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Lane 1:
The required speed to qualify is \frac{100\mathrm{m}}{11.29\mathrm{s}}=8.86\mathrm{ms}^{-1}. This is 8.86\times\frac{3600}{1600}\mathrm{mph} = 19.9\mathrm{mph}. This is a similar speed to a bus travelling in a built-up area, but of course the bus will continue at this speed for much longer!
Lane 2:
Shelly-Ann Fraser-Pryce travelled at a speed of \frac{100\mathrm{m}}{10.75\mathrm{s}}=9.30\mathrm{ms}^{-1}. If she had continued at this speed until the athlete running at the qualifying speed had finished (an extra 11.29\mathrm{s}-10.75\mathrm{s} = 0.54\mathrm{s}), she'd have ran an additional 5.02\mathrm{m}.
Well done to Annabelle from Saint Martns, Brendan from Colfe's School and Zoe, Abigail, Ella and Amy from Queen Elizabeth High School who found this!
Lane 3:
When Usain Bolt crossed the finish line in 9.58\mathrm{s}, Shelly-Ann Fraser-Pryce would still have 10.75\mathrm{s}-9.58\mathrm{s} = 1.17\mathrm{s} to run, meaning she would be 1.17\mathrm{s}\times9.30\mathrm{ms}^{-1}=10.88\mathrm{m} behind.
(These calculations assume they run at their average speed throughout the race, which is certainly not the case, as the athletes take several seconds to get to full speed. However, these figures should give an acceptable approximation.)
Lane 4:
Your 200\mathrm{m} time depends on lots of factors including age, gender and fitness, but a good ballpark figure would be 30\mathrm{s}. After Usain Bolt finishes in his world record time of 19.19\mathrm{s}, you'd still have \frac{10.81}{30}\times200\mathrm{m}=72\mathrm{m} left to run! (Again, we're assuming constant speed throughout the race.)
Well done to Richard from Wilson's School who managed to get up to this part!
Lane 5:
Suppose on the flat rowing lake, the crew take 210\mathrm{s} (3\colon 30) to row 1\mathrm{km}. Overall, the time for their 2\mathrm{km} race is 420\mathrm{s} (7\colon 00). This corresponds to a speed of 4.76\mathrm{ms}^{-1}, which we're assuming is the same in both directions and throughout the race.
Suppose on the River Thames, the tide makes the boat go x\; \mathrm{ms}^{-1} faster than before on the downstream section, and x\; \mathrm{ms}^{-1} slower than before on the upstream section. The overall time for this race is therefore \left(\frac{1000\mathrm{m}}{(4.76-x)\mathrm{ms}^{-1}}+\frac{1000\mathrm{m}}{(4.76+x)\mathrm{ms}^{-1}}\right)\mathrm{s}.
If x=1, the overall time on the Thames is 439.3\mathrm{s}, 19.3\mathrm{s} slower than the race on the lake. However, if x=0.2, the overall time is 392.8\mathrm{s}, 27.2\mathrm{s} faster than the race on the lake.
Lane 6:
Velodromes for the Olympics are allowed to measure 250\mathrm{m}, 333.3\mathrm{m} or 400\mathrm{m} in length. We'll assume 400\mathrm{m} to simplify the calculations. Assuming half of this distance is along the home and back straights along which there's no difference which line you take, there's 200\mathrm{m} around the 180^{\circ} turns where the position on the track is important. If we assume the cyclist on the blue line travels 400\textrm{m per lap}, then the radius of the semicircular sections is \frac{100}{\pi}\mathrm{m} =31.8\mathrm{m}. Cyclist B will therefore travel 200\mathrm{m}+2\times\pi\times(31.8-1)\mathrm{m}=393.7\mathrm{m} and cyclist C 200\mathrm{m}+2\times\pi\times(31.8+2)\mathrm{m}=412.6\mathrm{m}.
Track cyclists try to cycle as close to the inside of the track as possible in order to minimise the distance travelled. However, if they do this they can end up 'boxed-in' by other cyclists; this is one of the other factors that should be considered.
Lane 7:
Assuming a lane is 1\mathrm{m} across, the sound from the starting pistol only has to travel 8\mathrm{m} to reach the athlete in lane 8. This takes approximately \frac{8\mathrm{m}}{340\mathrm{ms}^{-1}} = 0.02\mathrm{s}, much less than an average human reaction time, so doesn't give anyone a significant advantage. Firing the gun from the centre of the track by lane 4 would halve the time before the sound has reached everyone, but this isn't necessary.
Lane 8:
The speed of sound is around 340\mathrm{ms}^{-1} (amongst other things, it depends on altitude and air temperature). Estimating that the crowd is 100\mathrm{m} from the podium, the sound will take \frac{100\mathrm{m}}{340\mathrm{ms}^{-1}}=0.29\mathrm{s} to reach them.
A geostationary satellite used for communications has an altitude of around 36000\mathrm{km}. Radio signals (which move at the speed of light) will take around 2\times\frac{3.6\times10^7\mathrm{m}}{3\times10^8\mathrm{ms}^{-1}} = 0.24\mathrm{s} to travel to and return from the satellite.
It seems the two delays are similar, although this calculation neglects any delays in the process of transmitting the signal, either on earth or by the satellite, and also neglects any time spent by the digital radio decoding the signal.
Can you prove that the sum of the distances of any point inside a square from its sides is always equal (half the perimeter)? Can you prove it to be true for a rectangle or a hexagon?
The Earth is further from the Sun than Venus, but how much further? Twice as far? Ten times?