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'Speed-time Problems at the Olympics' printed from http://nrich.maths.org/

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Lane 1:

The required speed to qualify is $\frac{100\mathrm{m}}{11.32\mathrm{s}}=8.83\mathrm{ms}^{-1}.$ This is $8.83\times\frac{3600}{1600}\mathrm{mph} = 19.9\mathrm{mph}$. 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 travelled at a speed of $\frac{100\mathrm{m}}{10.78\mathrm{s}}=9.27\mathrm{ms}^{-1}.$ If she had continued at this speed until the athlete running at the qualifying speed had finished (an extra $11.32\mathrm{s}-10.78\mathrm{s} = 0.54\mathrm{s}$), she'd have ran an additional $5.01\mathrm{m}.$

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 would still have $10.78\mathrm{s}-9.58\mathrm{s} = 1.2\mathrm{s}$ to run, meaning she would be $1.2\mathrm{s}\times9.27\mathrm{ms}^{-1}=11.1\mathrm{m}$ 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.