The design of tracks shown in Diagram 1 is the track that is usually used in the tracks tournament. There are another two designs of tracks which has a measurement of 400m in Lane 1 and width of the line is the same.

Activity 4: Further Exploration

Question 1:

Using your own creativity, create two designs of tracks which has a measurement of 400 m in Lane 1 and width of each lane is the same.

CONJECTURE

In Activity 4, the two designs of tracks will be designed and the distance of each lane, shortest distance between starting line and finishing line of each lane and the number of tins of oil paint and the cost needed to draw the tracks will be calculated to choose the best design for school tournament.

I think the design of Diagram 1 is most suitable for school tournament it will not occupy too much spaces of the field of our schools. The design of Diagram 1 has smooth turning points that do not cause major or minor injuries to runner. Besides, the selection of oval track design can save energy and also to avoid the unnecessary consumption of energy by athletes for the semicircular curve direction mutations. The oval-shaped track in the middle of the venue can also be used for other games and also facilitate spectators.

TRACK DESIGN 1 :

Presentation of data:

Lane Radius (3 Distance of Each Lane (3 Shortest Distance Ln-Ln-1

Finishing Line

decimal places) decimal places)

between starting line and finishing

line (2 decimal places)

150x 23.142

=31.827 8x 50=400.000 0.00 L1-L0=0.00

2 31.827+1.2=33.027

4x 50+ 2x 3.142x 33.027=407.542

407.542-400.00=7.54

L2-L1

=407.542-400.000=7.54

3 33.027+1.2=34.227

4x 50+ 2x 3.142x 34.227=415.082

415.082-400.00 =15.08

L3-L2

=415.082-407.542=7.54

4 34.227+1.2=35.427

4x 50+ 2x 3.142x 35.427=422.624

422.624-400.00 =22.62

L4-L3

=422.624-415.082=7.54

5 35.427+1.2=36.627

4x 50+ 2x 3.142x 36.627=430.164

430.164-400.00 =30.16

L5-L4

=430.164-422.624=7.54

6 36.627+1.2=37.827

4x 50+ 2x 3.142x 37.827=437.704

437.704-400.00 =37.70

L6-L5

=437.704-430.164=7.54

7 37.827+1.2=39.027

4x 50+ 2x 3.142x 39.027=445.246

445.246-400.00 =45.25

L7-L6

=445.246-437.704=7.54

8 39.027+1.2=40.227

4x 50+ 2x 3.142x 40.227=452.786

452.786-400.00 =52.79

L8-L7

=452.786-445.246=7.54

A)FINDING TOTAL DISTANCE OF DESIGN 1

Total distance of laneThe distance of the 8 lane (obtained from table 4) are

Total distance of the running track= total distance of lane + distance of running track

added400.000 + 407.542 + 415.082 + 422.624 + 430.164 + 437.704 + 445.246 + 452.786=3411.148m

Distance of running track borderThe distance of the running track border is same as the distance of the 9th laneThus, using π = 3.142, the formula below is used:To find radius, r ,formula used:

where, a= 31.827 d = 1.2

T9 = 31.827 + (9 – 1)(1.2) = 41.427mGiven that each of the straight lane is 100 meter, and both of the semicircle (curve lane) in a same lane are in same length too, thus we can use the formula below:D =2 (100) + 2πr (Distance of semicircle <curve lane> )

(Distance of straight lane)

Distance of running track border =2 (100) + 2 (3.142) (41.427)= 460.327m

Distance of finishing lineSince there are 8 lanes with 1.2m wide each, the distance of finishing line is:8 x 1.2 = 9.6m

Distance of starting lineAlthough there are 8 starting line, there is 1 line overlap with the finishing line. So, we assume that there are only 7 starting line. Since the width of each lane is 1.2m, the total distance of starting line is: 7 x 1.2 = 8.4m

Total distance of running track = 3411.148 + 460.327 + 9.6 + 8.4= 3889.475m

B) FINDING NUMBER OF TIN OF OIL PAINT & COST NEEDED

i) Number of tin of oil paint needed100 m = 2 liter if oil paint

Tn = a + (n-1)d

Formula used = Distance x 2

100

= 3889.475 x 2

100 =77.7895 liter of oil paint (3 decimal places)

5 liter = 1 tin

Formula used = Volume

5

= 77.78955

=15.5579 tin of oil paint

Thus, the minimum of oil paint needed is 16 tins.

ii) Cost needed to draw all the eight tracks1 tin = RM10.0016 tins = RM10 X 16 = RM160.00

Track Design 2:

Thus, the number of tin of oil pain needed is 16 and

the cost needed to draw

Presentation of data:

LaneRadius (3 decimal places)

Distance of Each Lane (3 decimal

places)

