Monday, September 28, 2009

Period 3 - 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 the homework from today's entry in your homework log.)
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18 comments:

  1. StephanieYSeptember 28, 2009 at 8:12 PM

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  2. StephanieYSeptember 28, 2009 at 8:18 PM

    Assigned Point of Race: 2 seconds

    Calculations:

    t(s)= 2.0 s

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

    D(2.0)= 0.006163(2.0)^4-0.191217(2.0)^3+2.09964(2.0)^2+2.34451(2.0)-0.65173

    D(2.0)= 11.004722 m

    t(s)= 2.1 s

    D(2.1)= 0.006163(2.1)^4-0.191217(2.1)^3+2.09964(2.1)^2+2.34451(2.1)-0.65173

    D(2.1)= 11.880151 m

    AROC [2.0,2.1] = Δd/Δt
    = 11.880151m – 11.004722m / 2.1s-2.0s
    = 8.75429 m/s

    t(s)= 2.01 s

    D(2.01)= 0.006163(2.01)^4-0.191217(2.01)^3+2.09964(2.01)^2+2.34451(2.01)-0.65173

    D(2.01)= 11.091289 m/s

    AROC [2.0,2.01] = Δd/Δt
    = 11.091289 m – 11.004722m / 2.01s-2.0s
    = 8.6567 m/s

    t(s)= 2.001 s

    D(2.001)= 0.006163(2.001)^4-0.191217(2.001)^3+2.09964(2.001)^2+2.34451(2.001)-0.65173

    D(2.001)= 11.013369 m/s

    AROC [2.0,2.001] = Δd/Δt
    = 11.091289 m – 11.004722m / 2.001s-2.0s
    = 8.647 m/s

    ReplyDelete
  3. SabrinaDSeptember 28, 2009 at 8:40 PM

    Assigned point of race: 0.5 seconds

    S=0.5

    D(0.5)=0 .006163(0.5)^3-.191217(0.5)^2+2.34451(0.5)-0.65173

    D(0.5)=1.0219181 m/s

    Example 1: [0.5,0.6]

    S=0.6

    D(0.6)= 0 .006163(0.6)^3-.191217(0.6)^2+2.34451(0.6)-0.65173

    D(0.6)=1.4703423 m/s

    AROC[0.5,0.6] = 1.4703423 m/s-1.0219181 m/s
    0.6 s-0.5 s
    AROC = 4.484242 m

    Example 2: [0.5,0.51]
    S=0.51

    D(0.51)= 0 .006163(0.51)^3-.191217(0.51)^2+2.34451(0.51)-0.65173

    D(0.51)= 1.0651383 m/s

    AROC[0.5, 0.51] = 1.0651383 m/s-1.0219181 m/s
    0.51 s – 0.5 s

    AROC= 4.32202 m

    Example 3: [0.5, 0.501]
    S=0.501

    D(0.501)=0 .006163(0.501)^3-.191217(0.501)^2+2.34451(0.501)-0.65173

    D(0.501)= 1.0262237 m/s
    AROC[0.5,0.501] = 1.0262237 m/s-1.0219181 m/s
    0.501 s- 0.5s
    AROC= 4.3056 m

    Therefore, the instantaneous velocity at 0.5 seconds is approximately 4.305 m

    ReplyDelete
  4. Kelly CSeptember 28, 2009 at 8:44 PM

    Assigned Point of Race: 7.5 s


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

    D(7.5) = 0.006163(7.5)^4 - 0.191217(7.5)^3 + 2.09964(7.5)^2 + 2.34451(7.5) - 0.65173
    D(7.5) = 73.86729 m

    D(7.6) = 0.006163(7.6)^4 - 0.191217(7.6)^3 + 2.09964(7.6)^2 + 2.34451(7.6) - 0.65173
    D(7.6) = 75.063188 m

    D(7.51) = 0.006163(7.51)^4 - 0.191217(7.51)^3 + 2.09964(7.51)^2 + 2.34451(7.51) - 0.65173
    D(7.51) = 73.986991 m

    D(7.501) = 0.006163(7.501)^4 - 0.191217(7.501)^3 + 2.09964(7.501)^2 + 2.34451(7.501) - 0.65173
    D(7.501) = 73.879261 m


    AROC (7.5-7.6) = (75.063188-73.86729)/(7.6-7.5)
    AROC (7.5-7.6) = 11.95898 m/s

    AROC (7.5-7.51) = (73.986991-73.86729)/(7.51-7.5)
    AROC (7.5-7.51) = 11.9701 m/s

    AROC (7.5-7.501) = (73.879261-73.86729)/(7.501-7.5)
    AROC (7.5-7.501) = 11.971 m/s


    Therefore, the instantaneous rate of change for 7.5 seconds is about 11.971 m/s.

