Questions: Extended Response Question (8 marks) Question 1 A traditional athletics track is 400 metres in length and consists of two straights (of 100 m each) and two curves. You have the job of designing the athletics track for the 2032 Brisbane Olympics. a) Calculate the internal width of the track. b) A runner in lane 2 will need a head start to make the race fair. The length of the outer lanes will be longer than the first lane and the width of the lane is 1.5 m. Give your answers to the nearest centimetre. c) The painters tasked with marking the lanes have asked for all distances to be given in scientific notation to the nearest centimetre. Restate your values from Question 2 in the form requested by the painters. (3+3+2=6 marks)

Solution Steps

The total length of the track is 400 meters, consisting of two straights of 100 meters each and two curves. The length of the straights is 2 * 100 = 200 meters. Therefore, the combined length of the curves is 400 - 200 = 200 meters. Each curve is a semicircle, so the total length of the curves (200 meters) is the circumference of a full circle.

Using the formula for the circumference of a circle, \( C = 2\pi r \), we can find the radius \( r \):

\[ 2\pi r = 200 \] \[ r = \frac{200}{2\pi} \] \[ r \approx \frac{200}{6.2832} \] \[ r \approx 31.83 \text{ meters} \]

The internal width of the track is twice the radius (since it spans the diameter of the circle):

\[ \text{Internal width} = 2r \] \[ \text{Internal width} \approx 2 \times 31.83 \] \[ \text{Internal width} \approx 63.66 \text{ meters} \]

The width of each lane is 1.5 meters. The radius for lane 2 is:

\[ r_2 = r + 1.5 \] \[ r_2 \approx 31.83 + 1.5 \] \[ r_2 \approx 33.33 \text{ meters} \]

The circumference for lane 2 is:

\[ C_2 = 2\pi r_2 \] \[ C_2 \approx 2\pi \times 33.33 \] \[ C_2 \approx 209.44 \text{ meters} \]

The head start needed for lane 2 is the difference in the lengths of the two lanes:

\[ \text{Head start} = C_2 - C \] \[ \text{Head start} \approx 209.44 - 200 \] \[ \text{Head start} \approx 9.44 \text{ meters} \]

Convert the internal width and head start to scientific notation:

• Internal width: \( 63.66 \text{ meters} \approx 6.37 \times 10^1 \text{ meters} \)
• Head start: \( 9.44 \text{ meters} \approx 9.44 \times 10^0 \text{ meters} \)

Final Answer