19485 m/s

    AROC [5.5, 5.51] = 49.704884m-49.582987m / 5.51s - 5.5s
    ≈12.1897 m/s

    AROC [5.5, 5.501] = 49.595176m-49.582987m / 5.501s - 5.5s
    ≈12.189 m/s


    ∴ I would approximate the slope of the tangent at 5.5 seconds to be about 12.189 m/s.

    ReplyDelete
  16. ElsaWSeptember 28, 2009 at 9:21 PM

    Assigned Point of Race: 8.5s

    D(8.5)= 0.006163(8.5)^4-0.191217(8.5)^3+ 2.09964(8.5)^2+2.34451(8.5)- 0.65173
    D(8.5)=85.71570

    TOV

    8.5=85.71570
    8.6= 86.88782
    8.51= 85.83300
    8.501= 85.72743

    AROC[8.5,8.6]= 86.88782-85.71570/8.6-8.5
    = 11.7212 m/s
    AROC[8.5,8.51]= 85.83300-85.71570/8.51-8.5
    = 11.73 m/s
    AROC[8.5,8.501]=85.72743-85.71570/8.501-8.5
    = 11.73 m/s
    Therefore i would approx the slope of the tangent at 8.5s to be about 11.73 m/s

    ReplyDelete
  17. GregoryLSeptember 28, 2009 at 9:36 PM

    D(6.5)= 61.785197
    D(6.6)= 63.002361
    D(6.51)= 61.907471
    D(6.501)= 61.797919

    AROC(6.6,6.5)= 63.002361-61.785197/6.6-6.5
    = 12.17164 m/s

    AROC(6.5,6.51)=61.907471-61.785197/6.51-6.5
    = 12.2274 m/s

    AROC(6.501,6.5)=61.797919-61.785197/6.501-6.5
    = 12.722 m/s

    The instantaneous velocity for 6.5 secs would be 12.722 m/s.

    ReplyDelete
  18. taylorSeptember 28, 2009 at 9:40 PM

    At 2 seconds...

    D(5)=0.006163(2)^4-0.191217(2)^3+2.09964(2)^2+2.34451(2)-0.65173= 11.004722

    TOV

    2 = 11.004722
    2.1 = 11.880151
    2.01 = 11.091289
    2.001 = 11.013369

    AROC [2,2.1]
    = (11.880151-11.004722)/(2.1-2)
    = 8.75429

    AROC [2,2.01]
    = (11.091289-11.004722)/(2.01-2)
    = 8.6567

    AROC [2, 2.001]
    = (11.013369-11.004722)/(2.001-2)
    =8.647

    Therefore, i would approximate the slope of the tangent 2s to be about 8.65 m/s.

    ReplyDelete
  19. TeresaSeptember 28, 2009 at 9:41 PM

    Assigned Point of Race: 5.5s

    These points are the calculated results in the TOV:
    (5.5, 4958298681), (5.6, 50.80247161), (5.51, 49.70488391), (5.501, 29.59517596), (5.5001, 49.58420572), (5.50001, 49.5831087)

    The average rate of change for each time interval:
    AROC[5.5, 5.6] = (Y2-Y1)/(X2-X1)
    AROC[5.5, 5.6] = (50.80247161m- 49.58298681m) / (5.6s-5.5s)
    AROC[5.5, 5.6] = 12.194848m/s

    AROC[5.5, 5.51] = (Y2-Y1)/(X2-X1)
    AROC[5.5, 5.51] = (49.70488391m- 49.58298681m) / (5.51s-5.5s)
    AROC[5.5, 5.51] = 12.18971m/s

    AROC[5.5, 5.501] = (Y2-Y1)/(X2-X1)
    AROC[5.5, 5.501] = (49.59517593m- 49.58298681m) / (5.501s-5.5s)
    AROC[5.5, 5.501] = 12.18915m/s

    AROC[5.5, 5.5001] = (Y2-Y1)/(X2-X1)
    AROC[5.5, 5.5001] = (49.58420572m- 49.58298681m) / (5.5001s-5.5s)
    AROC[5.5, 5.5001] = 12.1891m/s

    AROC[5.5, 5.50001] = (Y2-Y1)/(X2-X1)
    AROC[5.5, 5.50001] = (49.5831087m- 49.58298681m) / (5.50001s-5.5s)
    AROC[5.5, 5.50001] = 12.189m/s

    Therefore, I would approximate the slope of the tangent at 5.5 seconds to be about 12.189m/s.

    ReplyDelete
  20. AlexandriaCSeptember 28, 2009 at 9:46 PM

    Assigned Point of Race: 8.0s

    D(8)=0.006163(8)^4-0.191217(8)^3+ 2.09964(8)^2+2.34451(8)- 0.65173
    D(8)=79.821854

    Table of Values:
    8=79.821854
    8.1=81.00532
    8.01=79.94031
    8.001=79.83370

    AROC[8,8.1]=81.00532-79.821854/8.1-8
    =11.83466 m/s

    AROC[8,8.01]=79.94031-79.821854/8.01-8
    =11.8456 m/s

    AROC[8,8.001]=79.83370-79.821854/8.001-8
    =11.846 m/s

    Therefore I would approximate the slope of the tangent at 8s to be about 11.846 m/s.