    ReplyDelete
  5. corradocSeptember 28, 2009 at 8:53 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
  6. Nara SSeptember 28, 2009 at 9:07 PM

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

    Assigned Point of Race: 7.5

    ToV
    t(s) = D(m)
    7.5 = 73.86729
    7.51 = 73.986991
    7.501 = 73.8797261
    7.5001 = 73.868487

    AROC [7.5,7.51] = 73.986991 - 73.86729 m / 7.51 – 7.5 s
    = 0.119701 m / 0.01 s
    = 11.9701 m/s

    AROC [7.5,7.501] = 73.879726 - 73.86729 m / 7.501 – 7.5 s
    = 0.011971 m / 0.001 s
    = 11.971 m/s

    AROC [7.5,7.5001] = 73.868487 - 73.86729 m / 7.5001 – 7.5 s
    = 0.001197 m / 0.0001 s
    = 11.97 m/s

    Therefore the instantaneous velocity at 7.5 s is approximately 11.97 m/s

    ReplyDelete
  7. Joseph RaymundoSeptember 28, 2009 at 9:21 PM

    Instantaneous Velocity at: 1 Second

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

    Time(s) = 1s

    D(1) = 0.006163(1)^4-0.191217(1)^3+2.09964(1)^2+2.34451(1)-0.65173
    D(1)= 3.607366m

    D(1.1)= 0.006163(1.1)^4-0.191217(1.1)^3+2.09964(1.1)^2+2.34451(1.1)-0.65173
    D(1.1)= 4.4763653m

    AROC[1.0,1.1]= Δd/Δt
    = (4.476365-3.607366)/(1.1-1)
    = 8.156281 m/s

    D(1.01)= 0.006163(1.01)^4-0.191217(1.01)^3+2.09964(1.01)^2+2.34451(1.01)-0.65173
    D(1.01)= 3.667474m

    AROC[1.0,1.01]= Δd/Δt
    = (3.667474-3.607366)/(1.01-1)
    = 6.0108 m/s

    D(1.001)= 0.006163(1.001)^4-0.191217(1.001)^3+2.09964(1.001)^2+2.34451(1.001)-0.65173
    D(1.001)= 3.613362m

    AROC[1.0,1.001]= Δd/Δt
    = (3.613362-3.607366)/(1.001-1.0)
    = 5.996m/s

    Therefore the instantaneous velocity at 1 second is approximately 5.996m/s

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  8. BStubbsSeptember 28, 2009 at 9:30 PM

    Tangent for 9sec
    TOV
    T --> D
    9sec 91.55786m
    9.1sec 92.721571m
    9.01sec 91.674289m
    9.001sec 91.569504m

    AROC[9,9.1]=(92.72157m-91.5579m)/(9.1s-9.0s)
    =1.163711m/0.1s
    =11.63711m/s

    AROC[9,9.01]=(91.67429m-91.5579m)/(9.01s-9s)
    =0.116429/0.01s
    11.6429m/s

    AROC[9,9.001]=(91.5695m-91.5579m)/(9.001s-9s)
    =0.011644m/0.001s
    =11.644m/s

    Therefore the Tangent is at approximately 11.645m/s.

    ReplyDelete
  9. ShaniseASeptember 28, 2009 at 9:33 PM

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  10. ShaniseASeptember 28, 2009 at 9:38 PM

    This comment has been removed by the author.

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  11. ShaniseASeptember 28, 2009 at 9:40 PM

    Assigned Point of Race: 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(9.51)=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[9.5,9.51] = (Y2-Y1)/(X2-X1)
    = (97.482985m- 97.36697m) / (9.51s-9.5s)
    = (0.116015m)/ (0.01s)
    = 11.6015m/s

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

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

    Therefore the instantaneous velocity at 9.5 seconds is approximately 11.6m/s.

