## 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.)

1. 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.

2. 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

3. 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.

4. 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.

5. 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.

6. 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.

7. This comment has been removed by the author.

8. 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

9. 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

10. 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

11. 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

12. 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.

13. 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

:)

14. 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

15. 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.

16. 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

17. 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.

18. 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.

19. 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.

20. 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.

21. 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.

22. 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

23. 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.

24. 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

25. 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

26. 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.