    ReplyDelete
  21. MattLSeptember 28, 2009 at 9:49 PM

    I was assigned: 4 seconds

    1. (4, 31.66039)
    2. (4.1, 32.818359)
    3. (4.01, 31.775839)
    4. (4.001, 31.671931)



    AROC for:

    1.
    = (32.818359 - 31.66039)/(4.1-4)
    = 11.57969m/s


    2.
    = 31.775839-31.66039/4.01-4
    = 11.5449 m/s

    3.

    = 31.671931-31.66039/4.001-4
    = 11.541 m/s

    therefore i would approximate the instantaneous slope at 4 seconds to be 11.541m/s.

    ReplyDelete
  22. CurtisMSeptember 28, 2009 at 9:57 PM

    Assigned Time: 7.5 seconds

    Table of Values
    7.5 = 73.86729031 s
    7.6 = 75.06318768 s
    7.51 = 73.98699107 s
    7.501 = 73.87926149 s

    AROC = (v2 - v1 )/ (t2 - t1)

    AROC (7.6-7.5) = 11.9589737 s
    AROC (7.51 - 7.5) = 11.970076 s
    AROC (7.501 - 7.5) = 11.97118 s

    The approximate IROC at 7.5s is 11.9712 seconds

    ReplyDelete
  23. SharonSeptember 28, 2009 at 10:19 PM

    D(3) = 20.614904
    D(3.1) = 21.666412
    D(3.01) = 20.719426
    D(3.001) = 20.62535

    AROC [3,3.1]
    = (21.666412 - 20.614904)/(3.1-3)
    = 10.51508 m/s

    AROC [3,3.01]
    = (20.719426-20.614904)/(3.01-3)
    = 10.4522 m/s

    AROC [3,3.001]
    = (20.62535-20.614904)/(3.001-3)
    = 10.446 m/s

    I would approximate the slope of the tangent at 3 sec to be about 10.446m/s.

    ReplyDelete
  24. angus.lSeptember 28, 2009 at 11:01 PM

    Assigned Point of Race: 7 s

    D(s) = 0.006163x^4 - 0.191217x^3 + 2.09964x^2 + 2.34451x - 0.65173

    D{7} = 0.006163{7}^4 - 0.191217{7}^3 + 2.09964{7}^2 + 2.34451{7} - 0.65173
    D{7}= 67.852132m

    D{7.1}= 0.006163{7.1}^4 - 0.191217{7.1}^3 + 2.09964{7.1}^2 + 2.34451{7.1} - 0.65173
    D{7.1} = 68.715988 m

    D{7.01} = 0.006163{7.01}^4 - 0.191217{7.01}^3 + 2.09964{7.01}^2 + 2.34451{7.01} - 0.65173
    D{7.01} = 67.97298365 m

    D{7.001} = 0.006163{7.001}^4 - 0.191217{7.001}^3 + 2.09964{7.001}^2 + 2.34451{7.001} - 0.65173
    D{7.001} = 67.8642181 m

    AROC [7,7.1] = (68.715988-67.852132)/(7.1-7)
    AROC [7,7.1] = 8.63856m/s

    AROC [7,7.01] = (67.97298365-67.852132)/(7.01-7)
    AROC [7,7.01] = 12.085165 m/s

    AROC [7,7.001] = (67.8642181-67.852132)/(7.001-7)
    AROC [7,7.001] = 12.0861 m/s


    I would approximate the slope of the tangeat at 7 second to be about 12.0861m/s

    ReplyDelete
  25. vanessaeSeptember 28, 2009 at 11:10 PM

    For 2s

    D(2)=0.006163(2)^4-0.191217(2)^3+2.09964(2)^2+2.34451(2)-0.65173
    =11.004722

    D(2)=11.004722
    D(2.1)=11.880151
    D(2.01)=11.091289
    D(2.001)=11.013369

    AROC[2,2.1]
    =(11.880151-11.004722)/(2.1-2)
    =8.75429

    AROC[2,2.01]
    =(11.091289-11.004722)/(2.01-2)
    =8.6567

    AROC[2,2.001]
    =(11.013369-11.004722)/(2.001-2)
    =8.647

    Therefore, the slope of the tangent at 2 seconds will be approximately 8.65m/s

    ReplyDelete
  26. catherine. tSeptember 29, 2009 at 9:48 PM

    Assigned point of race: 1.5 seconds

    D(s) = 0.006163x^4 - 0.191217x^3 + 2.09964x^2 + 2.34451x - 0.65173

    D(1.5)=0.006163(1.5)^4-0.191217(1.5)^3+2.09964(1.5)^2+2.34451(1.5)-0.65173
    = 6.975068
    D(1.6)=0.006163(1.6)^4-0.191217(1.6)^3+2.09964(1.6)^2+2.34451(1.6)-0.65173
    = 7.731729
    D(1.51)=0.006163(1.51)^4-0.191217(1.51)^3+2.09964(1.51)^2+2.34451(1.51)-0.65173
    = 7.049559
    D(1.501)=0.006163(1.501)^4-0.191217(1.501)^3+2.09964(1.501)^2+2.34451(1.501)-0.65173
    = 6.982505

    AROC [1.5, 1.6]= (7.731729-6.975068)/(1.6-1.5)
    AROC [1.5, 1.6]= 7.56661m/s
    AROC [1.5, 1.51]= (7.049559-6.975068)/(1.51-1.5)
    AROC [1.5, 1.51]= 7.4491m/s
    AROC [1.5, 1.501]= (6.982505-6.975068)/(1.501-1.5)
    AROC [1.5, 1.501]= 7.437m/s

    Therefore the slope of the tangent at 1.5 seconds is approximately 7.44m/s.

    ReplyDelete

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