    ReplyDelete
  12. David L.September 28, 2009 at 9:41 PM

    Assigned Point of Race: 4.5 s

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

    D{6} = 0.006163{6}^4 - 0.191217{6}^3 + 2.09964{6}^2 + 2.34451{6} - 0.65173
    D{6}= 55.686746 m

    D{6.1}= 0.006163{6.1}^4 - 0.191217{6.1}^3 + 2.09964{6.1}^2 + 2.34451{6.1} - 0.65173
    D{6.1} = 56.90795133 m

    D{6.01} = 0.006163{6.01}^4 - 0.191217{6.01}^3 + 2.09964{6.01}^2 + 2.34451{6.01} - 0.65173
    D{6.01} = 55.80888071 m

    D{6.001} = 0.006163{6.001}^4 - 0.191217{6.001}^3 + 2.09964{6.001}^2 + 2.34451{6.001} - 0.65173
    D{6.001} = 55.69895957 m

    AROC [6,6.1] = (56.90795133-55.686746)/(6.1-6)
    AROC [6,6.1] = 12.2120533 m/s

    AROC [6,6.01] = (55.80888071-55.686746)/(6.01-6)
    AROC [6,6.01] = 12.213471 m/s

    AROC [6,6.001] = (73.879261-55.686746)/(6.001-6)
    AROC [6,6.001] = 12.21357 m/s


    Therefore, the instantaneous rate of change at 4.5 seconds is approximately 12.213 m/s.

    ReplyDelete
  13. Rachel.DSeptember 28, 2009 at 9:50 PM

    ASSIGNED POINT OF RACE: 0.5 seconds

    TABLE OF VALUES
    t D(t)
    0.5 1.0219181
    0.6 1.4703423
    0.51 1.0651303
    0.501 1.0262237

    INTERVAL #1
    AROC[0.5, 0.6]= Δd/Δt
    = (1.4703423-1.02191810) / (0.6-0.5)
    =4.484242 m/s

    INTERVAL #2
    AROC[0.5, 0.51]= Δd/Δt
    = (1.0651383-1.02191810) / (0.51-0.5)
    =4.32202 m/s

    INTERVAL #3
    AROC[0.5, 0.501]= Δd/Δt
    = (1.0262237-1.02191810 / (0.501-0.5)
    =4.3056 m/s

    Because the AROC value is decreasing as the interval decreases, we can assume that the instantaneous velocity at 0.5 seconds is approximately 4.305m/s.

    ReplyDelete
  14. Jay CeeSeptember 28, 2009 at 9:52 PM

    Assigned Point of Race: 2.5 Seconds

    D(2.5)= 0.006163(2.5)^4 - 0.141217(2.5)^3 + 2.09964(2.5)^2 + 2.34451(2.5)- 0.65173
    = 15.58527156 (approximately)

    t=2.6 s

    D(2.6)= 0.006163(2.6)^4 - 0.141217(2.6)^3 + 2.09964(2.6)^2 + 2.34451(2.6)- 0.65173
    = 21.23091368 (approximately)

    AROC[2.5,2.6]= (21.23091368-15.58527156)/(2.6-2.5)
    = 56.4564212

    t= 2.51 sec

    D(2.51)= 0.006163(2.51)^4 - 0.141217(2.51)^3 + 2.09964(2.51)^2 + 2.34451(2.51)- 0.65173
    = 15.61259171 (approximately)

    AROC[2.5,2.51](16.47244943-15.58527156)/(2.6-2.51)
    = 9.857531889

    t= 2.501 sec

    D(2.501)= 0.006163(2.501)^4 - 0.141217(2.501)^3 + 2.09964(2.501)^2 + 2.34451(2.501)- 0.65173
    = 15.59491504 (approximately)

    AROC[2.5,2.501](15.59491504-15.58527156)/(2.501-2.5)
    = 9.64348

    Jay Cee says @ 2.5 sec he ran 9.64 m/s!

    ReplyDelete
  15. kevin zSeptember 28, 2009 at 10:59 PM

    Assigned Point of Race: 2.5 Seconds

    t=2.5s
    D(2.5)= 0.006163(2.5)^4 - 0.141217(2.5)^3 + 2.09964(2.5)^2 + 2.34451(2.5)- 0.65173
    = 15.58527156 (approximately)

    t=2.6 s
    D(2.6)= 0.006163(2.6)^4 - 0.141217(2.6)^3 + 2.09964(2.6)^2 + 2.34451(2.6)- 0.65173
    = 21.23091368 (approximately)

    AROC[2.5,2.6]= (21.23091368-15.58527156)/(2.6-2.5)
    = 56.4564212

    t= 2.51 sec
    D(2.51)= 0.006163(2.51)^4 - 0.141217(2.51)^3 + 2.09964(2.51)^2 + 2.34451(2.51)- 0.65173
    = 15.61259171 (approximately)

