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Four vehicles travelled on a road. What can you deduce from the times that they met?

There and Back

Brian swims at twice the speed that a river is flowing, downstream from one moored boat to another and back again, taking 12 minutes altogether. How long would it have taken him in still water?


At Holborn underground station there is a very long escalator. Two people are in a hurry and so climb the escalator as it is moving upwards, thus adding their speed to that of the moving steps. ... How many steps are there on the escalator?

How Do You React?

Age 14 to 16
Challenge Level

Herschel of the European School of Varese sent us a very succinct solution.
Speed-time graph:

Since gravity exerts a constant force, the acceleration is also constant: $9.8ms^{-2}$. Therefore, speed is proportional to time and the speed-time graph shows a straight line (with a gradient of $9.8$).

Distance-time graph:

The distance travelled depends on the speed, which increases over time. Therefore, the distance travelled in the first second is less than that travelled during the second, which is in turn less than that travelled during the third. The graph is therefore a half-parabola with the gradient increasing over time. This means that it takes longer for the ruler to travel the first $15cm$ compared to the next $15 cm$, and a reading of $30 cm$ therefore indicates a reaction time that is slower, but not twice as slow!

To look at it algebraically, distance is calculated by the formula $d=\frac{1}{2}\times a\times t^2$. This means that the distance is NOT proportional to the reaction time, but instead to the SQUARE of the reaction time. Double the distance therefore means a reaction time that is 1.41 (i.e. the square-root of $2$) times slower rather than twice as slow. Rearranging the formula above to make $t$ the subject gives us $t=\sqrt{\frac{2d}{a}}$, which means the reaction time for $15cm$ is $\sqrt{0.03}=0.17$ seconds, while that for $30cm$ is $\sqrt{0.06}=0.24$ seconds - only $0.07$ seconds slower, or $41$%.

As for the "typical" $0.2$ seconds reaction time, the typical distance should be $0.5\times 9.8\times0.2^2=0.196m$ which is $19.6cm$. I found it hard to get consistant results for my own reactions - my distances ranged from $15cm$ ($0.17$ seconds) to $25cm$ ($0.23$ seconds). Incidentally, it's funny to note that a distance of $0.204$ meters ($20.4$ centimeters) corresponds precisely to a time of $0.204$ seconds!