Shortest Distance between starting line and finishing

line (2 decimal places)Ln-Ln-1

1400

2x 3.142=63.654400.000 0.00 L1-L0=0.00

2 63.654+1.2=64.854

2x 3.142x 64.854=407.543 407.543-400.00=7.54 L2-L1=407.543-400.000

=7.54

3 64.854+1.2=66.054

2x 3.142x 66.054=415.083 415.083-400.00=15.08 L3-L2=415.083-407.543

=7.54

4 66.054+1.2=67.254

2x 3.142x 67.254=422.624 422.624-400.00=22.62 L4-L3=422.624-415.083

=7.54

5 67.254+1.2=68.454

2x 3.142x 68.454=430.165 430.165-400.00=30.17 L5-L4=430.165-422.624

=7.54

6 68.454+1.2=69.654

2x 3.142x 69.654=437.706 437.706-400.00=37.71 L6-L5=437.706-430.165

=7.54

7 69.654+1.2=70.854

2x 3.142x 70.854=445.247 445.247-400.00=45.25 L7-L6=445.247-437.706

=7.54

8 70.854+1.2=72.054

2x 3.142x 72.054=452.787 452.787-400.00=52.79 L8-L7=452.787-445.247

=7.54

A)FINDING TOTAL DISTANCE OF DESIGN 2

Total distance of laneThe distance of the 8 lane (obtained from table 5) are addedTotal distance of the

running track= total distance of lane + distance of running track

400.000 + 407.543 + 415.083 + 422.624 + 430.165 + 437.706 + 445.247+ 452.787=3411.155m

Distance of running track borderThe distance of the running track border is same as the distance of the 9th laneThus, using π = 3.142, the formula below is used:To find radius, r ,formula used:

where, a=63.654 d = 1.2

T9 =63.654 + (9 – 1)(1.2) = 73.254mGiven that each of the straight lane is 100 meter, and both of the semicircle (curve lane) in a same lane are in same length too, thus we can use the formula below:D =2 (100) + 2πr (Distance of semicircle <curve lane> )

(Distance of straight lane)

Distance of running track border =2 (3.142) (73.254)= 460.328m

Distance of finishing lineSince there are 8 lanes with 1.2m wide each, the distance of finishing line is:8 x 1.2 = 9.6m

Distance of starting lineAlthough there are 8 starting line, there is 1 line overlap with the finishing line. So, we assume that there are only 7 starting line. Since the width of each lane is 1.2m, the total distance of starting line is: 7 x 1.2 = 8.4m

Total distance of running track = 3411.155 + 460.328+ 9.6 + 8.4= 3889.483m

B) FINDING NUMBER OF TIN OF OIL PAINT & COST NEEDED

i) Number of tin of oil paint needed100 m = 2 liter if oil paint

Tn = a + (n-1)d

Formula used = Distance x 2

100

= 3889.483 x 2

100 =77.79 liter of oil paint (2 decimal places)

5 liter = 1 tin

Formula used = Volume

5

= 77.795

=15.558 tin of oil paint

Thus, the minimum of oil paint needed is 16 tins.

ii) Cost needed to draw all the eight tracks1 tin = RM10.0016 tins = RM10 X 16 = RM160.00

Question 2:

Based on your track designed in question (a) and Diagram 1, which design would you choose for your school tournament?

Thus, the number of tin of oil pain needed is 16 and

the cost needed to draw

State your justification.

I will

choose field track question (a) of design 1 in for my school tournament.

This is because the track designed in Diagram 1 is 400 meters around the oval, which consists of two straight track and two semi-circular arcs connected by runways. This design gives full consideration to the human body’s exercise habits.

During the movement, inertia, the tendency of body to maintain its original state of motion, the action of external forces is needed to change the state of motion. If the runways design into a round, then athletes have to consume large strength for every moment to change the direction of motion. If it is designed to be a square, at the apex of the player becomes too large will not favor the continuity of the movement. If the site is designed to make linear range, then it is too big and not easy to watch the game.

Therefore, selection of oval track design, two straights can save energy, but also to avoid the unnecessary consumption of energy by athletes for the semicircular curve direction mutations. Meanwhile, the oval-shaped track in the middle of the venue can also be used for other games and also facilitate spectators.

The total distance of the running track of diagram 1 is 3411.148m; the total distance of the running track of design 1 is 3889.475m, while the total distance of the running track of design 2 is 3889.483m. The shorter total distance of the running track, the less the amount of oil paint needed. The total distance of the running track of diagram 1 (3411.148m) is shorter than total distance of the running track of design 1 and design 2 (3889.475m and 3889.483m). So, uses of diagram 1 can save more cost because this type of design needs less amount of oil paint if compared to design 1 and design 2.

From this justification, the conjecture is accepted and proved.

Diagram 1