    AROC[2.5,2.51](16.47244943-15.58527156)/(2.6-2.51)
    = 9.857531889

    t= 2.501 sec
    D(2.501)= 0.006163(2.501)^4 - 0.141217(2.501)^3 + 2.09964(2.501)^2 + 2.34451(2.501)- 0.65173
    = 15.59491504 (approximately)

    AROC[2.5,2.501](15.59491504-15.58527156)/(2.501-2.5)
    = 9.64348

    Final answer: 9.64348m/s

    ReplyDelete
  16. Michelle JSeptember 28, 2009 at 11:04 PM

    Usain's instantaneous velocity at 8 seconds:
    D(s)=0.006163x^4-0.191217x^3+2.09964x^2+2.34451x-0.65173

    D(8)=0.006163(8)^4-0.191217(8)^3+2.09964(8)^2+2.34451(8)-0.65173
    D(8)=79.794710
    D(8.1)= 0.006163(8)^4-0.191217(8)^3+2.09964(8)^2+2.34451(8)-0.65173
    D(8.1)= 80.977155m

    AROC [8,8.1]= (80.977155m-79.794710)/ (8.1-8)
    =11.82445 m/s


    D(8.01)=0.006163(8.01)^4-0.191217(8.01)^3+2.09964(8.01)^2+2.34451(8.01)-0.65173
    D(8.01)=79.913073

    AROC[8,8.01]= (79.913073-79.794710)/(8.01-8)
    = 11.8363 m/s

    D(8.001)= 0.006163(8.001)^4-0.191217(8.001)^3+2.09964(8.001)^2+2.34451(8.001)-0.65173
    D(8.001
    D(8.001)= 79.806555

    AROC[8,8.001]=(79.806555-79.794710)/(8.001-8)
    =11.845 m/s

    Therefore, at 8 seconds Usain's instantaneous velocity was approximately 11.845 m/s

    ReplyDelete
  17. Brenna FSeptember 28, 2009 at 11:31 PM

    Instantaneous Speed at: 2.5 Seconds

    t = 2.5 s

    D(2.5)= 0.006163(2.5)^4 - 0.141217(2.5)^3 + 2.09964(2.5)^2 + 2.34451(2.5)- 0.65173
    = 15.58527156 m

    t = 2.6 s

    D(2.6)= 0.006163(2.6)^4 - 0.141217(2.6)^3 + 2.09964(2.6)^2 + 2.34451(2.6)- 0.65173
    = 16.558367 m

    AROC[2.5,2.6]= (16.558367-15.58527156)/(2.6-2.5)
    = 9.73095000 m/s

    t = 2.51 s

    D(2.51)= 0.006163(2.51)^4 - 0.141217(2.51)^3 + 2.09964(2.51)^2 + 2.34451(2.51)- 0.65173
    = 15.681787 m

    AROC[2.5,2.51](15.681787-15.58527156)/(2.51-2.5)
    = 9.651500000 m/s

    t= 2.501 s

    D(2.501)= 0.006163(2.501)^4 - 0.141217(2.501)^3 + 2.09964(2.501)^2 + 2.34451(2.501)- 0.65173
    = 15.59491504 m

    AROC[2.5,2.501](15.59491504-15.58527156)/(2.501-2.5)
    = 9.643480000 m/s

    Therefore the approximate instantaneous speed at 2.5 seconds is 9.643 m/s.

    ReplyDelete
  18. AdeeeeeeeSeptember 28, 2009 at 11:58 PM

    Assigned Point of Race: 5 seconds

    Calculations:

    t(s)= 5.0 s

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

    D(5.0)= 0.006163(5.0)^4-0.191217(5.0)^3+2.09964(5.0)^2+2.34451(5.0)-0.65173

    D(5.0)= 43.51157m

    t(s)= 5.01s

    D(5.01)= 0.006163(5.01)^4-0.191217(5.01)^3+2.09964(5.01)^2+2.34451(5.01)-0.65173

    D(5.01)= 43.632397 m

    AROC [5.0, 5.01] = Δd/Δt
    =43.632397m – 43.51157m / 5.01-5.0s
    = 12.0827 m/s

    t(s)= 5.001s

    D(5.001)= 0.006163(5.001)^4-0.191217(5.001)^3+2.09964(5.001)^2+2.34451(5.001)-0.65173

    D(5.001)= 43.523651 m/s

    AROC [5.0, 5.001] = Δd/Δt
    = 43.523651 m – 43.51157m / 5.001-5.0
    = 12.081 m/s

    Therefore, the AROC from 5.0-5.01 and 5.0-5.001 will be 12.0827 m/s and 12.081 m/s respectively.

    ReplyDelete

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