## 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.)

1. This comment has been removed by the author.

2. 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

3. 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

4. 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.

5. 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.

6. 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

7. 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

8. 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.

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10. This comment has been removed by the author.

11. 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.

12. 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.

13. 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.

14. 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!

15. 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

16. 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

17. 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.

18. 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.